Class 8 Maths - Factorisation

Factorise $ab({x^2} + {y^2}) - xy({a^2} + {b^2})$ .

$ab( x^2 + y^2) – xy (a^2 +b^2)$

$= ab x^2 + aby^2 – xya^2 – xy b^2$

$= ( abx^2 – xya^2) + ( aby^2 – xyb^2)$

$= ax( bx – ay) + by (ay – bx)$

$= (bx-ay) ( ax-by)$

Divide the polynomial $3x^{4} -4x^{3} -3x - 1$ by $x -1$ .

Quotient

$= 3x^3 - x^2 - x - 4$

Remainder

$= -5$

973215_1053628_ans_e934a9a41f4d4850b16e41929b78ef00.png

Find the root of given equation ${x}^{2}+6x+5$

$\begin{array}{l} { x^{ 2 } }+6x+5 \\ ={ x^{ 2 } }x+5x+5 \\ =x\left( { x+1 } \right) +5\left( { x+1 } \right) \\ =\left( { x+1 } \right) \left( { x+5 } \right) \end{array}$

Factorize $x\left( {x + z} \right) - y\left( {y + z} \right)$

$\begin{aligned}x\left( {x + z} \right) - y\left( {y + z} \right) &= {x^2} + xz - {y^2} - yz\\& = {x^2} - {y^2} + xz - yz\\& = \left( {x - y} \right)\left( {x + y} \right) + z\left( {x - y} \right)\\& = \left( {x - y} \right)\left( {x + y + z} \right)\end{aligned}$

Factorise the expression and divide them as directed.
$(5p^2-25p+20)\div (p-1)$ .

$5{p^2}-25{p}+20$

$=5{p^2}-5{p}-20{p}+20$

$=5{p}(p-1)-20(p-1)$

$=(p-1)(5p-20)$

Now,

$(5 p^2-25{p}+20)\div (p-1)$

$=\dfrac{(p-1)(5{p}-20)}{(p-1)}$

$=5{p}-20$

$=5\left(p-4\right)$

Factousation, $16{p^2}{q^2}{r^3} - 24{p^3}{q^2}{r^2} + 8{p^3}{q^2}{r^2}$

1336581_1245237_ans_80325eea4f54452abd3d6d89d0f8039b.PNG

factorise
$3 x ^ { 2 } + 6 x ^ { 3 }$

$3x^2+6x^3\\=3x^2(1+2x)$

Factorise:
$ab^{2}+b(a-1)-1$

Factorise

$ab^2+b(a-1)-1$

$\Rightarrow ab^2+ba-b-1$

$\Rightarrow ab(b+1)-1(b+1)$

$\Rightarrow (b+1)(ab-1)$

Hence, this is answer.

1195706_1242172_ans_78c118dcb49c4c4f86fc18637cf54694.png

Factorize:
$4 a ^ { 2 } - 8 a b$

$4a^2-8ab\\=4a(a-2b)$

Factorize
$x ^ { 2 } + 7 x + 12 = 0$

$x^2+7x+12\\=x^2+4x+3x+12\\=x(x+4)+3(x+4)\\=(x+4)(x+3)$

Factorize:
$2 x ^ { 3 } b ^ { 2 } - 4 x ^ { 5 } b ^ { 4 }$

$2x^3b^2-4x^5b^4\\=2x^3b^2(1-2x^2b^2)$

Factorize: $15 x ^ { 4 } y ^ { 3 } - 20 x ^ { 3 } y$

$15x^4y^3-20x^3y\\=5x^3y(3xy^2-4)$

Factorize:. $15 x + 5$

15x+5=5(3x+1)

Factorise completely by removing a monomial factor.
5m+5n

1334600_1265735_ans_8ba6814cf19c4d938122423d07cb7437.PNG

Factorise completely by removing a monomial factor.
5a+8b

1334589_1265746_ans_151bf29c82d041f08554eb54d969cb69.PNG

Factorise completely by removing a monomial factor.
2x-4

1334598_1265726_ans_aae4c4f1ebc54b7c917ee790e3404c96.PNG

Factorize:
$a ^ { 3 } b - a ^ { 2 } b ^ { 2 } - b ^ { 3 }$

$a^3b-a^2b^2-b^3\\=b(a^3-a^2b-b^2)$

Factorise completely by removing a monomial factor.
5x+10

1334594_1265705_ans_00f5b24ac65043a2ba88d85b82d60b28.PNG

Solve : $(x^3+2x^2+3x) \div 2x$

1338767_1264402_ans_9df468182216446db192fe24e9dffe8c.jpg

Factorise completely by removing a monomial factor.
-3m-15n

1334591_1265764_ans_82528a8c281e45fbb70a9617a606e2c4.PNG

Factorise completely by removing a monomial factor.
7x-14y

1334590_1265754_ans_24d904e859674039933f4ec5e8d991e9.PNG

Factorise completely by removing a monomial factor.
$3y-9$

1334587_1265693_ans_dc0e90f81d1b4efa9c25c459f153bca5.PNG

Factorize:
$6 x ^ { 2 } y + 9 x y ^ { 2 } + 4 y ^ { 3 }$

$6x^2y+9xy^2+4y^3\\=y(6x^2+9xy+4y^2)$

Factorise completely by removing a monomial factor.
-7p-14q

1334592_1265772_ans_24d31c5518ba484c9036735eca2ee9ed.PNG

Factorise completely by removing a monomial factor.
ax+bx

$ax+bx$

$(a+b)x$ .

1200452_1265820_ans_356bc57c39ce46fe887d1a072db2b1b0.jpg

Factorise completely by removing a monomial factor.
$4 + 12{x^2}$

$4+12x^2=4(1+3x^2)$ .
1209973_1265838_ans_48d4b63120994163be433c9505f9d88a.jpg

Factorise completely by removing a monomial factor.
${x^2}{y^{}} + x{y^2}$

Given binomial is

$x^2y+xy^2$

Taking

$xy$ common from both terms, we get:

$xy(x+y)$

$\Rightarrow (xy)(x+y)$

Hence, the given binomial can be represented as the product of a monomial and a binomial as:

$x^2y+xy^2=(xy)(x+y)$ .

Factorize:
$6{x^2} - 11x$

$6x^2-11x$

$=x(6x-11)$ .

Factorise :
$a{x^2} + a{b^3}$

$ax^2+ab^3=a(x^2+b^3)$ .

Factorise completely by removing a monomial factor
${x^2}{y^2} + {x^2}$

$x^2y^2+x^2=x^2(y^2+1)$ .
1209979_1265886_ans_11d8734964ff4a889cbf380588d2f98d.jpg

divide
$({y^2} + 10y + 24) \div (y + 4)$

$(y^2+10y+24)/(y+4)=[(y+4)(y+6)]/(y+4)=y+6$

Factorise completely by removing a monomial factor
ax+ay+az

1334607_1265863_ans_6139a9ceb549424492beba18921d6448.PNG

Factories:-
$6n+12n^2$

We have,

$6n+12^2$

On factorization we get,

$6n(1+2n)$

Hence, this is the answer.

Factories:-
$6n+12n^2+21n$

We have,

$6n+12n^2+21n$

$=12n^2+27n$

On factorization we have,

$=3n(4n+9)$

Hence, this is the answer.

Factorize $x^3+13x^2+32x+20$ , if it is given that $x+2$ is its factor.

$x^3 + 13x^2 + 32x + 20$

$=x^3+2x^2+11x^2+22x+10x+20$

$=x^2(x+2)+11x(x+2)+10(x+2)$

$= (x+2)(x^2+11x+10)$

$= (x+2)[x^2+x+10x+10]$

$= (x+2)[x(x+1)+10(x+1)]$

$=(x+2)(x+1)(x+10)$

Factorise
$12{x^2} - 9x$

1335914_1270540_ans_b145de79cbe04b019d74961005a75935.PNG

$12xy(9x^{2}-16y^{2})\div 4xy(3x+4y)$

$12xy(9x^2-16y^2)\div 4xy(3x+4y)$

$\dfrac{12xy(9x^2-16y^2)}{4y(3x+4y)}$

$\dfrac{3((3x)^2-(4y)^2)}{3x+4y}=\dfrac{3(3x+4y)(3x-4y)}{(3x+4y)}$

$\Rightarrow 3(3x-4y)$

$\Rightarrow 9x-12y$ .

1204164_1393810_ans_410bf84426b84c64a1c8a1e7cee38bcd.JPG

Factorize: ${x^3} + 6{x^2}y + 9x{y^2}$ .

Now,

${x^3} + 6{x^2}y + 9x{y^2}$

$=x(x^2+6xy+9y^2)$

$=x(x+3y)^2$ .

Factorize:
$(a^2-b^2)x^2+2ax+1$ .

Now,

$(a^2-b^2)x^2+2ax+1$

$=(a^2x^2+2ax+1-b^2x^2$

$=(ax+1)^2-(bx)^2$

$=(ax+bx+1)(ax-bx+1)$ .

Factorise :
1. $2x^{2}-x-3$
2. $-2x^{2}+3x+9$

$(1)$

$2x^2-x-3$

$\Rightarrow 2x^2-3x+2x-3$

$\Rightarrow x(2x-3)+1(2x-3)$

$\Rightarrow (x+1)(2x-3)$

$(2)$

$-2x^2+3x+9$

$\Rightarrow -2x^2+6x-3x+9$

$\Rightarrow -2x(x-3)-3(x-3)$

$\Rightarrow -(x-3)(2x+3)$ .

1222818_1394245_ans_3b8ed597284147e7bdb8440260cd3b63.jpg

Solve:
$z^2-16z+55=0$ .

Now,

$z^2-16z+55=0$

or,

$z^2-11z-5z+55=0$

or,

$(z-5)(z-11)=0$

or,

$z=5,11$ .

Factorize: ${x}^{2}+10x+24$

Given,

${x}^{2}+10x+24$

$={x}^{2}+6x+4x+24$

$=x\left(x+6\right)+4\left(x+6\right)$

$=\left(x+6\right)\left(x+4\right)$

$(4x^4-5x^3-7x+1)\div (4x-1)$ .

quotient

$={x}^{3}-{x}^{2}-\cfrac{x}{4}-\cfrac{29}{16}$

remainder

$=\cfrac{-13}{16}$

We know , $$dividend=divisor\times quotient +remainder

$4{x}^{4}-5{x}^{3}-7x+!=(4x-1)({x}^{3}-{x}^{2}-\cfrac{x}{4}-\cfrac{29}{16}-\cfrac{13}{16}$

1343488_1251168_ans_b4a4a9fbbc3e4b06a1c866fa66192d47.PNG

Divide $\left( { 2y }^{ 3 }+{ 4y }^{ 2 }+3 \right) \div { 2y }^{ 2 }$

$\therefore (2y^{3}+4y^{2}+3)\div 2y^{2}$

$=y+2+\cfrac{3}{2y^{2}}$

1380961_1240116_ans_0c667084a1004e5fb21044fa041463e7.png

Factorise : $150 -6x^2$

Given :

$150 -6x^2$

By taking

$6$ as common

$150 -6x^2$

$= 6 (25 -x^2 )$

We can write it as

$150 -6x^2$

$=6 (5^2 -x^2 )$

Based on equation

$a^2 -b^2 = (a+b)(a-b)$

We get,

$150 -6x^2$

$= 6 (5+x)(5-x)$

Factorise : $8ab^2 - 18a^3$

Given :

$8ab^2 -18 a^3$

By taking

$2a$ as common

$8ab^2 -18 a^3$

$= 2a ( 4b^2 -9a^2)$

We can write it as

$8ab^2 -18 a^3$

$=2a[ (2b)^2 -(3a)^2]$

Based on equation

$a^2 -b^2 = (a+b)(a-b)$

We get,

$8ab^2 -18 a^3$

$= 2a(2b+3a)(2b-3a)$

Factorise : $x^2 +\dfrac {1}{x^2} - 2 -3x +\dfrac {3}{x}$

Given :

$x^2 +\dfrac {1}{x^2} - 2 -3x +\dfrac {3}{x}$

We can further write it as

$x^2 +\dfrac {1}{x^2} - 2 -3x +\dfrac {3}{x}$

$= \left( x -\dfrac {1}{x}\right)^2 - 3\left(x - \dfrac {1}{x} \right)$

By taking the common terms out

$x^2 +\dfrac {1}{x^2} - 2 -3x +\dfrac {3}{x}$

$= \left( x -\dfrac {1}{x} \right)\left(x - \dfrac {1}{x} -3\right)$

Factorise : $a (a-2b-c) +2bc$

Given :

$a (a-2b-c) +2bc$

By multiplying the terms

$a (a-2b-c) +2bc$

$=a^2 -2ab-ac +2bc$

By taking the common terms out

$a (a-2b-c) +2bc$

$= a(a-2b) -c(a-2b)$

So we get,

$a (a-2b-c) +2bc$

$= (a-c)(a-2b)$

Factorise : $4a^2 -9b-2bc -c^2$

Given :

$4a^2 -9b-2a -3b$

We can write it as

$4a^2 -9b-2a -3b$

$= (2a)^2 -(3b)^2 -(2a+3b)$

Based on the equation

$a^2 -b^2 = (a+b)(a-b)$

We get,

$4a^2 -9b-2a -3b$

$= ( 2a+3b)(2a-3b)-(2a+3b)$

By taking

$( 2a+ 3b)$ as common

$4a^2 -9b-2a -3b$

$= (2a +3b)(2a-3b -1 )$

Factorise the following expressions:
(i) $32a^2b - 72b^3$
(ii) $9 (a +b)^3 - 25 ( a+b)$

$(i) 32a^2b-72b^3$

$=8b(4a^2-9b^2)$

$=8b[(2a)^2-(3b)^2]$

$=8b(2a+3b)(2a-3b)$

$(ii) 9(a+b)^3-25(a+b)$

$=(a+b)[9(a+b)^2-25]$

$=(a+b)[{ \{ 3(a+b) }\} ^{ 2 }-(5)^{ 2 }]$

$=(a+b)[(3a+3b)^2-(5)^2]$

$=(a+b)[(3a+3b+5)(3a+3b-5)]$

$=(a+b)(3a+3b+5)(3a+3b-5)$

$18 m +16 n$

${\textbf{Step - 1: Perform factorisation using suitable method}}{\text{.}}$

${\text{We have to find the factorisation of 18m + 16n}}{\text{.}}$

${\text{So let's factorise 18m and 16n first}}{\text{.}}$

${\text{18m}} = 2 \times 3 \times 3 \times {\text{m}}{\text{.}}$

${\text{16n}} = 2 \times 2 \times 2 \times 2 \times {\text{n}}{\text{.}}$

${\textbf{Step - 2: Find the factors of given, using common factor method}}{\text{.}}$

${\text{So here the common factors is only 2}}{\text{.}}$

${\text{The factorised form of 18m + 16n is like , }}2 \times \left( {9{\text{m}} + 8{\text{n}}} \right).$

$\mathbf{{\textbf{Hence, the required factorised form is 2}} \left( {9{\textbf{m + 8n}}} \right).}$

$20x^2 -45y^2$

$20x^2 -45y^2$
take out common in all terms,

$5(4x^2 -9y^2)$

$5( (2x)^2 - (3y)^2)$
we know that

$a^2 -b^2 = (a+b)(a-b)$

$5(2x +3y)(2x -3y)$

$x^3 -25x$

$x^3 -25x$
Take out common in all terms

$x(x^2 -25)$
Above terms can be written as,

$x(x^2 -5^2)$
We know that

$a^2 -b^2 = (a+b)(a-b)$

$x(x +5)(x-5)$

$14 (a-3b)^3 - 21p (a -3b)$

$14 (a-3b)^3 - 21p (a -3b)$
Take out common in all terms,
Then,

$7(a-3b)[2(a-3b)^2-3p]$
Therefore, HCF of

$14 (a-3b)^3$ and

$21p(a-3b)$ is

$7(a-3b).$

Factorise $27a^3b^3 - 18 a^2b^3 + 75a^3b^2$

$27a^3b^3 - 18 a^2b^3 + 75a^3b^2$

Minimum power of

$a$ in every term is

$2$

Minimum power of

$b$ in every term is

$2$

$3$ can be taken common from

$27,18\ \& \ 75$ .

Taking out common factor in all the terms, we get:

$3a^2b^2(9a-6b+25a)$

$10 (2p +q)^3 -15b( 2p +q)^2 + 35(2p +q)$

$10 (2p +q)^3 -15b( 2p +q)^2 + 35(2p +q)$
Take out common in all terms,
Then,

$5(2p+q)[2a(2p+q)^2-3b(2p+q)+7]$
Therefore,HCF of

$10a(2p+q)^3,15b(2p+q)^2$ and

$35(2p+q)$ is

$5(2p+q).$

$150 -6a^2$

$150 -6a^2$
Take out common in all terms,

$6( 25 - a^2 )$

$6 ( 5^2 -a^2)$
We know that

$a^2 -b^2 = (a+b)(a-b)$
So

$6 ( 5+a)(5-a)$

$32x^2 -18y^2$

$32x^2 -18y^2$
take out common in all terms

$2( 16x^2 -9y^2 )$

$2( (4x)^2 - (3y)^2)$
We know that

$a^2 -b^2 = (a+b)(a-b)$

$2 (4x +3y)(4x-3y)$

$3a(x^2 +y^2) +6b ( x^2 +y^2 )$

$3a(x^2 +y^2) +6b ( x^2 +y^2 )$
Take out common in both terms,
Then,

$3(x^2+y^2)(a+2b)$
Therefore,HCF of

$3a(x^2+y^2) \quad and \quad 6b(x^2+y^2) \quad is \quad 3(x^2+y^2)$ .

$x(x^2+y^2-x^2) +y(-x^2-y^2 +z^2) -z (x^2 +y -z^2)$

$x(x^2+y^2-x^2) +y(-x^2-y^2 +z^2) -z (x^2 +y -z^2)$
Take out common in all terms,
Then,

$(x^2+y^2-z^2)[x-y-z]$
Therefore,HCF of

$x(x^2+y^2-z^2),y(-x^2-y^2+z^2)$ and

$z(x^2+y-z^2)$ is

$(x^2+y^2-z^2)$ .

$\pi a^5 - \pi^3 ab^2$

$\pi a^5 - \pi^3 ab^2$
Take out common in all terms,

$\pi a ( a^4 - \pi^2 b^2)$

$\pi a(a^2)^2 - ( \pi b)^2 )$
We know that

$a^2 -b^2 = (a+b)(a-b)$

$\pi a(a^2 +\pi b) (a^2 - \pi b )$

Factorise $m^2+m+1=0$ .

${ m^{ 2 } }+m+1=0 \\ factorise\, \, this\, \, quadratic\, \, equation\, \, by\, \, \\ { { Discriminant\ method } }{ { . } } \\ { { D=1-4 } }\times { { 1=-3 } } \\ \Rightarrow { { Roots } }\, \, { { = } }\frac { { { { -1 } }\pm \sqrt { -3 } } }{ { 2\times 1 } } \, \, \, \, \, \, \left[ { \because \sqrt { -1 } =i } \right] \\ \Rightarrow m=\frac { { -1\pm \sqrt { -3 } } }{ 2 } \\ \Rightarrow m=-1\pm \sqrt { 3 } i \\ \therefore m=\sqrt { 3 } i-1\, \, or\, \, \left( { -\sqrt { 3 } i-1 } \right) \, \,$

Factorize:
${a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3}+8$

1650984_1669429_ans_29a85c9886c74b66a2a906a89366445e.jpg

Factorise:
$x^{2}+20x-69$

1295051_1255879_ans_329b2e2a32cf4f4dbdd33e82a05cf50b.jpg

Divide $\sqrt{2}a^3+3\sqrt{2}a^2+6a$ by $2a$ .

1304522_1581637_ans_07b39c834ab04dad9cfc4e9426a7e30c.jpg

Factorize $\dfrac {1}{6}a^{2}-a+\dfrac {4}{3}$ .

$\dfrac { 1 }{ 6 } { a }^{ 2 }-a+\dfrac { 4 }{ 3 }$

$=\dfrac{a^2-6a+8}{6}$

$=\dfrac{a^2-4a-2a+8}{6}$

$=\dfrac{a\left( a-4\right) -2\left( a-4\right)}{6}$

$=\dfrac{\left( a-2\right) \left( a-4\right)}{6}$

Hence, the answer is

$\dfrac{\left(a-2\right)\left( a-4\right)}{6}.$

1187413_1218545_ans_7b253e6b9e484f098194b69e7558583f.png

Factorise:
$4u^2+8u$

1338070_1278097_ans_0b238c44bb1c4ed3b9bc0c403212231d.PNG

Factorise
$- 3{x^2} + 3xy - 4xza$

1353301_1270623_ans_0421011b480c4972b03f8c21606a6e42.PNG

Factorise:
$\left( {a - b} \right) + {\left( {a - b} \right)^2}$

1352016_1270773_ans_0a2ae66120d2411c8aef37c3f7ad8a98.PNG

Factorise:
$- 5{a^2} + 4a + 5\left( { - {a^2} + 5} \right)$

1338039_1276519_ans_95828ee1c2394ca49b7ded58097f372c.PNG

Factorize $x^{3}+13x^{2}+32x+20$

1304726_1271480_ans_298e5677f6d44260ba4634e1a64a2ad7.JPG

Resolve $2{b}^{2}-5b-3$ in to factors.

Given,

$2{b}^{2}-5b-3$

$=2{b}^{2}-6b+b-3$

$=2b\left(b-3\right)+1\left(b-3\right)$

$=\left(b-3\right)\left(2b+1\right)$

Factorise: ${ 7x }^{ 3 }z-{ 21x }z^{ 2 }$

$7{x}^{3}z-21x{z}^{2}$

$=7xz\left({x}^{2}-3z\right)$

${ ab }^{ 2 }+(a-1)b-1$

${ ab }^{ 2 }+(a-1)b-1$
Above qyestion can be written as,

$ab^2+ab-b-1$
Take out common in all terms,

$ab(b+1)-1(b+1)$

$(b+1)(ab-1)$

Factorise: ${ x }^{ 2 }+18x+45$

$\begin{array}{l} { x^{ 2 } }+18x+45 \\ ={ x^{ 2 } }+15x+3x+45 \\ =x\left( { x+15 } \right) +3\left( { x+15 } \right) \\ =\left( { x+15 } \right) \left( { x+3 } \right) \end{array}$

Factorize $p^2+1.5p+0.5$ .

1318507_1586943_ans_5b3cd5e8ed1e413396cedf2dd2244eaf.jpg

Find the quotient and remainder on dividing the polynomials $a{x^2} + \left( {b + ac} \right)x + bc\;by\;x + c$

1211384_1505694_ans_4c8eb1a5af14450680717dfab468f453.jpg

Resolve into factors $3x^2-6x$

Given that:

$3x^2-6x$

$= 3\times x\times x-2\times 3\times x$ , factor each monomial

$=3x(x-2)$ , factor out common factor

$3x$

The factors form of

$3x^2-6x$ is

$3x.(x -2)$

Divide : $a^2 + 7a +12$ by $(a+4)$ .

$a^2+7a+12 = a^2+4a+3a+12$

$=a(a+4)+3(a+4)$

Now dividing it by

$(a+4)$ :

$\Rightarrow \dfrac{a(a+4)+3(a+4)}{(a+4)} = a+3$

Find the quotient and remainder.
$(2p^{2}+7p-9)\div (p-6)$

1294378_1609041_ans_0bae36a272094f2992519bf3b42ba931.jpg

Factorise

$18a^3b^3−27a^2b^3+36a^3b^2$

let’s take HCF of above equationBy taking

$9a^2b^2$ as a common factor for the above equation we get

$18a^3b^3-27a^2b^3+36a^3b^2=9a^2b^2(2ab - 3b +4a)$

Factorise

$12x^2y^3 - 21 x^3y^2$

let’s take HCF of above equationBy taking

$3x^2y^2$ as a common factor for the above equation we get,

$12x^2y^3$ -

$21x^3y^2 = 3x^2y^2(4y-7x)$

Factorise

$15ab^2 – 20a^2b$

let’s take HCF of above equationBy taking 5ab as a common factor for the above equation we get,

$15ab^2 - 20a^2b =5ab(3b - 4a)$

Factorise the following:
$27 - 125x^{3} - 135x + 225x^{2}$

$27 - 125x^{3} - 135x + 225x^{2}$

$= (3)^{3} - (5x)^{3} - 3(3)(5x)[3 - 5x]$

$= (3 - 5x)^{3}$

$[\because (a - b)^{3} = a^{3} - b^{3} - 3ab(a - b)]$

Resolve in to factors $x^2+xy$

Given that:

$x^2+xy$

$=x\cdot x+x\cdot y$ , factor each terms

$=x(x+y)$ , factor out common factor

$x$

Hence the factorisation of

$x^2+xy$ is

$x(x+y)$

Factorise $9{x}^{2}+12xy$ .

We write

$9{x}^{2}+12xy=(3x)(3x)+(3x)(4y)$
Here

$3x$ is the common to both the terms. Hence

$(3x)(3x)+(3x)(4y)=3x(x+4y)$
Thus

$3,x$ and

$(3x+4y)$ are the factors of

$9{x}^{2}+12xy$

Factorise the following:
$125x^{3} + 64y^{3} - 300x^{2}y + 240xy^{2}$

$125x^{3} - 64y^{3} - 300x^{2}y + 240xy^{2}$

$= (5x)^{3} - (4y)^{3} - 3(5x)(4y) [5x - 4y]$

$= (5x - 4y)^{3}$

$[\because (a - b) = a^{3} - b^{3} - 3ab(a - b)]$

$5x^2-10x$

We can see that

$5x$ is the HCF of

$5x^2$ and

$10x$ .

We can remove them from both the terms.

Thus we get,

$5x^2 - 10x$

$= (5x)(x) - (5x) \times 2$

$= (5x)(x - 2)$ .

$6xy + 9y^2$

Factor out

$3y$ from the expression,

$6xy+9y^2=3y(2x+3y)$

So, the factors of the given expression are equal to

$3,y$ and

$(2x+3y).$

Divide $-8x^{9}$ by $2x^{7}$

We divide

$-8$ by

$2$ and

$x^{9}$ by

$x^{7}$ . Thus

$\dfrac {-8x^{9}}{2x^{7}} = \left (\dfrac {-8}{2}\right )\left (\dfrac {x^{9}}{x^{7}}\right ) = -4x^{9 - 7} = -4x^{2}$ .

Find the common factors of the terms $8x$ and $24$ .

We need to find common factors of

$8x$ and

$24$ .

The factor of

$8x=2\times 2\times 2\times x$

and that of

$24=3\times 2\times 2\times 2$

Therefore, the common factors are

$2\times 2\times 2=8$ .

Factorise: $25-50p-100q$ .

We observe that

$25-50p-100q=25(1-2p-4q)$ . Hence

$25$ and

$(1-2p-4q)$ are the factors of

$25-50p-100q$ .

Factorise: $2p(a-b)+3q(5a-5b)+4r(2b-2a)$ .

$2p(a-b)+3q(5a-5b)+4r(2b-2a)$

$=2p(a-b)+3q\times 5(a-b)-4\times r\times 2(a-b)$

$=(a-b)(2p+15q-8r)$
Thus

$(a-b)$ and

$(p+15p-8r)$ are the factors of the given expression.

$25a^2 b + 35ab^2$

Given expression is

$25a^2b+35ab^2$

It can be factorized as

$25a^2b+35ab^2=5ab(5a+7b)$ .

Factorise the following:
$64a^{3} + 21b^{3}$

$64a^{3} + 27b^{3}$

$= (4a)^{3} + (3b)^{3}$

$= (4a + 3b)[(4a)^2 - 4a.3b + (3b)^2]$

$[\because a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})]$

$= (4a + 3b)[16a^{2} - 12ab + 9b^{2}]$

Factorise $6ab - b^2 - 2bc + 12 ac$

Given expression is

$6ab-b^2-2bc+12ac$

It can be factorised as,

$6ab-b^2-2bc+12ac$

$=6ab-b^2+12ac-2bc$

$=b(6a-b)+2c(6a-b)$

$=(6a-b)(b+2c)$

Thus factors are

$(6a-b)$ and

$(b+2c)$ .

$7p^2 + 49pq$

We need to factorise

$7p^2+49pq$

We can write it as

$7p^2+49pq=7p(p+7q)$
Thus factors are

$7,p,(p+7q)$ .

Factorise:
$a^4-343a$

$a^4-343a$

$=a(a^3-343)$

Using,

$a^3-b^3=(a-b)(a^2+b^2+ab)$

$=a(a^3-7^3)$

$=a(a-7)(a^2+7^2+7a)$

$=a(a-7)(a^2+49+7a)$

Factorise: $4ab{ x }^{ 2 }+8abx+12aby$

Since in the equation

$4abx^2+8abx+12aby$ ,

$4ab$ is the common term, therefore, can be factorised as follows:

$4abx^2+8abx+12aby=4ab(x^2+2x+3y)$

factorise : ${ 27p }^{ 3 }\left( { 4q-2r }^{ 3 } \right) +{ 64q }^{ 3 }\left( { 2r-3p }^{ 3 } \right) +{ 8r }^{ 3 }\left( { 3p-4q }^{ 3 } \right)$

2039421_1285022_ans_5e387b322e074cfa9f92d7f2c53e6829.jpg

Find the factors of $y^{2}-7y+12$ .

${y}^{2}-7y+12$

$y^2-3y-4y+12$

$y(y-3)-y(y-3)$

$(y-3)(y-4)$

$\therefore$

$y=3,4$

$(p^{2}+5p+4)\div(p+1)$

2045403_1571400_ans_bfcdb54b8ac743a7af3c90e91a76e44e.jpg

$36a^2b - 60a^2bc$

We need to factorise

$36a^2b-60a^2bc$

$=36a^2b-60a^2bc$

$=3\times 4a^2b(3-5c)$

Factorise the following:
$2x^3y^2-4x^2y^3+8xy^4$

To factorise:

$2x^3y^2-4x^2y^3+8xy^4$

Now,

$2x^3y^2-4x^2y^3+8xy^4$

$=2xy^2\times x^2-2xy^2\times 2xy+2xy^2\times 4y^2$

$=2xy^2(x^2-2xy+4y^2)$

Hence, this is the answer.

Factorize the following.
$20a^{12}b^2-15a^8b^4$ .

1682352_1673422_ans_2c352bbee4b14e13b2078f9882583616.jpg

$4m^2, 6m^2, 8m^3$

We need to find common factors of

$4m^2, 6m^2, 8m^3$

The factor of

$4m^2=2\times 2\times m\times m$ ,

$6m^2=3\times 2\times m\times m$

and that of

$8m^3=2\times 2\times 2\times m\times m\times m$

Therefore, the common factors are

$2\times m\times m=2m^2$ .

Factorize the following.
$2a^4b^4-3a^3b^5+4a^2b^5$ .

1682457_1673431_ans_7551428d0779489eb4f2871342b9dcd4.jpg

Factorize the following.
$28a^2+14a^2b^2-21a^4$ .

1682468_1673432_ans_d304553947514524a92c5de4de5e17b2.jpg

Factorize the following expression.
$abx^2+(ay-b)x-y$ .

1687283_1673560_ans_0acb2b358e8c4d53af9f1007295eee66.jpg

Factorize the following expression.
$a(a-2b-c)+2bc$ .

1687341_1673573_ans_b9a879aa9e0d4c4fa55912e91815c2c9.jpg

Factorize the following algebraic expression.
$6x(2x-y)+7y(2x-y)$ .

1686882_1673450_ans_d38cd98cf31646439fd9ea37bba3aff8.jpg

Factorize the following.
$16m-4m^2$ .

1682608_1673440_ans_3d4d426f4f184777a831733064261b01.jpg

Factorize the following.
$9x^2y+3axy$ .

Take

$3xy$ common from both the terms

$9x^2y+3axy=3xy(3x+a)$

Factorize the following expression.
$a(a+b-c)-bc$ .

1687344_1673576_ans_fe906399bb12496799c2fe5870406fa5.jpg

Factorize the following expression.
$16(a-b)^3-24(a-b)^2$ .

1687300_1673567_ans_251ffb479d21482683be0eb6460cb541.jpg

Find the common factors of the given terms: $7xy, 35x^2y^3$

We need to find common factors of

$7xy, 35x^2y^3$

The factor of

$7xy=7\times x\times y$

and that of

$35x^2y^3=7\times 5\times x\times x\times y\times y\times y$

Therefore, the common factors are

$7\times y\times x=7xy$ .

$3ax - 6xy + 8by - 4bx$

$3ax-6xy+8by-4bx$

$=3x(a-2y)-2b(a-2y)$

$=(3x-2b)(a-2y)$

Thus the factors are

$(3x-2b),(a-2y)$

$5x^2 - 25xy$

We need to factorise

$5x^2-25xy$

We can write it as

$5x^2-25xy=5x(x-5y)$
Thus factors are

$5,x,(x-5y)$ .

Factorize the following expression.
$a^2x^2+(ax^2+1)x+a$ .

1687315_1673570_ans_ece42bbd75e140438f82834d3a3027e2.jpg

Factorize the following expression.
$x^2+xy+xz+yz$ .

1687179_1673533_ans_0d79d5a4c3b74e06a6e0e08bcb9e1230.jpg

Factorize the following.
$x^2yz+xy^2z+xyz^2$ .

1682622_1673444_ans_406ab5a59e684d6e963930d513e68763.jpg

Factorize the following algebraic expression.
$49-a^2+8ab-16b^2$ .

1588648_1673722_ans_02bda3f37c004635b1c2a39b6cb0edc4.jpg

Factorize the following expression.
$ab-a-b+1$ .

1687366_1673581_ans_12c15203e26049a7969b6323ec86d3ab.jpg

Factorize the following expression.
$x-y-x^2+y^2$ .

1694019_1673649_ans_9bbc40d6abe74b9fb88ff11f0b41513a.jpg

Factorize the following expression.
$x^4-625$ .

1694015_1673643_ans_3861a4b50fc34493878f234dcb2b4117.jpg

Factorize the following expression.
$x^2+y-xy-x$ .

1687369_1673584_ans_e93d217eec404dbc8dc0611d996ee4e1.jpg

$12x^2y, 18xy^2$

The expansion of

$12x^2y$ is

$3\times 2\times 2\times x\times x\times y$

And that of

$18xy^2$ is

$3\times 3\times 2\times x\times y\times y$

By factorization, we see that the term

$2,3,x,y$ are common to both.

Hence, the common factors of the given terms is

$6xy$ .

Find the common factors of the given terms: $15p, 20qr, 25rp$ .

The factor of

$15p=3\times 5\times p$

$20qr=4\times 5\times q\times r$

$25pr=5\times 5\times p\times r$

The common factors is

$5$

$ut + at^2$

We need to factorise

$ut+at^2$

We can write it as

$ut+at^2=t(u+at)$
Thus factors are

$t,(at+u)$ .

$x^3 + 2x^2 + 5x + 10$

$x^3+2x^2+5x+10$

$=x^2(x+2)+5(x+2)$

$=(x^2+5)(x+2)$
the factors are

$(x^2+5),(x+2)$

$25x^2 + 9y^2 - 30xy$

Using the identity

$(a^2+b^2-2ab)=(a-b)^2$

We can write,

$(25x^2+9y^2-30xy)$ as

$=(5x)^2+(3y)^2-2\times 5x\times 3y$

$=(5x-3y)^2$

Hence, factors are

$(5x-3y)(5x-3y)$ .

Factorize the following quadratic polynomial by using the method of completing the square.
$p^2+6p+8$ .

1590495_1673866_ans_5d747843cf3c4b2095b94c7e913f0717.jpg

Factorize the following expressions:
$17l^{2} + 85m^{2}$

Given polynomial is

$17l^2+85m^2$

Here we can take

$17$ common form both the terms.

Thus we have

$17l^2+85m^2=17(l^2+5m^2)$

Hence, factors are

$17,(l^2+5m^2)$ .

Factorize the following expression.
$4(xy+1)^2-9(x-1)^2$ .

1694026_1673654_ans_9d6e4fdad620439c8c35217dcd6af907.jpg

Factorize the following expression.
$p^2q^2-p^4q^4$ .

1694008_1673639_ans_d53c1bd5f02e425ba03f23765f801bef.jpg

Factorize the following expressions:
$-12y + 20y^{3}$

Given polynomial is

$-12y+20y^3$

Here we can take

$-4y$ common from both the terms.

Thus we have

$-12y+20y^3=-4y(3-5y^2)$

Hence, factors are

$-4y,(3-5y^2)$

Factorize the following expressions:
$5a^{2} + 35a$

Given polynomial is

$5a^2+35a$

Here we can take

$5a$ common form both the terms.

Thus we have

$5a^2+35a=5a(a+7)$

Hence, factors are

$5,a,(a+7)$ .

Factorize the following expressions:
$pq - pqr$

Given polynomial is

$pq-pqr$

Here we will take

$pq$ common from both the terms.

Thus we have

$pq-pqr=pq(1-r)$

Hence, factors are

$p,q,(1-r)$ .

Factorize: $6xy - 4y + 6 - 9x$

Given polynomial is

$6xy-4y+6-9x$

It can be factorised as

$6xy-4y+6-9x=2y(3x-2)-3(3x-2)$

$=(3x-2)(2y-3)$

Thus factors are

$(3x-2),(2y-3)$ .

Factorize the following expressions:
$18m^{3} - 45mn^{2}$

Given polynomial is

$18m^3-45mn^2$

Here we can take

$9m$ common form both the terms.

Thus we have

$18m^3-45mn^2=9m(2m^2-5n^2)$

Hence, factors are

$3,3m,(2m^2-5n^2)$

Factorize:
$mx - my - nx + ny$

We need to factorise

$mx-my-nx+ny$

Thus we have

$mx-my-nx+ny=m(x-y)-n(x-y)$
Factor out the term

$x-y$ , we get

$=(x-y)(m-n)$

Factorize:
$ax^{3} - bx^{2} + ax - b$

Given polynomial is

$ax^3-bx^2+ax-b$

It can be factorised as

$ax^3-bx^2+ax-b=x^2(ax-b)+1(ax-b)$

$=(ax-b)(x^2+1)$

Thus factors are

$(ax-b),(x^2+1)$ .

Factorize:
$2m^{3} + 3m - 2m^{2} - 3$

Given polynomial is

$2m^3+3m-2m^2-3$

It can be factorised as

$2m^3+3m-2m^2-3=m(2m^2+3)-1(2m^2+3)$

$=(m-1)(2m^2+3)$

Thus factors are

$(m-1),(2m^2+3)$ .

Factorize:
$2x + 3xy + 2y + 3y^{2}$

Given polynomial is

$2x+3xy+2y+3y^2$

It can be factorised as

$2x+3xy+2y+3y^2=x(2+3y)+y(2+3y)=(x+y)(2+3y)$

Thus factors are

$(x+y),(2+3y)$ .

Factorize:
$p^{2} - 6p + 8$

We need to factorise

$p^2-6p+8$

It can be written as

$p^2-6p+8=p^2-4p-2p+8$

$=p(p-4)-2(p-4)$

$=(p-4)(p-2)$

Factorize the polynomial:
$a^{2} + 13a + 12$

$a^2+13a+12$

$=a^2+12a+a+12$

$=a(a+12)+1(m+12)$

$=(a+12)(a+1)$
factors are

$(a+12),(a+1)$

Factorize:
$a^{2} + 11b + 11ab + a$

Given polynomial is

$a^2+11b+11ab+a$

It can be rearranged as

$a^2+11ab+a+11b$

It can be factorised as

$a^2+11ab+a+11b=a(a+11b)+1(a+11b)$

$=(a+11b)(a+1)$

Thus the factors are

$(a+11b)(a+1)$ .

Factorize:
$15b^{2} - 3bx^{2} - 5b + x^{2}$

Given polynomial is

$15b^2-3bx^2-5b+x^2$

It can be factorised as

$15b^{2} - 3bx^{2} - 5b + x^{2}=3b(5b-x^2)-1(5b-x^2)$

$=(3b-1)(5b-x^2)$

Thus factors are

$(3b-1),(5b-x^2)$ .

Factorize:
$169p^{2} - 625q^{2}$

Given expression is

$169p^2-625q^2$

We know

$a^2-b^2=(a-b)(a+b)$

By using this identity, we have

$169p^2-625q^2=(13p)^2-(25q)^2$

$=(13p-25q)(13p+25q)$
Thus factors are

$(13p-25q),(13p+25q)$ .

Factorize:
$a^{2}x^{2} + axy + abx + by$

Given polynomial is

$a^2x^2+axy +abx+by$

It can be factorised as

$a^2x^2+axy+abx+by=ax(ax+y)+b(ax+y)$

$=(ax+y)(ax+b)$

Thus factors are

$(ax+y),(ax+b)$ .

Work out the following divisions:
$(9x^{5} - 15x^{4} - 21x^{2}) \div (3x^{2})$

$(9x^{5} - 15x^{4} - 21x^{2}) \div (3x^{2})$

$=\dfrac{9x^5-15x^4-21x^2}{3x^2}$

$=\dfrac{9x^5}{3x^2}-\dfrac{15x^4}{3x^2}-\dfrac{21x^2}{3x^2}$

$=3x^3-5x^2-7$

Work out the following divisions:
$(5x^{3} - 4x^{2} + 3x) \div (2x)$

$\dfrac{5x^3-4x^2+3x}{2x}$

$=\dfrac{5x^3}{2x}-\dfrac{4x^2}{2x}+\dfrac{3x}{2x}$

$=\dfrac{5}{2}x^2-2x+\dfrac{3}{2}$

Factorize the polynomial:
$x^{2} - 14xy + 24y^{2}$

$x^2-12xy+24y^2$

$=x^2-12xy-2xy+24y^2$

$=x(x-12y)-2y(x-12y)$

$=(x-2y)(x-12y)$
factor are,

$(x-2y),(x-12y)$

Factorize the polynomial:
$m^{2} - 21m - 72$

$m^2-21m-72=m^2-24m+3m-72=(m-24)(m+3)$
factors are

$(m-24),(m+3)$

Work out the following divisions:
$(8x^{4}yz - 4xy^{3}z + 3x^{2}yz^{4})\div (xyz)$

$\dfrac{8x^4yz-4xy^3z+3x^2yz^4}{xyz}$

$=\dfrac{8x^4yz}{xyz}-\dfrac{4xy^3z}{xyz}+\dfrac{3x^2yz^4}{xyz}$

$=8x^3-4y^2+3xz^3$

Simplify the following expressions:
$(x^{2} + 7x + 10)\div (x + 2)$

$=\dfrac{x^2+7x+10}{x+2}$

$=\dfrac{x^2+5x+2x+10}{x+2}$

$=\dfrac{x(x+5)+2(x+5)}{x+2}$

$=\dfrac{(x+5)(x+2)}{x+2}$

$=x+5$

Work out the following divisions:
$4x^{2}y - 28xy + 4xy^{2} \div (4xy)$

$\dfrac{4x^2y-28xy+4xy^2}{-4xy}$

$=\dfrac{4x^2y}{-4xy}-\dfrac{28xy}{-4xy}+\dfrac{4xy^2}{-4xy}$

$=-x+7-y$

$=7-x-y$

Work out the following divisions:
$5y^{3} - 4y^{2} + 3y \div y$

$\dfrac{5y^3-4y^2+3y}{y}$

$=\dfrac{5y^3}{y}-\dfrac{4y^2}{y}+\dfrac{3y}{y}$

$=5y^2-4y+3$

Simplify:
$(7m^{2} - 6m)\div m$

$\dfrac{7m^2-6m}{m}$

$=\dfrac{7m^2}{m}-\dfrac{6m}{m}$

$=7m-6$

Factorize the polynomial:
$x^{2} - 5x + 6$

$x^2-5x+6$

$=x^2-3x-2x+6$

$=x(x-3)-2(x-3)$

$=(x-2)(x-3)$

factors are

$(x-3),(x-2)$

Factorize the following:
$2m^{2} - 10mn - 2m + 10n$

We need to factorise

$2m^2-10mn-2m+10n$

It can be written as

$2m^2-10mn-2m+10n=2m(m-5n)-2(m-5n)$

$=(m-5n)(2m-2)$

$=2(m-5n)(m-1)$

Factorise:
$4a - 8b + 5ax - 10 bx$

$4a - 8b + 5ax - 10bx$

$= (4a - 8b) + (5ax - 10 bx)$

$= 4(a - 2b) + 5x(a - 2b)$

$= (a - 2b) (4 + 5x)$

Resolved into factor
$ap+ap^{2}$

$\begin{array}{l} ap+a{ p^{ 2 } }=a\left( { p+{ p^{ 2 } } } \right) \\ =ap\left( { 1+p } \right) \end{array}$

Hence, the factor is

$ap\left( { 1+p } \right)$

Solve $2x^2 - 5x =0$

Given:

$2x^{2}-5x=0$

$=x\left (2x-5\right )=0$

Case 1:

$x=0$

Case 2:

$2x-5=0$

$x=\dfrac 52$

Therefore

$x=0, \dfrac 52$

Factorize:
${ ab }^{ 2 }-ab-{ a }^{ 2 }b+{ a }^{ 2 }$

$a{ b }^{ 2 }-ab-{ a }^{ 2 }b+{ a }^{ 2 }\\ \Rightarrow ab(b-1)-{ a }^{ 2 }(b-1)\\ \Rightarrow (ab-{ a }^{ 2 })(b-1)=a(b-a)(b-1)$

Factorisation of 721 ?

Factorization of

$721$

$\Rightarrow 7\times 103$

Hence, this is the answer.

Factorize of $a^{2}+10\ a+24$

$a^{2}+10a+24$

Split

$24$ as such that sum or difference of factor is equal to

$10$

$24 = 6\times 4$

$= 6+4 = 10$

$a^{2}+6a+4a+24$

$a(a+6)+4(a+6)$

$= (a+6)(a+4)$

Divide the following:
$({ x }^{ 3 }-64)\div (x-4)$

1339265_1261037_ans_6903b396b56f448eac1c1479323011f8.jpg

Divide, Write the quotient and the remainder.
$40a^{3}\div (-10a)$

$40{a}^{3}\div\left(-10a\right)$

$=\dfrac{40{a}^{3}}{-10a}$

$=-4{a}^{2}$

Divide: $3{a}^{3}-9{a}^{2}b-6a{b}^{2}$ by $-3a$

given,

$3{a}^{3}-9{a}^{2}b-6a{b}^{2}\div\,-3a$

$=\dfrac{3{a}^{3}-9{a}^{2}b-6a{b}^{2}}{-3a}$

$=\dfrac{-3a\left(-{a}^{2}+3ab+2{b}^{2}\right)}{-3a}$

$=-{a}^{2}+3ab+2{b}^{2}$

$=2{b}^{2}+3ab-{a}^{2}$

Factorize:
$99x^2-202xy+99y^2$ .

Now,

$99x^2-202xy+99y^2$

$=100x^2-200xy+100y^2-(x^2+2xy+y^2)$

$=(10x-10y)^2-(x+y)^2$

$=(9x-11y)(11x-9y)$ .

Simplify :
$3x(2+5x)-6(1-2x)$

$3x(2+5x)-6(1-2x)$

$=6x+15x^2-6+12x$

$=15x^2+18x-6$

Factorise
$6{ x }^{ 2 }-5x-6$

$6x^2-5x-6$

$6x^2+4x-9x-6$

$2x(3x+2)-3(3x+2)$

$\Rightarrow (2x-3)(3x+2)$ .

1205708_1453800_ans_ef8af3a9acd34aad93dc09b943533306.jpg

Factories the expressions.
$y(y+z)+9(y+z)$

$y(y+z)+9(y+z)$

$\Rightarrow (y+z)(y+9)$ .

1199969_1445638_ans_01ed311dfe4842c88af8a74e3c05e289.jpg

Factorize:
$x ^ { 2 } - 23 x + 132$

$\begin{aligned}{}{x^2} - 23x + 132 &= {x^2} - 12x - 11x + 132\\& = x(x - 12) - 11(x - 12)\\ &= (x - 11)(x - 12)\end{aligned}$

Factorise :
$6x^{2}+5x-6$

$6 x^2+5 x-6=6 x^2+9 x-4 x-6=3 x(2 x+3)-2(2 x+3)=(3 x-2)(2 x+3)$

Factorise the following expression.
${q^2} - 10q + 21$

Now,

${q^2} - 10q + 21$

$=q^2-7q-3q+21$

$=(q-3)(q-7)$ .

Factorise the expression
$ax^{2}+bx$

$a {x}^{2} + bc$

$= x \left( ax + b \right)$

Hence

$a {x}^{2} + bx = x \left( ax + b \right)$ .

Factorise:
$q^2-10q+21$

${q}^{2} - 10q + 21$

$= {q}^{2} - 7q - 3q + 21$

$= q \left( q - 7 \right) - 3 \left( q - 7 \right)$

$= \left( q - 3 \right) \left( q - 7 \right)$

Solve : $3\left( x-4 \right) ^{ 2 }-5\left( x-4 \right)$

$3(x-4)^2-5(x-4)$

$\Rightarrow (x-4)[3(x-4)-5]$

$\Rightarrow (x-4)[3x-12-5]$

$\Rightarrow (x-4)(3x-17)$ .

1205728_1453886_ans_f1003bf30f19400e828a6bdb64743200.jpg

Solve it :-
$x^2 + (a + \frac{1}{a}) x + 1 = 0$

1326044_1566063_ans_3c035f1fbe2045ecaf5004792501a7f5.jpeg

Factorize; ${ g }^{ 2 }-10 g+21.$

1318684_1490219_ans_1a3a655a0e954885b812e03e347fdd96.JPG

Factorise:
$\sqrt {2}x-\sqrt {3}x^{2}$

$\sqrt{2}{x} - \sqrt{3} {x}^{2}$

$= x \left( \sqrt{2} - \sqrt{3} x \right)$

Hence

$\sqrt{2}{x} - \sqrt{3} {x}^{2} = x \left( \sqrt{2} - \sqrt{3} x \right)$

Factorise : ${ a }^{ 2 }-\left( b+5 \right) a+5b$

$a^2-(b+5)a+5b$

$\Rightarrow a^2-ab-5a+5b$

$\Rightarrow a(a-b)-5(a-b)$

$\Rightarrow (a-5)(a-b)$ .

1205718_1453810_ans_e457d2820f2342dca2bd9275f97af0a9.jpg

Find
$\sqrt { 3} { x }^{ 2 } -2x\sqrt { 3 }$

$\sqrt{3}x^2-2x\sqrt{3}$

$\Rightarrow \sqrt{3}x(x-2)$ .

1205730_1453890_ans_74bd55cc5dc841c8959b3bd280f712c8.jpg

Factorize
$21 py^2 -56py$

$21 py^2 -56py$
Take

$7py$ common in both terms,
Thus,

$7py(3y-8)$

Divide :
$x^{5} - 15x^{4} - 10x^{2}$ by $-5x^{2}$

$x^{5} - 15x^{4} - 10x^{2}$ by

$-5x^{2}$
It can be written as

$= (x^{5} - 15x^{4} - 10x^{2}) / -5x^{2}$
Separating the terms

$= x^{5} / - 5x^{2} - 15x^{4} / -5x^{2} - 10x^{2} / -5x^{2}$

$= -1 / 5x^{3} + 3x^{2} + 2$

Divide :
$8m - 16$ by $-8$

$8m - 16$ by

$-8$
We can write it as

$= (8m - 16) / -8$
Separating the terms

$= \dfrac{8m}{-8}-\dfrac{16}{-8}$

$=-m+2$

Divide :
$4a^{2} - a$ by $-a$

$4a^{2} - a$ by

$-a$
We can write it as

$= (4a^{2} - a) / - a$
Separating the trms

$= 4a^{2} / -a - a. -a$

$= -4a + 1$

Divide :
$3y^{3} - 9ay^{2} - 6ab^{2}y$ by $- 3y$

$3y^{3} - 9ay^{2} - 6ab^{2}y$ by

$- 3y$
It can be written as

$= 3y^{3} / -3y - 9ay^{2} / - 3y - 6ab^{2}y / -3y$

$= - y^{2} + 3ay^{2} + 2ab^{2}$

Divide :
$10x^{3}y - 9xy^{2} - 4x^{2}y^{2}$ by $xy$

$10x^{3}y - 9xy^{2} - 4x^{2}y^{2}$ by

$xy$
It can be written as

$= (10x^{3}y - 9xy^{2} - 4x^{2}y^{2}) / xy$
Separating the terms

$= 10x^{3}y / xy - 9xy^{2} / xy - 4x^{2}y^{2} / xy$

$= 10x^{2} - 9y - 4xy$

Divide :
$8x + 24$ by $4$

$8x + 24$ by

$4$
We can write it as

$= (8x + 24) / 4$
Separing the terms

$= 8x / 4 + 24 / 4$

$= 2x + 6$

$4x^3 -6 x^2$

$4x^3 -6 x^2$
Take out common in both terms,
Then,

$2x^2(2x-3)$
Therefore,HCF of

$4x^3\quad and \quad 6x^2 \quad is \quad 2x^2.$

Factorise the expression $m^2 - 4m - 21$

Comaring

$m^2 - 4m - 21$ with the identity

$x^2 + (a + b) x + ab$ we not that ab = - 21, and a + b = - 4. So, (-7) + 3 = 4 and (-7)(3) = -21
Hence,

$m^2 - 4m - 21 = m^2 - 7m + 3m - 21$

$= m(m - 7) + 3(m - 7)$

$= (m - 7)(m + 3)$
Therefore

$m^2 - 4m - 21 = (m - 7)(m + 3)$

Factorise : $14{x^2} + 9x + 1$

$14x^2+7x+2x+1$

$\Rightarrow 7x\left( { 2x+1 } \right) +1\left( { 2x+1 } \right)$

$\Rightarrow \left( { 7x+1 } \right) \left( { 2x+1 } \right)$

Hence, the answer is

$\left( { 7x+1 } \right) \left( { 2x+1 } \right).$

Solve
$3x^{4}-4x^{3}$

$3{ x }^{ 4 }-4{ x }^{ 3 }={ x }^{ 3 }\left( 3x-4 \right)$

Facrorise $m(m-1)-n(n-1)$

$m(m-1)-n(n-1)$

${ m }^{ 2 }-m-{ n }^{ 2 }+n$

$\left( { m }^{ 2 }-{ n }^{ 2 } \right) -\left( m-n \right)$

$\left( m+n \right) \left( m-n \right) -\left( m-n \right)$

$=\left( m-n \right) \left( m+n-1 \right)$

Identify the terms and factors in the following expression.
i) $5ab^2 + 7a^2b$
ii) $12xyz - 6xy$

$5ab^2 + 7a^2b$

$=ab(5b+7a)$ .

ii)

$12xyz - 6xy$

$=xy(12z-6)$ .

Divide $(4{x}^{2}-17xy+4{y}^{2} ) by ( x-4y)$

$\begin{array}{l} 4{ x^{ 2 } }-17xy+4{ y^{ 2 } }=4{ x^{ 2 } }-16xy-xy+4{ y^{ 2 } } \\ \Rightarrow 4x\left( { x-4y } \right) -y\left( { x-4y } \right) \\ \Rightarrow \left( { x-4y } \right) \left( { 4x-y } \right) \\ =\left( { 4x-y } \right) \end{array}$

Factorize:
$p^2+p-(a+2)(a+1)$ .

Now,

$p^2+p-(a+2)(a+1)$

$=p^2+p\{(a+2)-(a+1)\}-(a+2)(a+1)$

$=p(p+a+2)-(a+1)(p+a+2)$

$=(p+a+2)(p-a-1)$ .

Factorize:
$6(x-y)^2-(x-y)-15$ .

Now,

$6(x-y)^2-(x-y)-15$

$=6(x-y)^2-10(x-y)+9(x-y)-15$

$=2(x-y)(3x-3y-5)+3(3x-3y-5)$

$=(3x-3y-5)(2x-2y+3)$ .

Factorise
$5y^{2}+5y-10$

Consider

$5y^2+5y-10$

$5y^2+5y-10=5(y^2+y-2)$

$=5(y-1)(y+2)$

$\therefore 5y^2+5y-10=5(y-1)(y+2)$

Factorise $64xy+8xy^{2}$

Simplifying $64xy + 8x{y^2}$

$\Rightarrow 64xy + 8x{y^2} = 8xy\left( {8 + y} \right)$

Divided $(32{x}^{4}{y}^{3}-16{x}^{3}{y}^{4})$ by $(-8{x}^{2}y)$

$\dfrac{32x^4y^3 - 16x^3y^4}{-8x^2y}$

Taking

$16x^3y^3$ common from numerator,

$\Rightarrow \dfrac{16x^3y^3(2x - y)}{-8x^2y}$

$\Rightarrow$

$2xy^2(y - 2x)$

$\Rightarrow 2xy^3 - 4x^2y^2$

Factorise:
$24x^3 – 36x^2y$

Let’s take HCF of the equation $$24x^3 - 36x^2y

By taking

$12x^2$ as a common factor for the above equation we get,

$24x^3 - 36x^2y = 12x^2(2x - 3y)$

$38x^{3}y^{2}z + 19xy^{2}$ is equal to __________ .

1975368_1792776_ans_a1e08ab85c07496c8c91bdd016eb669c.jpg

Factorise
$12x^2-27$

By simplifying further the above equation is

$3(4x^2 - 9)$
By using the formula

$a^2 - b^2 = (a+b) (a-b)$
Now solving for the above equation

$3(4x^2 - 9) = 3 ((2x)^2 - (3)^2 = 3(2x+3) (2x - 3)$

Factorise

$x(a-3) + y(3-a)$

$x(a-3) +y(a-3)$

By taking

$(a-3)$ as a common factor for the above equation we get,

$x(a-3) +y(a-3) = (a-3) (x+y)$

Factorise

$x^3(2a-b) + x^2(2a-b)$

By taking

$2a-b$ as a common factor for the above equation we get,

$x^3(2a-b) + x^2(2a-b)=(2a-b)(x^3+x^2)$

Factorise
$16p^3-4p$

$16p^3-4p$ can be written as

$4p (4p^2 -1)$
By using the formula

$a^2 - b^2 = (a +b) (a -b)$
Now solving for the above equation

$(4p^2 - 1) =4p((2p)^2-(1)^2)$

$=4p (2p +1)(2p -1)$

$9+6z+9a^2$

we rewrite it as

$z^2 + 6z + 9$

We solve by using the formula

$(a +b)^2 = a^2 + b^2 + 2ab$

We get,

$z^2 +(3)^2 + 2(z) (3) = (z + 3)^2$

Factorise
$3x^5-48x^3$

$3x^5-48x^3$ can be written as

$3x^3(x^2-16)$
By using the formula

$a^2 - b^2 = (a + b) (a - b)$
Now solving for the above equation

$3x^3(x^2 - 16) = 3x^3((x)^2 - (4)^2) = 3x^3(x + 4) (x - 4)$

Factorise
$20a^2-45b^2$

By simplifying further the above equation is

$5(4a^2 - 9b^2)$
By using the formula

$a^2 - b^2 = (a + b)$
Now solving for the above equation

$5 (4a^2 - 9b^2) = 5((2a)^2 - (3b)^2)=5(2a +3b) (2a -3b)$

Perform the following division:
$(ax^{3}-bx^{2}+cx) \div (-dx)$

$\dfrac{ax^{3}-bx^{2}+cx}{-dx}$

$=\dfrac{ax^{3}}{(-dx)}+\dfrac{bx^{2}}{dx}+\dfrac{cx}{(-dx)}$

$=-\dfrac{a}{d}x^{2}+\dfrac{b}{d}x-\dfrac{c}{d}$

Factorise the following polynomials :
(i) $a^2b+ab^2 -abc - b^2c +axy +bxy$
(ii) $ax^2 -bx^2 + ay^2 -by^2 + az^2 - bz^2$

$(i) a^2b+ab^2-abc-b^2c+axy+bxy$

$ab(a+b)-bc(a+b)+xy(a+b)$

$(a+b)(ab-bc+xy)$

$(ii) ax^2-bx^2+ay^2-by^2+az^2-bz^2$

$x^2(a-b)+y^2(a-b)+z^2(a-b)$

$(a-b)(x^2+y^2+z^2)$

$8x^2 - 6x^2 + 10x$

$8x^2 - 6x^2 + 10x$
Tke out common in both terms,
Then,

$2x(4x^2-3x+5)$
Therefore,HCF of

$8x^3,6x^2 \quad and \quad 10x \quad is \quad 2x.$

Divide:
$14p^2q^3 - 32p^3q^2 + 15pq^2 - 22p + 18q$ by $-2p^2q$ .

$\frac{14p^2q^3 - 32p^3q^2 + 15pq^2 - 22p + 18q}{-2p^2q}$
=

$\frac{14p^2q^3}{-2p^2q} - \frac{32p^3q^2}{-2p^2q} + \frac{15pq^2}{-2p^2q} - \frac{22p}{-2p^2q} + \frac{18q}{-2p^2q}$
=

$7q^2 + 16pq - \frac{15q}{2p} + \frac{11}{pq} - \frac{9}{p^2}$

$28 p^2q^2 r - 42 pq^2r^2$

$28 p^2q^2 r - 42 pq^2r^2$
Tke out common in both terms,
Then,

$14pq^2r(2p-3r)$
Therefore, HCF of

$28p^2q^2r \quad and \quad 42 pq^2r^2 \quad is \quad 14pq^2r$

Factorise the following :
$12x^3 -14x^2 -10x$

$12x^3 -14x^2 -10x$

$2x(6x^2-7x-5)$ ........[ Now,as

$6 \times (-5)=-30 \Rightarrow -30=-10 \times 3 \quad and \quad -7=-10+3]$

$2x(6x^2+3x-10x-5)$

$2x \{3x (2x+1)-5(2x+1)\}$

$2x(2x+1)(3x-5)$

Factorise the following :
$10x^2 -18x^3 +14x^4$

$10x^2 -18x^3 +14x^4$

HCF of

$10,18,14=2$

So,

$10x^2-18x^3+14x^4$

$2x^2(5-9x+7x^2)$

Factorise by taking out the common factors:
$2(2x - 5y)( 3x + 4y) - 6(2x - 5y) (x - y)$

$2(2x - 5y)( 3x + 4y) - 6(2x - 5y) (x - y)$

Taking

$(2x - 5y)$ common from both terms

$= (2x - 5y)[2(3x + 4y) - 6(x - y)]$

$=(2x - 5y)(6x + 8y - 6x + 6y)$

$=(2x - 5y)(8y + 6y)$

$=(2x - 5y)(14y)$

$=(2x - 5y) 14y$

Solve:
$x^2(a-b)-y^2(a-b)+z^2(a-b)$

Simplifying we get

$x^2(a-b)-y^2(a-b)+z^2(a-b)=(a-b)(x^2-y^2+z^2)$

Factorise:
$3a^5-108a^3$

$3a^5-108a^3=3a^3(a^2-36)$

$3a^5-108a^3=3a^3(a^2-6^2)$

$3a^5-108a^3=3a^3(a-6)(a+6)$

$[\therefore a^2-b^2=(a+b)(a-b)]$

Factorise:
$50a^3-2a$

$50a^3-2a=2a(25a^2-1)$

$50a^3-2a=2a((5a)^2-1^2)$

$50a^3-2a=2a(5a+1)(5a-1)$

Resolve in to factors: $5 - 10m - 20n.$

Given expression is

$5-10m-20n$ .

We can rewrite it as:

$5-5\times 2\times m-5\times 4\times n$

$=5-5(2m)-5(4m)$

Taking

$5$ common from all the terms, we get:

$5(1-2m-4n)$

Hence,

$5-10m-20n=5(1-2m-4n)$ .

Factorise :
$4x^2+\dfrac{1}{4x^2}+1$

$4x^2+\dfrac{1}{4x^2}+1$

$=4x^2+\dfrac{1}{4x^2}+2-1$

$=4x^2+\dfrac{1}{4x^2}+2 \times 2x \times \dfrac{1}{2x}-1$

$=(2x + \dfrac{1}{2x})^2-(1)^2$

$=(2x + \dfrac{1}{2x}-1)(2x + \dfrac{1}{2x}+1)$

Solve:
$4x(3x-2y)-2y(3x-2y)$

Simplifying we get

$4x(3x-2y)-2y(3x-2y)=(3x-2y)(4x-2y)$

$=(3x-2y)\times 2(2x-y)$

$=2(3x-2y)(2x-y)$

Factorise :
$24a^3-37a^2-5a$

$24a^{3} + 37a^{2} - 5a$

$=a[24a^2+37a-5]$

$=a[24a^2-3a+40a-5]$

$=a[3a(8a-1)+5(8a-1)]$

$=a(3a+5)(8a-1)$

Find and correct errors of the following mathematical expressions:
$\dfrac{3x^{2}+1}{3x^{2}} =1+1 =2$

The given statement is incorrect.
After separating the denominators, we get

$\dfrac{3x^{2}+1}{3x^{2}} =1+\dfrac{1}{3x^{2}}$

The above statement is the correct one.

Resolve in to factors: $(1.6)a^2 - (0.8)a$

Given that :

$(1.6)a^2-(0.8)a$

$=(0.8\cdot 2)(a\cdot a)-(0.8)a$ , factor each monomial

$=(0.8a)(2a)-(0.8a)$

$=0.8a(2a-1)$ , factor out common factor

$(0.8)a$

Factors of

$(1.6)a^2-(0.8)a$ are

$0.8a , (2a - 1)$

Resolve in to factors : $x^2+xy$

Given that:

$x^2+xy$

$= x\times x+x\times y$ , factor each monomial

$=x(x+y)$ , take out common factor

$x$

Factors of

$x^2+xy$ are

$x, (x+y)$

Divide the given polynomial by the given monomial : $(5x^2 - 6x)\div 3x$

Given polynomial is

$(5x^{ 2 }-6x)$ and the given monomial is

$3x$

The required division is

$=\dfrac{(5x^{ 2 }-6x)}{3x}$

$=\dfrac{x(5x-6)}{3x}$

$=\dfrac{(5x-6)}{3}$

$5y^2 -20y - 8z+ 2yz$

$5y^2-20y -8z+2yz$

$=5y^2-20y+2yz-8z$

$=5y(y-4)+2z(y-4)$

$=(y-4)(5y+2z)$

$6xy-4y+6 -9x$

$=6xy-9x-4y+6$

$=3x(2y-3)-2(2y-3)$

$=(2y-3)(3x-2)$

$y(y + z)+ 9(y + z)$

$=(y + z)(y + 9)$

Divide $64y^3-1000$ by $8y -20$ .

Here, Dividend

$= 64{ y }^{ 3 }1000$

Divisor

$= 8y-2$

Hence, quotient

$= 8{ y }^{ 2 }+20y+50$

Write down all possible factors of $3x^2y$

We have,

$3x^2y= 1 \times 3 \times xy =3 \times x^2y = 3x \times xy$

$=3xy \times x = x^2 \times 3y = y \times 3x^2$
Thus, the possible factors of

$3x^2y$ are

$1, 3, x, y, 3x, 3y, xy, 3xy, x^2y, 3x^2y$

$(n^{3} - n)$ is divisible by $3$ . Explain the reason

$n^3 - n = n (n^2 - 1) = n (n - 1) (n + 1)$

Whenever a number is divided by 3, the remainder obtained is either

$0$ or

$1$ or

$2$ .
∴

$n = 3p$ or

$3p + 1$ or

$3p + 2,$ where

$p$ is some integer.
If

$n = 3p$ , then

$n$ is divisible by

$3$ .
If

$n = 3p + 1,$ then

$n – 1 = 3p + 1 –1 = 3p$ is divisible by

$3$ .
If

$n = 3p + 2,$ then

$n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1)$ is divisible by

$3$ .
So, we can say that one of the numbers among

$n, n – 1$ and

$n + 1$ is always divisible by

$3$ .
⇒

$n (n – 1) (n + 1)$ is divisible by

$3$ .

$m^2 - mn + 4m - 4n$

$m^2-mn+4m-4n$

$=m(m-n)+4(m-n)$

$=(m+4)(m-n)$
The factors are

$(m+4),(m-n)$

$x^2 - 36$

We need to factorise

$x^2-36$

The given expression can be factorised by using property

$(a^2-b^2)=(a+b)(a-b)$

It can be written as

$x^2-36=x^2-6^2$

Here

$a=x, b=6$

Hence,

$(x^2-36)=(x+6)(x-6)$

$x^2 - 81$

We have to factorise

$x^2-81$

We will use identity

$a^2-b^2=(a-b)(a+b)$

It can be written as

$x^2-81=(x+9)(x-9)$

Thus factors are

$(x+9),(x-9)$ .

$36x^2 + 96xy + 64y^2$

Using identity,

$a^2+b^2-2ab=(a-b)^2$

$36x^2+96xy+64y^2=(6x)^2+2\times 6x\times 8y+(8y)^2$

$=(6x+8y)^2$

$=(6x+8y)(6x+8y)$

Hence, the factors are

$(6x+8y)(6x+8y)$ .

$m^2 - 121$

We have the identity

$(a-b)(a+b)=a^2-b^2$

We need to simplify by using identity

$m^2-121=(m-11)(m+11)$

The factors are

$m-11,m+11$ .

$25m^2 - 40mn + 16n^2$

Using the identity

$(a^2+b^2-2ab)=(a-b)^2$

We can write it as

$(25m^2+16n^2-40mn)$ as

$=(5m)^2+(4n)^2-2\times 5m\times 4n$

$=(5m-4n)^2$

Hence, the factors are

$(5m-4n)(5m-4n)$ .

$a^2 + 10 a + 25$

By using identity

$(a+b)^2=a^2+b^2+2ab$

It can be written as

$a^2+10a+25$

$a^2+5^2+2\times a\times 10=(a+5)^2=(a+5)(a+5)$

Hence, factors are

$(a+5)(a+5)$ .

$49x^2 - 25y^2$

We need to factorise

$49x^2-25y^2$

The given expression can be factorised by using property

$(a^2-b^2)=(a+b)(a-b)$

It can be written as

$49x^2-25y^2=(7x)^2-(5y)^2$

Here

$a=7x, b=5y$

Hence,

$(49x^2-25y^2)=(7x+5y)(7x-5y)$

$x^4 - y^4$

We need to factorise

$x^4-y^4$

The given expression can be factorised by using property

$(a^2-b^2)=(a+b)(a-b)$

It can be written as

$x^4-y^4=(x^2)^2-(y^2)^2$

Here

$a=x^2, b=y^2$

Hence,

$(x^4-y^4)=(x^2+y^2)(x^2-y^2)=(x^2+y^2)(x+y)(x-y)$

Find the errors in the following mathematical sentence:
$(3x + 2) \div 3x = \dfrac{2}{3x}$

$\dfrac{(3x + 2)}{3x} = 1 + \dfrac{2}{3x}$ is the correct division of the given mathematical sentence not

$\dfrac{2}{3x}$ .

Factorise: $x^2 -ax - bx + ab$

We need to factorise

$x^2-ax-bx+ab$

It can be written as

$(x^2-ax-bx+ab)=x(x-a)-b(x-a)=(x-b)(x-a)$

Thus, the factors are

$x-a$ and

$x-b$

$x(y + z) - 5(y + z)$

Given expression is

$x(y+z)-5(y+z)$

We can take

$(y+z)$ common from the given expression.

Hence, we get

$x(y+z)-5(y+z)=(x-5)(y+z)$ .

$mn + m + n + 1$

We can factorise it

$mn+m+n+1$ as

$=m(n+1)+1(n+1)$

$=(m+1)(n+1)$

Therefore, factors are

$(m+1)(n+1)$ .

Factorise the polynomial: $lx^2 + mx$

We need to factorise

$lx^2+mx$

It can be written as

$lx^2+mx=x(lx+m)$

Thus factors are

$x,(lx+m)$ .

Check whether the given expression is correct?
$\dfrac{4x + 3}{3} =x + 1$

The given mathematical expression is incorrect, the correct division of the same would be:

$\dfrac{4x + 3}{3} = \dfrac{4}{3} x + 1$

$7y^2 + 35z^2$

We need to factorise

$7y^2+35z^2$

It can be written as

$7y^2+35z^2=7(y^2+5z^2)$

Thus factors are

$7,(y^2+5z^2)$ .

Verify whether the given mathematical statement is correct?
$3x + 5 \div 3 = 5$

The given mathematical statement is incorrect.

Correct statement is:

$(3x + 5) \div 3 = x + \dfrac{5}{3}$

$a^4 - (b + c)^4$

We need to factorise

$a^4-(b+c)^4$

The given expression can be factorised by using property

$(x^2-y^2)=(x+y)(x-y)$

It can be written as

$a^4-(b+c)^4=(a^2)^2-[(b+c)^2]^2$

Here

$x=a^2, y=(b+c)^2$

Hence,

$(a^4-(b+c)^4)=(a^2+(b+c)^2)(a^2-(b+c)^2)=(a^2+(b+c)^2)(a+b+c)(a-b-c)$

Hence, the factors are

$(a^2+(b+c)^2),(a+b+c),(a-b-c)$ .

Factorize the following expressions:
$15a^{2}b + 35ab$

Given polynomial is

$15a^2b+35ab$

Here we can take

$5ab$ common form both the terms.

Thus w ehave

$15a^2b+35ab=5ab(3a+7)$

Hence, factors are

$5,a,b,(3a+7)$ .

Factorize the following expressions:
$7x - 14y$

Given expression is

$7x-14y$

Here we have to take

$7$ common from both the terms.

Thus we have

$7x-14y=7x-(7\times 2y)=7(x-2y)$

Factors are

$7,(x-2y)$ .

Factorize the following expressions:
$3x - 45$

Given expression is

$3x-45$

We can factorised it as

$3x-45=3x-(3\times 15)=3(x-15)$

Thus factors are

$3,(x-15)$ .

Simplify:-
$2ab + 2b + 3a$

1713811_618137_ans_c360bf3962ac49b398e41218e00891ec.jpg

Solve: $(x^{6} - 3x^{4} + 2x^{2})\div 3x^{2}$

$(x^{6} - 3x^{2} + 2x^{2}) \div 3x^{2}$

$= \dfrac {x^{6} - 3x^{4} + 2x^{2}}{3x^{2}}$

$= \dfrac {x^{6}}{3x^{2}} - \dfrac {3x^{4}}{3x^{2}} + \dfrac {2x^{2}}{3x^{2}}$

$= \dfrac {1}{3}x^{4} - x^{2} + \dfrac {2}{3}$
Alternative Method:
We can find the common factor

$x^{2}$ and simplify it as

$\dfrac {x^{6} - 3x^{4} + 2x^{2}}{3x^{2}} = \dfrac {x^{2}(x^{4} - 3x^{2} + 2)}{3x^{2}}$

$= \dfrac {1}{3} (x^{4} - 3x^{2} + 2)$

$= \dfrac {x^{4}}{3} - x^{2} + \dfrac {2}{3}$

Factorize the following expressions:
$2x + 6$

Given expression is

$2x+6$

Here we can take

$2$ common.

Then we have

$2x+6=2(x+3)$

Factorize the following:
$x^{3} - 3x^{2} + x - 3$

We need to factorise

$x^3-3x^2+x-3$

Here we will apply method of taking common terms.

Thus

$x^3-3x^2+x-3=x^2(x-3)+1(x-3)=(x-3)(x^2+1)$

Thus factors are

$(x-3)(x^2+1)$ .

Factorize the following:
$2xy - 3ab + 2bx - 3ay$

We need to factorise

$2xy-3ab+2bc-3ay$

It can be factorised by taking common factors from two terms at a time.

It can be written as

$2xy-3ay+2bx-3ab=y(2x-3a)+b(2x-3a)$

$=(2x-3a)(y+b)$

Factorize the following expressions:
$4x^{2} + 20xy$

We need to factorise

$4x^2+20xy$

Here we will take

$4x$ common.

Thus

$4x^2+20xy=4x(x+5y)$

Hence, factors are

$4x, (x+5y)$ .

Simplify: $3(5y^{2} - 3y + 2)$

We need to simplify

$3(5y^2-3y+2)$

Here

$3$ is getting multiplied with the whole bracket.

Lets solve

$3(5y^2-3y+2)=3\times 5y^2-3\times 3y+3\times 2$

$=15y^2-9y+6$

Factorize the following expressions:
$3x^{2} - 12xy$

We need to factorise

$3x^2-12xy$

Here we can take

$3x$ common.

Thus we have

$3x^2-12xy=3x(x-4y)$

Hence, factors are

$3x, (x-4y)$ .

Solve: $(8x^{3} - 5x^{2} + 6x) \div 2x$

$(8x^{3} - 5x^{2} + 6x)\div 2x$

$= \dfrac {8x^{3} - 5x^{2} + 6x}{2x}$

$= \dfrac {8x^{3}}{2x} - \dfrac {5x^{2}}{2x} + \dfrac {6x}{2x}$

$= 4x^{2} - \dfrac {5}{2} x + 3$
Alternative Method
For

$8x^{3} - 5x^{2} + 6x$ , separating

$2x$ from each term we get,

$8x^{3} - 5x^{2} + 6x = 2x(4x^{2}) - 2x\left (\dfrac {5}{2}x\right ) + 2x(3)$

$= 2x\left (4x^{2} - \dfrac {5}{2}x + 3\right )$

$\dfrac {8x^{3} - 5x^{2} + 6x}{2x} = \dfrac {2x(4x^{2} - \dfrac {5}{2}x + 3)}{2x}$

$= 4x^{2} - \dfrac {5}{2}x + 3$

Solve: $(7x^{2} - 5x)\div x$

$(7x^{2} - 5x) \div x = \dfrac {7x^{2} - 5x}{x}$

$= \dfrac {7x^{2}}{x} - \dfrac {5x}{x}$

$= \dfrac {7\times x\times x}{x} - \dfrac {5\times x}{x}$

$= 7x - 5$
Alternative Method:

$\dfrac {7x^{2} - 5x}{x} = 7x^{2 - 1} - 5x^{1 - 1}$

$= 7x^{1} - 5x^{0} = 7x - 5 (1)$

$[\because a^{0} = 1]$

$= 7x - 5$

Factorise the following:
$6a^5 - 18a^3 + 42a^2$

$6a^5 - 18a^3 + 42a^2$

Highest common factor is

$6a^2$

$\therefore 6a^5 - 18a^3 + 42 a^2$

$= 6a^2(a^3 - 3a + 7)$

Factorise:
$2a^3 + 4a^2$

$2a^3 + 4a^2$

Highest common factor is

$2a^2$

$\therefore 2a^3 + 4a^2$

$= 2a^2 (a + 2)$ .

Find the remainder using remainder theorem, when:
$3x^3+4x^2-5x+8$ is divided by $x-1$

According to remainder theorem, when

$f(x)$ is divided by

$(x-1)$ , Remainder

$=f(1)$

$f(x) = 3x^3+4x^2-5x+8$

$f(1)=3(1)^3+4(1)^2-5(1)+8$

$=3+4-5+8=10$

$\therefore$ Remainder

$=10$

Find the remainder using remainder theorem, when:
$8x^4+12x^3-2x^2-18x+14$ is divided by $x+1$

According to remainder theorem when

$f(x)$ is divided by

$(x+1)$ , Remainder

$=f(-1)$

$f(x)=8x^4+12x^3-2x^2-18x+14$

$f(-1)=8(-1)^4+12(-1)^3-2(-1)^2-18(-1)+14$

$=8-12-2+18+14=26$

$\therefore$ Remainder

$=26$

Find the factors of $a(b - c)^{3} + b(c - a)^{3} + c(a - b)^{3}$ .

Given equation can be written as

$a{ \left( b-c \right) }^{ 3 }+b{ \left( c-a \right) }^{ 3 }+c{ \left( a-b \right) }^{ 3 }$

$=a(b-c)^3-b(b-c+a-b)^3+c(a-b)^3$

$=a(b-c)^3-b(b-c)^3-b(a-b)^3-3b(b-c)(a-b)(a-c)+c(a-b)^3$

$=(b-c)^3(a-b)+(a-b)^3(c-b)-3b(b-c)(a-b)(a-c)$

$=(a-b)(b-c)((b-c)^2-(a-b)^2-3b(a-c))$

$=(a-b)(b-c)((b-c+a-b)(b-c-(a-b))-3b(a-c))$

$=(a-b)(b-c)((a-c)(2b-c-a)-3b(a-c))$

$=(a-b)(b-c)(a-c)(2b-c-a-3b)$

$=(a-b)(b-c)(c-a)(a+b+c)$

Find the remainder using remainder theorem, when:
$4x^3-12x^2+11x-5$ is divided by $2x-1$

According to remainder theorem, when

$f(x)$ is divided by

$(2x-1)$ , Remainder

$=f(\dfrac{1}{2})$

$f(x)=4x^3-12x^2+11x-5$

$f(\dfrac{1}{2})=4(\dfrac{1}{2})^3-12 (\dfrac{1}{2})^2+11(\dfrac{1}{2})-5$

$=4\times \dfrac{1}{8}-12 \times \dfrac{1}{4}+11\times \dfrac{1}{2}-5$

$=\dfrac{1}{2}-3+\dfrac{11}{2}-5$

$=\dfrac{12}{2}-8$

$=6-8=-2$

$\therefore$ Remainder =

$=-2$

Determine whether $(x+1)$ is a factor of the polynomial:
$6x^4+7x^3-5x-4$

According to factor theorem, when

$f(x)$ is divided by

$(x+1)$ ,

$f(-1)=0$

$f(x)=6x^4+7x^3-5x-4$

$f(-1)=6(-1)^4+7(-1)^3-5(-1)-4$

$=6-7+5-4=0$

$\because f(-1)=0$ , we can say that

$(x+1)$ is a factor of the given polynomial

Determine whether $(x+1)$ is a factor of the polynomial:
$2x^4+9x^3+2x^2+10x+15$

According to factor theorem, when

$f(x)$ is divided by

$(x+1)$ ,

$f(-1)=0$

$f(x)=2x^4+9x^3+2x^2+10x+15$

$f(-1)=2(-1)^4+9(-1)^3+2(-1)^2+10(-1)+15$

$=2-9+2-10+15=0$

$\because f(-1)=0, (x+1)$ is a factor of the given polynomial.

Find the factors of $a^{4} (b^{2} - c^{2}) + b^{4}(c^{2} - a^{2}) + c^{4} (a^{2} - b^{2})$ .

The given equation is

$a^4(b^2-c^2)+b^4(c^2-a^2)+c^4(a^2-b^2)$
We can write the above equation as

$a^4(b^2-c^2)-b^4(b^2-c^2+a^2-b^2)+c^4(a^2-b^2)$

$=a^4(b^2-c^2)-b^4(b^2-c^2)-b^4(a^2-b^2)+c^4(a^2-b^2)$

$=(b^2-c^2)(a^4-b^4)-(a^2-b^2)(b^4-c^4)$

$=(b^2-c^2)(a^2-b^2)(a^2+b^2)-(a^2-b^2)(b^2-c^2)(b^2+c^2)$

$=(b^2-c^2)(a^2-b^2)(a^2+b^2-b^2-c^2)$

$=(b^2-c^2)(a^2-b^2)(a^2-c^2)$

$=-(b^2-c^2)(a^2-b^2)(c^2-a^2)$

$=-(b+c)(b-c)(a+b)(a-b)(c+a)(c-a)$

Factorise: ${x^2} + 7x - 18$

$x^{2}+7x-18$

$=x^{2}-2x+9x-18$

$=x(x-2)+9(x-2)$

$=(x-2)(x+9)$

$\therefore x^{2}+7x-18=(x-2)(x+9)$

Factorize:
$m^3-m$ .

Now,

$m^3-m$

$=m(m^2-1)$

$=m(m+1)(m-1)$ .

Simplify:
$2\left(a^2+\dfrac{1}{a^2}\right)-\left(a-\dfrac{1}{a}\right)-7$ .

Now,

$2\left(a^2+\dfrac{1}{a^2}\right)-\left(a-\dfrac{1}{a}\right)-7$

$=2\left(a-\dfrac{1}{a}\right)^2+4-\left(a-\dfrac{1}{a}\right)-7$ [ Since

$(a^2+b^2)=(a-b)^2+2ab$ ]

$=2\left(a-\dfrac{1}{a}\right)^2-\left(a-\dfrac{1}{a}\right)-3$

$=2\left(a-\dfrac{1}{a}\right)^2-3\left(a-\dfrac{1}{a}\right)+2\left(a-\dfrac{1}{a}\right)-3$

$=\left(2a-\dfrac{2}{a}-3\right)\left(a-\dfrac{1}{a}+1\right)$

${(y - (b - a))^2}$

$(y-(b-a))^2$

$y^2+(b-a)^2-2(b-a)y$

$\implies y^2-2(b-a)y+(b-a)^2$

Simplify:
$x(x-4)-y(y-4)$ .

Now,

$x(x-4)-y(y-4)$

$=x^2-y^2+4y-4x$

$=(x+y)(x-y)+4(y-x)$

$=(x-y)(x+y-4)$

Factorize
${x^2} - x + 10$

Let

$x^{2}-x+10=0\implies x=\dfrac{1\pm\sqrt{1+40}}{2}=\dfrac{1\pm\sqrt{41}}{2}$

$x^{2}-x+10=\bigg(x-\dfrac{1-\sqrt{41}}{2}\bigg)\bigg(x-\dfrac{1+\sqrt{41}}{2}\bigg)$

Simplify:
$x^6+6x^3+8$ .

Now,

$x^6+6x^3+8$

$=$

$x^6+4x^3+2x^3+8$

$=$

$x^3(x^3+4)+2(x^3+4)$

$=(x^3+4)(x^3+2)$

Simplify:
$a^2-3b^2-c^2-2ab+4bc$ .

Now,

$a^2-3b^2-c^2-2ab+4bc$

$=(a^2-2ab+b^2)-4b^2-c^2+4bc$

$=(a-b)^2-(2b-c)^2$

$=(a+b-c)(a-b+c)$ .

Divide ${x}^{3}-6{x}^{2}+11x-6$ by $x-2$ and verify the division algorithm

Let

$f(x)={ x }^{ 3 }-6{ x }^{ 2 }+11x-6$

and

$p(x)=x-2$

Dividing

$f(x)$ by

$p(x)$ we get

Quotient

$q(x)={ x }^{ 2 }-4x+3$ and Remainder

$r(x)=0$

$p(x)q(x)=(x-2)({ x }^{ 2 }-4x+3)\\ p(x)q(x)={ x }^{ 3 }-4{ x }^{ 2 }+3x-2{ x }^{ 2 }+8x-6\\ p(x)q(x)={ x }^{ 3 }-6{ x }^{ 2 }+11x-6$

$\Rightarrow p(x)q(x)=f(x)$

$\Rightarrow p(x)q(x)+0=f(x)$

$\Rightarrow p(x)q(x)+r(x)=f(x)$

Hence division algorithm verified.

Factorise: $12ax - 4ab + 18bx - 6b^2=0$

$12ax - 4ab + 18bx - 6b^2=0$

$= 4a(3x-b) + 6b(3x-b) \\ = (4a+6b)(3x-b) \\ = 2(2a +3b)(3x-b)$

Factorise $x^4-5x^2+6$ .

Now,

$x^4-5x^2+6$

$=x^4-3x^2-2x^2+6$

$=x^2(x^2-3)-2(x^2-3)$

$=(x^2-2)(x^2-3)$ .

Factorise
$6x^{2}+11x+6$

Let

$6x^2+11x+6$

Let us consider the

$a=6,\;b=11,\;c=6$

and,

$b^2-4ac$

$=11^2-4 \times 6 \times 6$

$=121-144$

$=-23$

Since

$b^2-4ac<0$ ,

Therefore,

the equation

$6x^2+11x+6$ has no real roots.

Factorise $8a^{2}-22ab+15b^{2}$ .

$8{a^2} - 22ab + 15{b^2}$

$= 8{a^2} - 12ab - 10ab + 15{b^2}$

$= 4a\left( {2a - 3b} \right) - 5b\left( {2b - 3a} \right)$

$= \left( {2a - 3b} \right)\left( {4a - 5b} \right)$

If a polynomial $f(x)$ is divided by ${x}^{2}-16$ then remainder is $5x+3$ , what will be the remainder when the same polynomial is divided by $(x+4)$ ?

Let the quotient be

$q(x)$

$f(x)=(x^2-16)+(5x+3)$

$\dfrac{f(x)}{(x+4)}=\dfrac{(x+4)(x-4)}{x+4}q(x)+\dfrac{5x+3}{x+4}$

Remainder=Remainder when

$5x+3$ is divided by

$x+4$

$=5(-4)+3=-17$

Factorize $125a^{3}+343b^{3}$ .

given,

$125{a^3} + 343{b^3}$

$=(5a)^3+(7b)^3$

Using identity,

${a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})$

$(5a)^3+(7b)^3=(5x+7y)(25x^2-35x+49y^2)$

$p(x)=x^{3}+3x^{2}+3x+1$ is divided by $x=-\dfrac{1}{2}$
$p(a) x=-\dfrac{1}{2}=0$
$x=\dfrac{1}{2}$
$p\left(\dfrac{1}{2}\right)= \left(\dfrac{1}{2}\right)^{3}+3\left(\dfrac{1}{2}\right)^{2}+3\left(\dfrac{1}{2}\right)+1$

$p(x)=x^3+3x^2+3x+1$

$g(x)=x-\frac{1}{2}$

$g(x)=0,x=\frac{1}{2}$

by factor theorem when

$p(x)$ is divided by

$g(x)$

Remainder

$=p(\frac{1}{2})$

$(\dfrac{1}{2})^3+3(\dfrac{1}{2})^2+3(\dfrac{1}{2})+1=\dfrac{27}{8}$

Factorize $4x^{2}-3x-7$

$4{x^2} - 3x - 7$

$= 4{x^2} + 4x - 7x - 7$

$= 4x\left( {x + 1} \right) - 7\left( {x + 1} \right)$

$= \left( {4x - 7} \right)\left( {x + 1} \right)$

Simplify:
$a^2y^2+ay-(a+2)(a+1)$ .

Now,

$a^2y^2+ay-(a+2)(a+1)$

$=a^2y^2+ay\{(a+2)-(a+1)\}-(a+2)(a+1)$

$=ay(ay+a+2)-(a+1)(ay+a+2)$

$=(ay+a+2)(ay-a-1)$ .

Answer any two of the following:
factorize $4{ \left( a-1 \right) }^{ 2 }-4\left( a-1 \right) =3$

Let

$a-1=x$

$4x^2-4x=3$

$4x^2+2x-6x-3=0$

$2x(2x+1)-3(2x+1)=0$

$(2x+1)(2x-3)=0$

Put

$x=a-1$

$(2a-1)(2a-5)=0$

Find the value of $P$ for which polynomial $x^{3}+x^{2}-3x-P$ is exactly divisible by polynomial $(x+3)$ .

Consider the given polynomial.

$\Rightarrow {{x}^{3}}+{{x}^{2}}-3x-P$

Therefore, consider the division shown below.

If $\left( x+3 \right)$ completely divides ${{x}^{3}}+{{x}^{2}}-3x-P$ , then

$-9-P=0$

$P=-9$

Hence, the required value of

$P$ is

$-9$ .

1025066_1057300_ans_a873b77698234e408229692ec36ae57f.PNG

Divide $4x^3 + 3x^2 - 2x + 8$ by $x - 2$

$x-2$

$\bigg )\ 4x^3+3x^2-2x+8\bigg($

$4x^2+11x+20$

$4x^3-8x^2$

-------------------------------

$11x^2-2x$

$11x^2-22x$

--------------------------------

$20x+8$

$20x-40$

-----------------------------------------

$48$

$\Rightarrow \dfrac{4x^3+3x^2-2x+8}{x-2}=4x^2+11x+20+\dfrac{48}{x-2}$

Divide $6{a}^{4}-2{a}^{2}-a$ by ${a}^{2}$

$=\dfrac{6{a}^{4}-2{a}^{2}-a}{{a}^{2}}$

$=\dfrac{6{a}^{4}}{{a}^{2}}-\dfrac{2{a}^{2}}{{a}^{2}}-\dfrac{a}{{a}^{2}}$

$=6{a}^{2}-2-\dfrac{1}{a}$

Factorise: ${x}^{2}+6x+5$ $= 0$

${x^2} + 6x + 5$

$= {x^2} + 1x + 5x + 5$

$= x\left( {x + 1} \right) + 5\left( {x + 1} \right)$

$= \left( {x + 1} \right)\left( {x + 5} \right) = 0$

$x = - 1, - 5$

Factorise $ab\left( {{x^2} + {y^2}} \right) - xy\left( {{a^2} + {b^2}} \right)$

$ab(x^2+y^2)-xy(a^2+b^2)$

$\Rightarrow$

$abx^2+aby^2- a^2xy-b^2xy$

$\Rightarrow$

$abx^2+b^2xy+aby^2-a^2xy$

$\Rightarrow$

$bx(ax-by)-ay(ax-by)$

$\Rightarrow$

$(ax-by)(bx-ay)$

Factorize ${x^2} - 2x - 8$

$x^2-2x-8$

We can Split the Middle Term of this expression to factorize it.

$\Rightarrow$

$x^2-4x+2x-8$

$\Rightarrow$

$x(x-4)+2(x-4)$

$\Rightarrow$

$(x-4)(x+2)$

$\therefore$

$x=4$ or

$x=-2$

Simplify:
$(a^2-b^2)(x^2-y^2)+4abxy$

Now,

$(a^2-b^2)(x^2-y^2)+4abxy$

$=(ax)^2+(by)^2+2abxy-(ay)^2+2abxy-(bx)^2$

$=\{(ax)^2+(by)^2+2abxy\}-\{(ay)^2-2abxy+(bx)^2\}$

$=(ax+by)^2-(ay-bx)^2$

$=(ax+by+ay-bx)(ax+by-ay+bx)$

Factorize $8x^2-128$ .

1317525_1064843_ans_a627f61655a3424690c4949031e3d968.jpg

Simplify:
$p^2+p-(a+1)(a+2)$ .

Now,

$p^2+p-(3a+1)(3a+2)$

$=p^2+\{(3a+2)-(3a+1)\}p-(3a+1)(3a+2)$

$=p(p+3a+2)-(3a+1)(p+3a+2)$

$=(p+3a+2)(p-3a-1)$

Divide $2x^3-9x^2+10x$ by $(2x+5)$ and also verify the result.

From the fig. we get,

Quotient

$=x^2-7x+\dfrac{45}{2}$

Remainder

$=-\dfrac{225}{2}$

Verify;-

$2x+5=0$

$x=-\dfrac{5}{2}$ satisfy the equation,

$=2x^3-9x^2+10x$

$=2\left ( \dfrac{-5}{2} \right )^{3}-9\left ( \dfrac{-5}{2} \right )^2+10\left ( \dfrac{-5}{2} \right )$

$=2\left ( \dfrac{-125}{8} \right )-9\left ( \dfrac{25}{4} \right )+10\left ( \dfrac{-5}{2} \right )$

$=\left ( \dfrac{-125}{4} \right )-9\left ( \dfrac{25}{4} \right )+10\left ( \dfrac{-5}{2} \right )$

$=\dfrac{-125}{4} - \dfrac{225}{4} -\dfrac{50}{2}$

$=-\dfrac{225}{2}$

Therefore,

$x=-\dfrac{5}{2}$ satisfy the equation.

994752_1072144_ans_f5da2c8a7bc84f7d98c13c4ef6349720.png

Simplify:
$x^2+4x-y^2+4y$

Now,

$x^2+4x-y^2+4y$

$=x^2-y^2+4(x+y)$

$=(x-y)(x+y)+4(x+y)$

$=(x+y)(x-y+4)$ .

Simplify: $\cfrac{\cfrac{y}{6} + \cfrac{2y}{3}}{y +\cfrac{2y-1}{3}}$ .

$=\cfrac{\cfrac{y}{6} + \cfrac{2y}{3}}{y +\cfrac{2y-1}{3}}$

$=\cfrac{\cfrac{5y}{6}}{\cfrac{5y-1}{3}}$

$=\cfrac{5y}{10y-2}$

Divide $3x^3+2x^2+13x+42$ by $(x+2)$ and also verify the result.

From the figure,

Quotient

$=3x^2-4x+21$

Remainder

$=0$

Verify;-

$x+2=0$

$\ x= -2$ , satisfy the given equation,

$3x^3+2x^2+13x+42$

$=3\left (-2\right )^3+2\left ( -2 \right )^2+13\left ( -2 \right )+42$

$=3\left (-8\right )+2\left ( 4\right )+13\left ( -2 \right )+42$

$=-24+8-26+42$

$=0$

Therefore,

$x=-2$ satisfy the given equation.

994767_1072132_ans_39faa3767bfb4d9da748c096f6151a73.png

Factorize:
$a^2+b^2$

Now,

$a^2+b^2$

$=a^2-(ib)^2$ [ Since

$i^2=-1$ ]

$=(a+ib)(a-ib)$ .

Simplify: $x^2 + 6x + 9 - 4y^2$ .

We have,

$x^2 + 6x + 9 - 4y^2$

$=(x+3)^2-(2y)^2$

Now use

$(a^2-b^2)=(a-b)(a+b)$

$=(x+2y+3)(x-2y+3)$ .

Factorize : $(x+y) (2x+3y)- (x+y) (x+1)$

Given that:

$(x+y)(2x+3y)-(x+y)(x+1)$

$=(x+y)\ [(2x+3y)-(x+1)]$ [ Taking

$(x+y)$ common]

$=(x+y)\ [2x+3y-x-1]$

$=(x+y)\ [x+3y-1].$

This is the required solution.

Simplify: $\left(n^2 + \dfrac{1}{n^2}\right) - 4\left(n + \dfrac{1}{n}\right) + 6$

Now,

$\left(n^2 + \dfrac{1}{n^2}\right) - 4\left(n + \dfrac{1}{n}\right) + 6$

$=\left(n + \dfrac{1}{n}\right)^2-2.n.\dfrac{1}{n} - 4\left(n + \dfrac{1}{n}\right) + 6$ [ Since

$a^2+b^2=(a+b)^2-2ab$ ]

$=\left(n + \dfrac{1}{n}\right)^2- 4\left(n + \dfrac{1}{n}\right) + 4$

$=\left(n + \dfrac{1}{n}-2\right)^2$

Factorise: $8{y^2} - y$

We have

$8{{y}^{2}}-y$

$y\left( 8y-1 \right)$

Hence, this is the answer.

Factorize $6{ x }^{ 2 }+17x+5$ by using the Factor Theorem.

Let

$p(x)=6x^{2}+17x+5 = 6(x^2+\dfrac{17}{6}+\dfrac{5}{6})$

The term

$\dfrac{5}{6}$ can be written as a multiple of many terms. For example, one would be

$(\pm\dfrac{5}{2},\pm\dfrac{1}{3}),(\pm\dfrac{5}{6},\pm1),...……...$ .

Let

$p(\frac{-1}{3})=0$ then by factor theorem

$\dfrac{-1}{3}$ is the factor of

$P(x)$ then the other factor will be

$\dfrac{-5}{2}$ ;

Thus

$6x^{2}+17x+5$ can be factorized as

$(2x+5)(3x+1)$

Or,

$p(x)=6x^{2}+17x+5 = 6x^2+15x+2x+5$

$= 2x(3x+1)+5(3x+1)$

$= (2x+5)(3x+1)$

Divide $3{y^4} - 8{y^3} - {y^2} - 5y - 5$ by $y-3$ and find the quotient and the reminder.

Consider the given polynomial.

$3{{y}^{4}}-8{{y}^{3}}-{{y}^{2}}-5y-5$

So, quotient and remainder shown in figure.

Hence, quotient is $3{{y}^{3}}+{{y}^{2}}+2y+1$ and remainder is $-2$ .

996101_1074821_ans_8b3d673c75b64a19a5f19df3185e4fd6.png

Factorise.
$m^2-23m+120$ .

Consider the given expression.

$\Rightarrow {{m}^{2}}-23m+120$

$\Rightarrow {{m}^{2}}-15m-8m+120$

$\Rightarrow m\left( m-15 \right)-8\left( m-15 \right)$

$\Rightarrow \left( m-15 \right)\left( m-8 \right)$

Hence, this is the required factorization.

Factorise:
${x^2} + 8x + 16$

$x^{2}+8x+16$

$=x^{2}+4x+4x+16$

$=x(x+4)+4(x+4)$

$=(x+4)(x+4)$

$=(x+4)^{2}$

Solve: $16a^{2}-b^{2}+4a+b$

We have,

$16{{a}^{2}}-{{b}^{2}}+4a+b$

$={{\left( 4a \right)}^{2}}-{{b}^{2}}+\left( 4a+b \right)$

$=\left( 4a+b \right)\left( 4a-b \right)+\left( 4a+b \right)$

$=\left( 4a+b \right)\left( 4a-b+1 \right)$

Hence, this is the answer.

Find the remainder when $x^3+3x^2+3x+1$ is divided by $x+1$

Dividend

$={x}^{3}+3{x}^{2}+3x+1$

and Divisor

$=x+1$

Thus, remainder

$=0$

1263970_1089283_ans_24743e8eccbd4a5ebd5fb4647f59b1c6.PNG

Factorize by factorisation method $15x^5y^2+3x^3y+19$

To factorize by common factor method, we need to take out the common factor from each term of the expression.

$15x^5y^2+3x^3y+19=5\times3\times x^5\times y^2+3\times x^3\times y+19=3x^3y(5x^2y+1)+19$

Find the remainder when $x^3 + x^2 + x + 1$ is divided by $x - \frac{1}{2}$ , by using remainder theorem.

Now using reminder theorem,

$D=d\times Q+r$

$\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)=\left( {{x}^{2}}+\dfrac{3}{2}x+\dfrac{7}{4} \right)\left( x-\dfrac{1}{2} \right)+\dfrac{15}{8}$

Hence, this is the answer.
1025068_1098781_ans_7541a7e02c07418fba500b4ae2ff2b2c.png

Factorise:
$c+bc^2-ba^2-a$ .

To factorise:

$c+bc^2-ba^2-a$

Now,

$c+bc^2-ba^2-a$

$=c-a+b(c^2-a^2)$

$=(c-a)+b(c+a)(c-a)$

$=(c-a)[1+b(c+a)]$

$=(c-a)(1+bc+ab)$

Hence, factorised.

Factorise: $27{x^3} - 21{x^2} + 15{x^4}$

$27 x ^ { 3 } - 21 x ^ { 2 } + 15 x ^ { 4 }$

$= 15 x ^ { 4 } + 27 x ^ { 3 } - 21 x ^ { 2 }$

$= 3 x ^ { 2 } \left( 5 x ^ { 2 } + 9 x - 21 \right)$

Solve:
$8{a^2}b - 3ab + 5{b^2} \ by \ 6ab$

$\dfrac{8{a}^{2}b-3ab+5{b}^{2}}{6ab}$

$=\dfrac{b\left(8{a}^{2}-3a+5b\right)}{6ab}$

$=\dfrac{8{a}^{2}-3a+5b}{6a}$

Evaluate
$(5x^{3}-3x^{2})\div x^{2}$

Given

$(5x^3-3x^2)\div x^2=\dfrac{5x^3-3x^2}{x^2}$

$=\dfrac{5x^3}{x^2}-\dfrac{3x^2}{x^2}$

$=5x^2-3$

Simplify
$\left( {5{a^3}b - 7a{b^3}} \right) \div ab$

$\dfrac{(5a^{3}b-7ab^{3})}{ab}=5a^{2}-7b^{2}$

Simplify ${x^4} + 2{x^3}y - 2x{y^3} - {y^4}$

We have,

${{x}^{4}}+2{{x}^{3}}y-2x{{y}^{3}}-{{y}^{4}}$

$={{x}^{4}}-{{y}^{4}}+2{{x}^{3}}y-2x{{y}^{3}}$

$={{\left( {{x}^{2}} \right)}^{2}}-{{\left( {{y}^{2}} \right)}^{2}}+2xy\left( {{x}^{2}}-{{y}^{2}} \right)$

$=\left( {{x}^{2}}-{{y}^{2}} \right)\left( {{x}^{2}}+{{y}^{2}} \right)+2xy\left( {{x}^{2}}-{{y}^{2}} \right)$

$=\left( {{x}^{2}}-{{y}^{2}} \right)\left( {{x}^{2}}+{{y}^{2}}+2xy \right)$

$=\left( {{x}^{2}}-{{y}^{2}} \right){{\left( x+y \right)}^{2}}$

$=\left( x+y \right)\left( x-y \right){{\left( x+y \right)}^{2}}$

$={{\left( x+y \right)}^{3}}\left( x-y \right)$

Hence, this is the answer.

$24a^{3}b^{2} + 8a^{2}b^{2} + 12ab$ by $6ab$ .

$4{ a }^{ 2 }b+\cfrac { 4 }{ 3 } ab+2$

Divide ${2{x^3} - 4{x^2} - 3x - 1}$ by ${x + 2}$

Hence, this is the answer.
1161513_1100539_ans_03672181104d450cb5adaea5aba167a6.png

Divide $3x^{4}+2x^{2}-3$ by $(x+1)$ and find quotient and remainder.

We, know that

$P(x)=g(x).Q(x)+R(x)$

$\implies 3x^4+2x^2-3=(x+1)(3x^3-3x^2+5x-5)+2$

Hence, Quotient

$Q(x)=3x^3-3x^2+5x-5$

And, Remainder

$=2$

The volume of a cuboid is given by the expression $3x^{3}-12x$ . Find the possible expression for its dimensions.

Let us factorize the given expression,

$3{x}^{3}-12x =3x\left({x}^{2}-4\right)$

$=3x \left(x-2\right) \left(x+2\right)$

On comparing the above factorized form with the formula for volume of cuboid

$l\times b\times h.$ We can conclude that, the possible dimensions are,

$\left(3x\right), \left(x-2\right)$ and

$\left(x+2\right)$

Factorize: $12x^{2}-7x+1$

$12x^{2}-7x+1$

$=12x^{2}-4x-3x+1$

$=4x(3x-1) -1(3x-1)$

$=(4x-1)(3x-1)$

$\therefore (12x^{2}-7x+1)$ =

$(4x-1)(3x-1)$

Convert into factorials:
$3 \times 6 \times 9 \times 12 \times 15 \times 18$

$3\times 6\times 9\times 12\times 15\times 18=3^{6}(1\times 2\times 3\times 4\times 5\times 6)=3^{6}\times 6!$

Solve:-
$a^{2}-11a+30$ by $(a-5)$

$\dfrac{{a}^{2}-11a+30}{a-5}$

$=\dfrac{{a}^{2}-6a-5a+30}{a-5}$

$=\dfrac{a\left(a-6\right)-5\left(a-6\right)}{a-5}$

$=\dfrac{\left(a-6\right)\left(a-5\right)}{a-5}$

$=a-6$

Divide :
$15{p^3} \div 3p$

$\dfrac{15{p}^{3}}{3p}=\dfrac{5\times 3\times}{3p}=5\times{p}^{3-1}=5{p}^{2}$

$\left( { m }^{ 2 }-14m-32 \right) \div \left( m+2 \right)$

$(m^2-14m-32) \div (m+2)$

$=\dfrac{m^2-14m-32}{m+2}$

$=\dfrac{m^2-16m+2m-32}{m+2}$

$=\dfrac{m(m-16)+2(m-16)}{m+2}$

$=\dfrac{(m-16)(m+2)}{m+2}=m-16$

Factorise the expressions using the common factor method.
$8x^2y + 4x$

$8 {x}^{2} y + 4x$

$= \left( 4x \times 2xy \right) + 4x$

$= 4x \left( 2xy + 1 \right)$

Therefore,

$8 {x}^{2} y + 4x = 4x \left( 1 + 2xy \right)$

Solve :-
$\left( {3{x^2} + 5x - 9} \right) \times \left( {3x - 5} \right)$

$=(3x^{2}+5x-9)\times (3x-5)$

$=9x^{3}+15x^{2}-27x-15x^{2}-25x+45$

$=9x^{3}-52x+45$

Solve: $21 m ^ { 2 } \div 7 m$

$21m^2 \div 7m = \dfrac{21m^2}{7m} = 3m$

Divide. Write the quotient and the remainder.
(a) $21{m}^{2}\div 7m$
(b) $40{a}^{3}\div (-10a)$

$(i)$

$\left. 7m \right) \overline { 21{ m }^{ 2 } } \left( 3m \right. \\ \\ \quad -21{ m }^{ 2 }\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \qquad \quad 0$

$(ii)$

$\left. -10a \right) \overline { 40{ a }^{ 3 } } \left( -4a^{2} \right. \\ \\ \quad \quad-40{ a }^{ 3 }\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \qquad \quad 0$

Factories : $2xy+2y+3x+3$

Consider the given expression.

$2xy+2y+3x+3$

$\Rightarrow 2y(x+1)+3(x+1)$

$\Rightarrow (2y+3)(x+1)$

Hence, this is the answer.

Factorise:
$6{x}^{2}{y}^{3}-12{x}^{2}{y}^{2}+18{x}^{3}y$

$6x^2y(y^2-2y+3x)$

$=6x^2y(y(y-2)+3x)$

This

$x^2, y; (y^2-2y+3x)$ are its factors.

Divide the given polynomial by the given monomial.
$(p^{3}q^{6}-p^{6}q^{3})\div p^{3}q^{3}$

$(p^3 q^6 - p^6 q^3) \div p^3 q^3$

$= \dfrac{p^3 q^6}{p^3 q^3} - \dfrac{p^6 q^3} {p^3 q^3 }$

$= q^3 - p^3$ (Ans)

Factorize :
$7x(3x-y)+7y(3x-y)$

To factorise

$\rightarrow 7x(3x-y)+7y(3x-y)$

Take

$(3x-y)$ as common

$= (3x-y)(7x+7y)$

$= (3x-y).7(x+y) = 7(3x-y)(x+y)$

the product of two polynomials is $7.5{ a }^{ 3 }{ b }^{ 2 }-2.5ab+10{ a }^{ 2 }{ b }^{ 2 }$ . if one of them is 2.5ab,find the other polynomial.

Product

$(P)=7.5a^3b^2-2.5ab+10a^2b^2$

$P_1=2.5ab$

$P_2=$

$\cfrac{7.5a^3b^2-2.5ab+10a^2b^2}{2.5ab}\\ =3a^2b-1+4ab$

Solve $18a^{3}b-27a^{2}b$

$18a^3b-27a^2b\\=9a^2b(2a-3)$

Solve by factorization method: $\sqrt{3}x^2+10x+7 \sqrt{3}=0$

$\quad \sqrt { 3 } { x }^{ 2 }+10x+7\sqrt { 3 } =0\\ \Rightarrow \sqrt { 3 } { x }^{ 2 }+3x+7x+7\sqrt { 3 } =0\\ \Rightarrow \sqrt { 3 } x\left( x+\sqrt { 3 } \right) +7\left( x+\sqrt { 3 } \right) =0\\ \Rightarrow \left( x+\sqrt { 3 } \right) \left( \sqrt { 3 } x+7 \right) =0\\ \Rightarrow x=-\sqrt { 3 } \quad or\quad x=\frac { -7 }{ \sqrt { 3 } }$

Divide the polynomial ${x^3} - 2{x^2} - 4x - 1$ by $x - 1$ and find the
remainder.

$x^2-x-5$

______________

$x-1)x^3-2x^2-4x-1$

$x^3-x^2$

___________

$-x^2-4x$

$-x^2+x$

____________

$-5x-1$

$-5x+5$

___________

$-6$

___________

Remainder

$=-6$ .

Factorise
$10x{y^2} - 15{x^2}y$

1335917_1270571_ans_3f81b89490ba4555bc63cfbd8e34079e.PNG

Facrtorise
$9x^2-6x+1=0$

$9x^{2}-6x+1=0$

$\Rightarrow 9x^{2}-3x-3x+1=0$

$\Rightarrow 3x(3x-1)-1(3x-1)=0$

$\therefore (3x-1)^{2}=0$

so,

$x=\frac{1}{3},\frac{1}{3}$

1188926_1294174_ans_83c8a372380f453cae5d6f2f33aaca33.jpg

Factorise

$3x(5a - 6b) - 12{a^2}(5{a^{}} - 6b)$

1352020_1270743_ans_addcc164a41e4ba79fc5077d776ac59a.PNG

$-20 x^{4}$ by $5x$

$-20x^4$ by

$5x$

$=-4x^3$ .

$5x)-20x^4(-4x^3$

$-20x^4$

__________

$x$

__________

1144342_1269671_ans_85ac2e1d8b45414caa3ddc36f4ff2d58.jpg

Factorise $\left( {a - b} \right) + {\left( {a - b} \right)^2}$

$(a-b)+(a-b)^2=(a-b)[1+(a-b)]$

$=(a-b)(1+a-b)$

Factorize:-
$2x^2-7x-15$

We have,

$\begin{array}{l} 2{ x^{ 2 } }-7x-15 \\ =2{ x^{ 2 } }-10x+3x-15 \\ =2x\left( { x-5 } \right) +3\left( { x-5 } \right) \\ =\left( { x-5 } \right) \left( { 2x+3 } \right) \end{array}$

Hence, this is the answer.

Divide : $-36x^{4}/(-9x)$

$-36x^{ 4 }\div (-9x)$

$=4x^3$

Divide $2{x^2} - 7{x^3} + 17{x^2}+17x +5$ by $2{x^2}$

1213397_1306846_ans_c993135c41dc48e3894b059e5fac831f.JPG

Find the value of a, if ${\text{x}}\;{\text{ - }}\;{\text{a}}$ is factor of ${x^3} - {a^2}x + x + 2$ .

We have,

$p(x)=x^3-a^2x+x+2$

Since,

$(x-a)$ is the factor of this polynomial.

So,

$p(a)=0$

Therefore,

$a^3-a^2(a)+a+2=0$

$a^3-a^3+a+2=0$

$a+2=0$

$a=-2$

Hence, this is the answer.

${ 2x }^{ 2 }-5x-2xy+5y$

We have,

$2{{x}^{2}}-5x-2xy+5y$

$\Rightarrow x\left( 2x-5 \right)-y\left( 2x-5 \right)$

$\Rightarrow \left( 2x-5 \right)\left( x-y \right)$

Hence, this is the answer.

Factorise the following expression :
$7x-42$

$7x - 42$

$= 7\times(x - 6)$

Factorize:
$x^2-9x+20$

Now,

$x^2-9x+20$

$=x^2-4x-5x+20$

$=(x-4)(x-5)$ .

Find the factors of the polynomial given below.
$2m^{2}+5m+3$

$2m^{2} + 5m + 3$

$= 2m^{2} + 2m +3m +3$

$= 2m(m+1) + 3 (m+1)$

$= (2m+3) (m+1)$

$\therefore$ The factors are

$(2m+3)$ and

$(m+1)$

Solve the equation : $3 n ^ { 3 } + 4 n ^ { 2 } + n=0$

$3n^{3}+4n^{2}+n=0$

$\Rightarrow n[3n^{2}+4n+1]=0$

$\Rightarrow 3n^{2}+4n+1=0$ n=0

$\Rightarrow 3n^{2}+3n+n+1=0$

$\Rightarrow 3n[n+1]+1[n+1]=0$

$\Rightarrow (n+1)(3n+1)=0$

$\Rightarrow$ n=-1 or 3n=-1

$\Rightarrow$ n=-1 or

$n=-\frac{1}{3}$

$\therefore n=0, -1, -\frac{1}{3}$

1191059_1306801_ans_82bd6f07b5c94b7894c73485725406d8.jpeg

Find the product using the distributive law of multiplication.
$( x + 7 ) ( x - 2 )$

$(x+7)(x-2)$

$=x(x-2)+7(x-2)$

By distributive law

$=x^2-2x+7x-14$

$=x^2+5x-14$ .

Factorize:
$(2x^{3}+54)$

$\left( 2 {x}^{3} + 54 \right)$

$= 2 \left( {x}^{3} + 27 \right)$

$= 2 \left( {\left( x \right)}^{3} + {\left( 3 \right)}^{3} \right)$

$= 2 \left( x + 3 \right) \left( {x}^{2} - 3x + {3}^{2} \right) \quad \left( \because {a}^{3} + {b}^{3} = \left( a + b \right) \left( {a}^{2} - ab + {b}^{2} \right) \right)$

$\therefore \left( 2 {x}^{3} + 54 \right) = 2 \left( x + 3 \right) \left( {x}^{2} - 3x + 9 \right)$

Factories
$x^{2}+11x+18$

${x}^{2} + 11x + 18$

$= {x}^{2} + 9x + 2x + 18 \quad \left( \because 18 = 9 \times 2 \right)$

$= x \left( x + 9 \right) + 2 \left( x + 9 \right)$

$= \left( x + 2 \right) \left( x + 9 \right)$

Therefore,

${x}^{2} + 11x + 18 = \left( x + 9 \right) \left( x + 2 \right)$

Define factorisation.

Factorisation

$\rightarrow$ Factorisation means to find two or more expressions whose product is equal to the given expression.

Write the following polynomials in factored from:
(i) $90p^{3}q^{3}r^{3}+18p^{3}q^{3}r$
(ii) $48pqr+96qr^{3}$
(iii) $21x^{2}z^{2}+30y^{3}z$
(iv) $36x^{2}y^{3}z^{2}+72x^{2}y^{3}z^{3}$

(A)

$90 p^3 q^3 r^3 + 18 p^3 q^3 r = 18 p^3 q^3 r (5 r^2+1)$

(B)

$48 pqr + 96 pr^3 = 48 pr (q+2r^2)$

(C)

$21x^2z^2 + 30 y^3 z= 3z (7x^2z+ 10 y^3)$ .

1172977_1346875_ans_2cc4f86e4526453ea33b579ab38019d2.jpg

Solve :
$6xy-4y+6-9x$

Given,

$6xy - 4y + 6 - 9x$

$= 6xy - 9x - 4y + 6$

$= 3x \left( 2y - 3 \right) - 2 \left( 2y - 3 \right)$

$= \left( 2y - 3 \right) \left( 3x - 2 \right)$

Therefore,

$6xy - 4y + 6 - 9x = \left( 2y - 3 \right) \left( 3x - 2 \right)$

Carry out the following division :
$28 x ^ { 4 } \div 56 x$

$28{x^4}\div 56{x}=\dfrac{28}{56}\times \dfrac{x^4}{x}=\dfrac{x^3}{2}$

Simplify
$21b-60+7b-20b$

$21b-60+7b-20b$

$\Rightarrow 21b+7b-20b-60$

$\Rightarrow 28b-20b-60$

$\Rightarrow 8b-60$

$\Rightarrow 4[2b-15]$

Factorise : $16w^{3}-u^{4}w^{3}$

$16w^{3}-u^{4}w^{3}$

$=w^{3}(16-u^{4})$

$=w^{3}(4^{2}-(u^{2})^{2})$

$=w^{3}(4+u^{2})(4-u^{2})$

$=w^{3}(4+u^{2})(2^{2}-u^{2})$

$=w^{3}(4+u^{2})(2+u)(2-u)$

Factorize : $\dfrac { 36{m}^{2} } { 289 } - 81$

$\begin{array}{l} We\, have \\ \dfrac { { 36 {m}^{2}} }{ { 289 } } { }-81 \\ ={\left( { \dfrac { 6m }{ { 17 } } } \right) }^{2}-{ \left( 9 \right) ^{ 2 } } \\ =\left( { \dfrac { 6 m}{ { 17 } } +9 } \right) \left( { \dfrac { 6m }{ { 17 } } -9 } \right) \\ =\left( { \dfrac { { 6m+153 } }{ { 17 } } } \right) \left( { \dfrac { { 6m-153 } }{ { 17 } } } \right) \\ Which\, is\, the\, required\, answer. \end{array}$

Factories : $p ^ { 2 } + 6 p + 8$

We have,

$\begin{array}{l} { p^{ 2 } }+6p+8 \\ ={ p^{ 2 } }+4p+2p+8 \\ =p\left( { p+4 } \right) +2\left( { p+4 } \right) \\ =\left( { p+4 } \right) \left( { p+2 } \right) \end{array}$

Hence, this is the answer.

Factorize:
$x^{1}+x^{2}\times 25$

$x+25{x}^{2}=x\left(1+25x\right)$

Factorise.
$4a^{2}+7a-2$

$4{a}^{2}+7a-2$

$=4{a}^{2}+8a-a-2$

$=4a\left(a+2\right)-1\left(a+2\right)$

$=\left(a+2\right)\left(4a-1\right)$

Hence the factors are

$\left(a+2\right)\left(4a-1\right)$

Express the following as the product of exponent through prism factorization
$1156$

1156

$=2\times 578$

$=2\times2\times289$

$=2\times2\times17\times17$

1156=

$2^{2}\times17^{2}$

1211658_1362857_ans_8827674e579743bab84c1ffbb7327f4f.JPG

Factorize :
$a^{2}-b^{2}-a-b$

given,

$a^{2}-b^{2}-a-b$

$=(a^{2}-b^{2})-(a+b)$

$=(a+b)[(a-b)-1]$

$=(a+b)(a-b-1)$

Factorise.
$11{x}^{2}+17x+6$

$11{x}^{2}+17x+6$

$=11{x}^{2}+11x+6x+6$

$=11x\left(x+1\right)+6\left(x+1\right)$

$=\left(x+1\right)\left(11x+6\right)$

Hence the factors are

$\left(x+1\right)\left(11x+6\right)$

Factorise: $25m^{2}-70mn+49n^{2}$

$25{m}^{2}-70mn+49{n}^{2}$

$={\left(5m\right)}^{2}-2\times 5m\times 7n+{\left(7n\right)}^{2}$ is of the form

${a}^{2}-2ab+{b}^{2}={\left(a-b\right)}^{2}$

$={\left(5m-7n\right)}^{2}$

Factorise.
$4x^{2}+3x-7$

$4{x}^{2}+3x-7$

$=4{x}^{2}+7x-4x-7$

$=x\left(4x+7\right)-1\left(4x+7\right)$

$=\left(4x+7\right)\left(x-1\right)$

Hence the factors are

$\left(4x+7\right)\left(x-1\right)$

Factorise: $2x^{2}-3x+1$ .

Given: a quadratic equation;

$2{x}^{2}-3x+1$

$=2{x}^{2}-2x-x+1$

$=2x\left(x-1\right)-1\left(x-1\right)$

$=\left(2x-1\right)\left(x-1\right)$

Factorise.
$9x^{2}-6x+1$

$9{x}^{2}-6x+1$

$=9{x}^{2}-3x-3x+1$

$=3x\left(3x-1\right)-1\left(3x-1\right)$

$=\left(3x-1\right)\left(3x-1\right)$

Factos of

$9{x}^{2}-6x+1=\left(3x-1\right)\left(3x-1\right)$

Factorise:
$8xy+yz$

$8xy+yz$

$=y\left(8x+z\right)$

$(6x^{3}+11x^{2}-10x-7)\div (2x+1)$

1199865_1381949_ans_853fe51c8e7a44a1a45c279a5c7a4531.JPG

Factorise.
$9x^{2}-8x-1$

$9{x}^{2}-8x-1$

$=9{x}^{2}-9x+x-1$

$=9x\left(x-1\right)-1\left(x-1\right)$

$=\left(9x-1\right)\left(x-1\right)$

Factos of

$9{x}^{2}-6x+1=\left(9x-1\right)\left(x-1\right)$

Divide $4(2x^{2}+5x+3)$ by $2(2x+3)$

$\frac{4(2x^{2}+5x+3)}{2(2x+3)} =\frac{2(2x^{2}+5x+3)}{2x+3}$

$\frac{2(2x^{2}+3x+2x+3)}{(2x+3)} =\frac{2\left [ x(2x+3)+1(2x+3) \right ]}{2x+3}$

$\frac{2(2x+3)\left [ (x+1) \right ]}{(2x+3)} =2(x+1)$

$\Rightarrow 2(x+1)$

1202398_1390054_ans_f6088bb5896a4309be5769419c4b1b87.JPG

Factorise the expression
$am^{2}+bm^{2}+bn^{2}+an^{2}$

$am^{2}+bm^{2}+bn^{2}+an^{2}$

$\Rightarrow m^{2}(a+b)+n^{2}(b+a)$

$\Rightarrow m^{2}(a+b)+n^{2}(a+b)$

$\Rightarrow (m^{2}+n^{2})(a+b)$

Factorize the following:
$3x-9$

$3x-9$

$=3\left(x-3\right)$

Factorise: $14m-21$

$14m-21$

$=7\left(2m-3\right)$

Factorise: $15ab^{2}-20a^{2}b$

$15a{b}^{2}-20{a}^{2}b$

$=5ab\left(3b-4a\right)$

Factorise the expression
$(xy+y)+x+1$

$\left(xy+y\right)+x+1$

$=y\left(x+1\right)+\left(x+1\right)$

$=\left(x+1\right)\left(y+1\right)$

Solve: $(5p^{2}-25p+20)\div (p-1)$

$p-1) 5p^2-25p+20(5p-20$

$5p^2-5p$

_______________

$-20p+20$

________________

$0$ .

1214882_1396662_ans_8308a1dd715a4fd28dc5642d779906d5.jpg

Divide.
(a) $12{ x }^{ 3 }\quad by\quad 3x$

$\mathrm{(a)} 12{x^3}\div 3{x}=\dfrac{12 x^3}{3{x}}=\dfrac{12}{3}\times \dfrac{x^3}{x}=4{x^2}$

Carry out the following divisions
$-54l^{4}m^{3}n^{2}$ by $9l^{2}m^{2}n^{2}$

Now,

$\dfrac{-54l^4m^3n^2}{9l^2m^2n^2}$

$=-6l^2m$ .

Solve the following when $x^{3}+3x^{2}+3x+1$ is divisible by
$x$

$(x^3+3 x^2+3 x+1)\div x=\dfrac{x^3+3{x^2}+3 x+1}{x}$

$=\dfrac{x(x^2+3 x+3)}{x}+\dfrac{1}{x}$

$=x^2+3{x}+3+\dfrac{1}{x}$

$16(2p-3q)^{2}-4(2p-3q)$

$16(2 p-3 q)^2-4(2{p}-3{q})=4(2{p}-3{q})(4(2{p}-3{q})-1)$

$=(8{p}-12{q})(8{p}-12{q}-1)$

Factorise the following expression by finding common factor.
$p^{3}-16p^{2}=0$

$p^3-16p^2\\p^2(p-16)=0\\p^2=0\\p-16=0\\p=16$

Factorise ${ x }^{ 2 }+x-2$

${\textbf{Step - 1: Breaking the middle term of the quadratic equation}}{\text{.}}$

${\text{We have the expression as }}{x^2} + x - 2,{\text{so the middle term is }}x.$

${\text{We can write }}x{\text{ as }}2x - x.$

${\text{Now we have the expression as }}{x^2+x-2 = x^2} + 2x - x - 2$

${\textbf{Step - 2: Factorizing}}{\text{.}}$

${\text{On taking common from first two terms and common from last two term, we get}}$

$x^2+x-2 = x(x - 2) - (x - 2)$

$\Rightarrow x^2+x-2 = (x - 1)(x - 2)$

${\textbf{Therefore, }}\mathbf{{x^2} + x - 2 = 0{\text{ can be factorized into }}(x - 1)(x - 2) = 0.}$

Factorise:
$4x^{2}+9y^{2}+25z^{2}+12xy+30yz+20zx$

Now,

$4x^{2}+9y^{2}+25z^{2}+12xy+30yz+20zx$

$=4x^{2}+12xy+9y^{2}+25z^{2}+30yz+20zx$

$=(2x+3y)^2+(5z)^{2}+30yz+20zx$

$=(2x+3y)^2+(5z)^{2}+2\times 5z(3y+2x)$

$=(2x+3y+5z)^2$ .

Simplify:
$3{\left( {a - 2b} \right)^2} - 5\left( {a - 2b} \right)$

Now,

$3{\left( {a - 2b} \right)^2} - 5\left( {a - 2b} \right)$

$=(a-2b)\{3(a-2b)-5\}$

$=(a-2b)(3a-6b-5)$ .

Factorise
${ 5 b }^{ 2 }-6b+1$

$5{b}^{2}-6b+1$

$=5{b}^{2}-5b-b+1$

$=5b\left(b-1\right)-\left(b-1\right)$

$=\left(5b-1\right)\left(b-1\right)$

Factorise: ${ a }^{ 2 }{ x }^{ 2 }+\left( { ax }^{ 2 }+1 \right) x+a$

${a}^{2}{x}^{2}+\left(a{x}^{2}+1\right)x+a$

$={a}^{2}{x}^{2}+a{x}^{3}+x+a$

$=a{x}^{2}\left(a+x\right)+1\left(x+a\right)$

$=\left(x+a\right)\left(a{x}^{2}+1\right)$

Simplify:
$(x^2-5)(x+5)+25$

Now,

$(x^2-5)(x+5)+25$

$=x^3+5x^2-5x-25+25$

$=x^3+5x^2-5x$

$=x(x^2+5x-5)$ .

Simplify: $\left(3{x}^{2}-x\right)\div\left(-x\right)$ .

$=\dfrac{3{x}^{2}-x}{-x}$

$=\dfrac{x\left(3x-1\right)}{-x}$

$=\dfrac{3x-1}{-1}$

$=-\left(3x-1\right)$

$=-3x+1=1-3x$

Factorise the following expression.
${p^2} + 6p + 8$

Now,

$p^2+6p+8$

$=p^2+4p+2p+8$

$=(p+2)(p+4)$ .

Factorise:
2a+36c

$2a+36c=2\left(a+18c\right)$

On dividing $p\left(4 p^{2}-16\right)$ by $4 p(p-2),$ we get (a) $2 p+4$ (b) $p+2$

$p(4 p^2 - 16)=4 p(p^2- 4)=4 p(p^2-2^2)=4 p(p-2)(p+2)$

$p(4 p^2-16) \div 4 p(p-2)=\dfrac{p(4 p^2-16)}{4 p(p-2)}=\dfrac{4 p(p-2)(p+2)}{4 p(p-2)}=p+2$

The answer is

$(b)$

Find :
${ 9x }^{ 2 }-15x+6=?$

$9x^2-15x+6$

$3(3x^2-5x+2)$

$3(3x^2-3x-2x+2)$

$3(3x(x-1)-2(x-1))$

$3(3x-2)(x-1)$ .

1199875_1445106_ans_05fa614c45e745108cb2d6cbc370bb85.jpg

$36\left( x+4 \right) \left( { x }^{ 2 }+7x+10 \right) \div 9\left( x+4 \right)$

We need to divide the polynomial

$36(x+4)(x^2+7x+10)$ by the binomial

$9(x+4)$ .

Hence,

$\dfrac{36(x+4)(x^2+7x+10)}{9(x+4)}$

$=\dfrac{9(x+4)}{9(x+4)}\times 4(x^2+7x+10)$

$=4(x^2+7x+10)$

$=4(x^2+5x+2x+10)$

$=4[x(x+5)+2(x+5)]$

$=4[(x+2)(x+5)]$

$=4(x+2)(x+5)$

Hence,

$36(x+4)(x^2+7x+10)\div 9(x+4)=4(x+2)(x+5)$ .

In the division algorithm of polynomials the divisor is $(x+2)$ , quotient is $(x+1)$ and the remainder is $4$ . Find the dividend?

Given

$:b=x+2,q=x+1,r=4$

By Euclid's division lemma,

$a=bq+r$

$\Rightarrow a=\left(x+2\right)\left(x+1\right)+4$

$\Rightarrow a={x}^{2}+\left(2+4\right)x+2\times 1+4$

$\Rightarrow a={x}^{2}+6x+6$

$\therefore$ Dividend

$={x}^{2}+6x+6$

Factorise the given polynomial expression:
${a}^{2}{x}^{2}+(a{x}^{2}+1)x+a$

Given,

${a}^{2}{x}^{2}+(a{x}^{2}+1)x+a$

By Rearranging the terms we get,

${a}^{2}{x}^{2}+a+(a{x}^{2}+1)x$

$=a(a{x}^{2}+1)+(a{x}^{2}+1)x$

$=(x+a)(a{x}^{2}+1)$

$\therefore {a}^2{x}^2+(ax^2+1)+a=(x+a)(a{x}^{2}+1)$

Factorise : $27 a^3b^3 - 45 a^4 b^2$

1563616_1714930_ans_c138989cd4eb418d8061937fdee378f3.png

Factorise $14p^{2} + 21pq$ .

$14p^2+21pq\\7p (2p + 3q)$ .

Factorise the quadratic expression:
${x}^{2}+y-xy-x$

Given,

$x^2+y-xy-x$

By Rearranging the terms we get,

$x^2-x+y-xy$

$=x(x-1)-y(x-1)$

$=(x-1)(x-y)$

$\therefore x^2=y-xy-x=(x-1)(x-y)$

Factorise : $x^5 +x^2$

1563346_1715257_ans_8b093a2a9b9d496c816b202a5f581ca4.png

Factorise the given polynomial expression:
${x}^{3}+x-3{x}^{2}-3$

Given

$x^3+x-3x^2-3$

$=x^3-3x+x^2-3$

$=x^2(x-3)+1(x-3)$

$=(x-3)({x}^{2}+1)$

$\therefore x^3+x-3x^2-3=(x-3)(x^2+1)$

Factorise completely:
$18{x}^{2}-24x$

$18x^{2}-24x$ =

$6x(3x-4)$

Factorise : $x^3-x^2+ax+x-a-1$

1563619_1714964_ans_1cc80fd7c32b4b78b0b5661f96d2e1e0.png

Factorise : $32x^4 -500 x$

1563359_1715273_ans_6ba60cd1b04f40898a65cd3e918fb145.png

Factorise:
$6d^2e-9e^2$ .

Given expression:

$6d^2e-9e^2$

$= 3e(2d^2)-3e(3e)$ {Taking

$3e$ as common factor}

$=3e(2d^2-3e)$

Divide $5m^{3}-30m^{2}+45m$ by $5m$

$5m^3-30m^2+45m$ by

$5m$

On dividing polynomial by a monomial we have divide every variables of polynomial.

$\Rightarrow$

$\dfrac{5m^2-30m^2+45m}{5m}$

$=\dfrac{5m^3}{5m}-\dfrac{30m^2}{5m}+\dfrac{45m}{5m}$

$=\dfrac{5}{5}m^{3-1}-\dfrac{30}{5}m^{2-1}+\dfrac{45}{5}m^{1-1}$

$=m^2-6m+9$

Write the greatest common factor of the following terms $2xy$ , $-y^{2}$ , $2xy$

$y(2x, -y, 2x^{2})$

Greatest common factor is

$y$

Common factor of $ax^{2} + bx$ is _____ .

Factors of

$ax^{2} = a\times x \times x$

Factors of

$bx = b \times x$

Hence, the common factor of

$ax^2+bx$ is

$x$ .

Factorise the following expressions: $-xy - ay$

$—xy — ay$

$= y(-x-a)$

Factorise : $3a^7 b - 81 a^4b^4$

1563362_1715274_ans_2258f06443bc4b69a6c29be16f0a60cc.png

Factorise the following expressions: $ax^{2} - bx^{2} + cx$

$ax^{2} — bx^{2} + cx$

$=x(ax^{2}-bx+c)$

Factorise the following expressions: $6ab + 12bc$

$6ab + 12bc$

$= 6b(a+2c)$

$15ax^3 -9ax^2$

$15ax^3 -9ax^2$
Take out common in both terms,
Then,

$3ax^2(5x-3)$
Therefore,HCF of

$15 ax^3\quad and \quad 9ax^2 is \quad 3ax^2.$

If factorised form of $18mn + 10mnp$ is $amn (b + 5p)$ . Then product of $a$ and $b$ is

We have,

$18mn + 10mnp$

Factoring

$2mn$ out,

$=2mn(9+5p)$

$\therefore 18mn+10mnp =2mn(9+5p)$

comparing with

$a=2, b=9$

$ab=18.$

Factorise the following expressions: $l^{2}m^{2}n - lm^{2}n^{2} l^{2}mn^{2}$

Given expession:

$l^2m^2n-lm^2n^2l^2mn^2$

taking

$l^2m^2n$ common from the expression

$\Rightarrow l^2m^2n(1-lmn^3)$

the given expression can be factorised as

$l^2m^2n(1-lmn^3)$

Divide:
$5x^{2} - 3x$ by $x$

$\begin{aligned}{}\left( {5{x^2} - 3x} \right) \div x& = \frac{{5{x^2} - 3x}}{x}\\ &= \frac{{5{x^2}}}{x} - \frac{{3x}}{x}\\& = 5x - 3\end{aligned}$

Factorize : $a^{2}(b+c)^{2}+b^{2}(c+a)^{2}+c^{2}(a+b)+abc(a+b+c)+(a^{2}+b^{2}+c^{2})(bc+ca+ab)$

$f(a)=a^{2}(b+c)^{2}+b^{2}(c+a)^{2}+c^{2}(a+b)^{2}+abc(a+b+c)+(a^{2}+b^{2}+c^{2})(bc+ca+ab)$

$f(-b)=b^{2}(b+c)^{2}+b^{2}(c-b)^{2}-b^{2}c^{2}+(2b^{2}+c^{2})(bc-bc-b^{2})$

$=b^{2}(b^{2}+c^{2}+2b^{4}c^{2}+b^{2}-2bc)$

$=b^{2}\left [ 2b^{2}+2c^{2} \right ]$

$=2b^{4}+| 2b^{2}c^{2}-b^{2}c^{2}-2b^{4}-b^{2}c^{2}=0$

$\therefore (a+b)$ is a factor
Since the expression is cyclic,

$(a+b)(b+c)(c+a)$ are factors.

$\therefore k(a+b+c)(a+b)(b+c)(c+a)$
By putting

$a=0, b=1, c=2,$ we get

$k=1$

$\therefore (a+b+c)(a+b)(b+c)(c+a)$ are the factors.

Factorise $xy^2 - xz^2$ , Hence, find the value of
$40 \times 5.5^2 - 40 \times 4.5^2$

$Factorizing\quad xy^{ 2 }-xz^{ 2 }\\ =x(y^{ 2 }-z^{ 2 })\\ =x(y+z)(y-z),\\ i.e\quad value\quad of\quad 40\times 5.5^{ 2 }-40\times 4.5^{ 2 }\\ =40(5.5+4.5)(5.5-4.5)\\ =400$

What are the possible expressions for the dimension of the cuboids whose volumes are given as in image?

(i)

$3x^{2}-12x=3x(x-4)$

$=3\times x\times (x-4)$ ,

The possible values of dimensions are

$3,\ x$ and

$x-4.$

(ii)

$12ky^{2}+8ky-20k$

$=4k(3y^{2}+2y-5)$

$=4k(3y^{2}+5y-3y+5)$

$=4k\left \{ y(3y+5)-1(3y+5)\right \}$

$=4k(3y+5)(y-1)$

The possible values of dimensions are

$4k$ ,

$(3y+5)$ and

$(y-1).$

Factorise the following using appropriate identities:
(i) $9x^2+6xy+y^2$
(ii) $4y^2-4y+1$
(iii) $x^2-\displaystyle\frac{y^2}{100}$

(i)

$9x^{2}+6xy+y^{2}=(3x)^{2}+2(3x)(y)+(y)^{2}=(3x+y)^{2}$

(ii)

$4y^{2}-4y+1=(2y)^{2}-2(2y)(1)+1=(2y-1)^{2}$

(iii)

$x^{2}-\dfrac{y^{2}}{100}=(x)^{2}-\left ( \dfrac{y}{10} \right )^{2}=\left ( x-\dfrac{y}{10} \right )\left ( x+\dfrac{y}{10} \right )$

Find the common factors of the given terms.
(i) $12x, 36$

(ii) $2y, 22xy$

(iii) $14 pq, 28 p^{2}q^{2}$

(iv) $2x, 3x^{2}, 4$

(v) $6abc , 24 ab^{2} , 12a^{2}b$

(vi) $16x^{3}, -4x^{2}, 32x$

(vii) $10 pq, 20qr, 30rp$

(viii) $3x^{2}y^{3}, 10x^{3}y^{2}, 6x^{2}y^{2}z$

(i)

$12x = 2\times 2\times 3\times x$

$36 = 2\times 2\times 3\times 3$
Common factor

$= 2\times 3 =6$

(ii)

$2y = 2\times y$

$22y = 2\times 11\times y$
Common factor

$=2$

(iii)

$14 pq= 2\times 7\times p\times q$

$28 p^{2}q^{2} = 2\times 2\times 7\times p\times p\times q\times q$
Common factor

$= 14 pq$

(iv)

$2x = 2\times x$

$3x^{2} = 3\times x\times x$

$4 = 2\times 2$
Common factor

$= 1$

(v)

$6abc = 6\times a\times b\times c$

$24ab^{2} = 2\times 2\times 2\times 3\times a\times b\times b$

$12a^{2}b = 2\times 2\times 3\times a\times a\times b$
Common factor

$= 6ab$

(vi)

$16x^{3} = 2\times 2\times 2\times 2\times x\times x\times x$

$-4x^{2} = -2\times 2\times x\times x$

$32 x = 2\times 2\times 2\times 2\times 2\times x$
Common factor

$= 2x$

(vii)

$10pq = 2\times 5\times p\times q$

$20qr = 2\times 2\times 5\times q\times r$

$30rp = 2\times 3\times 5$
Common factor

$= 2\times 5 = 10$

(viii)

$3x^{2}y^{3} = 3\times x\times x\times y\times y\times y$

$10x^{3}y^{2} = 2\times 5\times x\times x\times x\times y\times y$

$6x^{2}y^{2} z = 2\times 3\times x\times x\times y\times y\times z$
Common factor =

$x^{2}y^{2}$

Factorise: ${ x }^{ 2 }+14x+45$

$x^2+14x+45$

$= x^2+9x+5x+45$

$= (x^2+9x) +(5x+45)$

$= x(x+9)+5(x+9)$

$=(x+9)(x+5)$

Factorise: $36{ x }^{ 2 }+25+60x$

$36x^2+25+60x$ or , $36x^2+60x +25$

$= 36x^2+30x +30x+25$

$= (36x^2+30x) +(30x+25)$

$= 6x(6x+5)+5(6x+5)$

$=(6x+5)(6x+5) = (6x+5)^2$

Find and correct errors of the following mathematical expressions:
$\dfrac{4x+5}{4x} =5$

The given statement is incorrect.
After separating the denominators, we get

$\dfrac{4x+5}{4x} = 1+ \dfrac{5}{4x}$ which is the correct statement.

Factorise
(i) $x^{2}+xy +8x+8y$
(ii) $15xy-6x+5y-2$
(iii) $ax+by-ay-by$
(iv) $15pq+15+9q+25p$
(v) $z-7+7xy-xyz$

(i)

$x^{2}+xy +8x+8y$

$= x(x+y)+8(x+y)$

$= (x+8)(x+y)$

(ii)

$15xy-6x+5y-2$

$= 3x(5y-2)+1(5y-2)$

$= (3x+1)(5y-2)$

(iii)

$ax+by-ay-by$

$= x(a+b) -y(a+b)$

$= (x-y) (a+b)$

(iv)

$15pq+15+9q+25p$

$= 3q (5p+3)+5(5p+3)$

$= (3q+5)(5p+3)$

(v)

$z-7+7xy-xyz$

$= z(1-xy)-7(1-xy)$

$= (z-7) (1-xy)$

Factorise
(i) $4p^{2} - 9q^{2}$
(ii) $63a^{2}-112b^{2}$
(iii) $49x^{2}-36$
(iv) $16x^{5}-144x^{3}$
(v) $(l+m)^{2}-(l-m)^{2}$
(vi) $9x^{2}y^{2}-16$
(vii) $(x^{2}-2xy+y^{2})-z^{2}$
(viii) $25a^{2}-4b^{2}+28bc-49c^{2}$

(i)

$4p^{2} - 9q^{2}$

$= (2p)^{2} - (3q)^{2}$

$= (2p+3q) (2p-3q)$

(ii)

$63a^{2}-112b^{2}$

$=7[ (3a)^{2} - (4b)^{2} ]$

$= 7(3a+4b) (3a-4b)$

(iii)

$49x^{2}-36$

$= [(7x)^{2} - (6)^{2}]$

$= (7x-6) (7x+6 )$

(iv)

$16x^{5}-144x^{3}$

$= 16 x^{3} [ x^{2} - 3^{2} ]$

$= 16 x^{3} (x-3)(x+3)$

(v)

$(l+m)^{2}-(l-m)^{2}$

$= 2m\times 2l = 4lm$

(vi)

$9x^{2}y^{2}-16$

$= [ (3xy)^{2} - 4^{2} ]$

$=(3xy-4) (3xy+4)$

(vii)

$(x^{2}-2xy+y^{2})-z^{2}$

$= [(x-y)^{2} - z^{2} ]$

$= (x-y+z) (x-y-z)$

(viii)

$25a^{2}-4b^{2}+28bc-49c^{2}$

$= (5a)^{2}- [(2b-7c)^{2}]$

$= (5a+2b-7c) (5a-2b+7c)$

Factorise: $2a\left( x-y \right) +3b\left( 5x-5y \right) +4c\left( 2y-2x \right)$

$2a(x−y)+3b(5x−5y)+4c(2y−2x)$

$= 2a(x−y)+3b\times {5(x−y)}+4.2(x-yc)$

$= 2a(x−y)+15b(x−y)-8c(x-y)$

$= (2a+15b-8c)(x-y)$

Find and correct errors of the following mathematical expressions:
$\dfrac{7x+5}{5} =7x$

The given statement is incorrect.
We will get correct statement by separating denominators, which is

$\dfrac{7x+5}{5} = 1+ \dfrac{7x}{5}$

Divide the polynomial by the given monomial
(i) $(5x^{2}-6x) \div 3x$
(ii) $(3y^{8}-4y^{6}+5y^{4})\div y^{4}$
(iii) $8(x^{3}y^{2}z^{2}+x^{2}y^{3}z^{2} +x^{2}y^{2}z^{3})\div 4x^{2}y^{2}z^{2}$
(iv) $(x^{3}+2x^{2}+3x) \div 2x$
(v) $(p^{3}q^{6}-p^{6}q^{3}) \div p^{3}q^{3}$

(i)

$(5x^{2}-6x) \div 3x$

$= \dfrac{1}{3}(5x-6)$

(ii)

$(3y^{8}-4y^{6}+5y^{4})\div y^{4}$

$= 3y^{4}-4y^{2}+5$

(iii)

$8(x^{3}y^{2}z^{2}+x^{2}y^{3}z^{2} +x^{2}y^{2}z^{3})\div 4x^{2}y^{2}z^{2}$

$=2(x+y+z)$

(iv)

$(x^{3}+2x^{2}+3x) \div 2x$

$=\dfrac{1}{2}(x^{2}+2x+3)$

(v)

$(p^{3}q^{6}-p^{6}q^{3}) \div p^{3}q^{3}$

$= q^{3}- p^{3}$

Find the remiander if we divide $\left( 15{ y }^{ 4 }-16{ y }^{ 3 }+9{ y }^{ 2 }-\dfrac { 1 }{ 3 } y-\dfrac { 50 }{ 9 } \right)$ by $\left( 3y-2 \right)$

$\left(15 y^4 - 16 y^3 + 9y^2 - \dfrac{1}{3} y - \dfrac{50}{9} \right) = (3y - 2) y + R$

Put

$y = 2/3$ in the above equation we get

$D + R = \dfrac{-32}{9} \Rightarrow R = -\dfrac{-32}{9}$

Factorise: $7xa-70xb$

Since in the expression

$7xa-70xb$ ,

$7x$ is the common term, therefore, can be factorised as follows:

$7xa-70xb=7x(a-10b)$

$4x^2, \ 6xy, \ 8y^2x$

The given terms are

$4x^2, 6xy$ and

$8y^2x$

To find out: The common factor of the given terms.

We can factorise each of the given terms as:

$4x^2=2\times 2\times x\times x$

$6xy=3\times 2\times x\times y$

$8y^2x=2\times 2\times 2\times y\times y\times x$

We see

$2$ and

$x$ common in each terms.

Hence, common factor

$=2\times x=2x$

Therefore, the common factor of the given terms is

$2x$ .

Factorise: $\sqrt { 3 } { y }^{ 2 }+11y+6\sqrt { 3 }$

$\sqrt{3}y^2+11y+6\sqrt{3}$

$= \sqrt{3}y^2+9y+2y+6\sqrt{3}$

$= (\sqrt{3}y^2+9y) +(2y+6\sqrt{3})$

$= \sqrt{3}y(y+3\sqrt{3})+2(y+3\sqrt{3})$

$=( y+3\sqrt{3})( \sqrt{3}y +2)$

Factorise: $13{ m }^{ 2 }+156{ n }^{ 2 }$

Since in the expression

$13m^2+156n^2$ ,

$13$ is the common term, therefore, can be factorised as follows:

$13m^2+156n^2=13(m^2+12n^2)$

$a^3 - a^2 b^2 - ab + b^3$

We need to factorise

$a^3-a^2b^2-ab+b^3$

Writing terms with common factors aside,

$(a^3-a^2b^2)-(ab-b^3)$

$a^2(a-b^2)-b(a-b^2)$

$=(a^2-b)(a-b^2)$
Thus factors are

$(a^2-b),(a-b^2)$ .

Factorise the expression $x^2 + 10 x + 25$

The given expression contains three terms and the first and third terms are perfect squares. That is

$x^2$ and 25

$(5^2)$ . Also the middle term contains the positive sign. This suggests that it can be written in the form of

$a^2 + 2ab + b^2$ , so

$x^2 + 10 x + 25 = (x)^2 + 2 (x) (5) + (5)^2$
We can compare it with

$a^2 + 2ab + b^2$ which in turn is equal to the LHS of the identity ie.

$(a + b)^2$ . Here a = x and b = 5
We have

$x^2 + 10 x + 25 = (x + 5)^2 = (x + 5)(x + 5)$

Factorise: $12{ \left( { a }^{ 2 }+7a \right) }^{ 2 }-8\left( { a }^{ 2 }+7a \right) \left( 2a-1 \right) -15{ \left( 2a-1 \right) }^{ 2 }$

$12(a^2+7a)^2−8(a^2+7a)(2a−1)−15(2a−1)^2$

Let $(a^2+7a)=x$ and $(2a-1) =y$

Then the previous equation becomes:

$12x^2-8xy-15y^2$

$= 12x^2-18xy+10xy-15y^2$

$= (12x^2-18xy)+(10xy-15y^2)$

$= 6x(2x-3y)+5y(2x-3y)$

$= (6x+5y)(2x-3y)$

Substituting the values of x and y:

$= {6(a^2+7a)+5(2a-1)}{2(a^2+7a)-3(2a-1)}$

$= (12a^2+42a+10a-5)(2a^2+14a-6a+3)$

$(12a^2+52a-5)(2a^2+8a+3)$

Factorise $48a^2 - 243b^2$

We see that the two terms are not perfect squares. But both has '3' as common factor.
That is

$48a^2 - 243 b^2 = 3[16a^2 - 81b^2]$

$= 3[(4a)^2 - (9b)^2]$ Again

$a^2 - b^2 = (a + b)(a - b)$

$= 3 [(4a + 9b)(4a - 9b)]$

$= 3(4a + 9b) (4a - 9b)$

Factorise the expression $4x^2 + 20x - 96$

We notice that 4 is the common factor of all the terms.
Thus

$4x^2 + 20x - 96 = 4[x^2 + 5x - 24]$
Now consider

$x^2 + 5x - 24$

$= x^2 + 8x - 3x - 24$

$= x(x + 8) - 3(x + 8)$

$=(x + 8)(x - 3)$
Therefore

$4x^2 + 20x - 96 = 4 (x + 8) (x - 3)$

Factorise:
$16z^2 - 48z + 36$

Taking common numerical factor from the given expression we get

$16z^2 - 48z + 36 = (4 \times 4z^2) - (4 \times 12z) + (4 \times 9) = 4(4z^2 - 12z + 9)$

Note that

$4z^2 = (2z)^2 ; 9 = (3)^2$ and

$12z = 2(2z) (3)$

as,

$a^2 - 2ab + b^2 = (a - b)^2$

$\therefore$

$4z^2 - 12z + 9 = (2z)^2 - 2(2z) (3) + (3)^2$

$= (2z - 3)^2$

By comparison,

$16z^2 - 48z + 36 = 4(4z^2 - 12z + 9) = 4(2z - 3)^2$

$= 4(2z - 3) (2z - 3)$

Factorise the expression $p^4 - 256$

$p^4 = (p^2)^2$ and

$256 = (16)^2$
Thus

$p^4 - 256 = (p^2)^2 - (16)^2$

$= (p^2 - 16)(p^2 + 16)$

$[\because p^2 - 16 = (p + 4)(p - 4)]$

$= (p + 4)(p - 4)(p^2 + 16)$

Divide $30(a^2bc + ab^2 c + abc^2)$ by $6abc$

$30(a^2bc + ab^2c + abc^2)$

$= 2 \times 3 \times 5 [(a \times a \times b \times c)+ (a \times b \times b \times c)+ (a\times b \times c \times c)]$

$= 2\times 3 \times 5 \times a\times b\times c (a + b + c)$
Thus

$30(a^2 bc + ab^2c + abc^2) \div 6abc$

$= \dfrac{2 \times 3 \times 5 \times abc (a + b + c)}{2 \times 3 \times abc}$

$= 5(a + b+ c)$
Alternatively

$30(a^2 bc + ab^2 c + abc^2) \div 6abc$

$= \dfrac{30 a^2 bc}{6abc} + \dfrac{30 ab^2c}{6abc} + \dfrac{30abc^2}{6abc}$

$= 5a + 5b + 5c$

$=5(a + b + c)$

$81x^2 - 198xy + 121y^2$

The given polynomial

$81x^2-198xy+121y^2$ can be factorized as follows:

$81x^{ 2 }-198xy+121y^{ 2 }\\ =(9x)^{ 2 }-(2\times 9x\times 11y)+(11y)^{ 2 }\\ =(9x-11y)^{ 2 }\quad \quad \quad \quad \quad \quad \quad \quad \quad (\because \quad (a-b)^{ 2 }=a^{ 2 }+b^{ 2 }-2ab)$

Hence,

$81x^{ 2 }-198xy+121y^{ 2 }=(9x-11y)^{ 2 }$ .

Factorise the expression $x^2 + 2xy + y^2 - 4z^2$

The first three terms of the expression is in the form

$(x + y)^2$ and the fourth term is a perfect square.
Hence,

$x^2 + 2xy + y^2 - 4z^2 = (x + y)^2 - (2z)^2$

$=[(x + y) + 2z][(x + y) - 2z]$

$= (x + y + 2z) (x + y - 2z)$

Find out the quotient and the remainder when
$P(x) = x^{3} + 4x^{2} - 5x + 6$ is divided by $g(x) = x + 1$

$x + 1)\overset {x^{2} + 3x - 8}{\overline {x^{3} + 4x^{2} - 5x + 6}}$

$\underset {(-)}{x^{3}} \underset {(-)}{+} 1x^{2}$

$+ 3x^{2} - 5x + 6$

$\underset {(-)}{+} 3x^{2} \underset {(-)}{+} 3x$

$-8x + 6$

$\underset {(+)} 8x \underset {(+)}{-} 8$

$14$
Therefore,

$q(x) = x^{2} + 3x - 8$ and

$R(x) = 14$ .

Divide $3y^{3} + 2y^{2} + y$ by $y$ .

Given polynomial is

$3y^{3}+2y^{2}+y$ .

Dividing it by

$y$ :

$(3y^{3}+2y^{2}+y)\div y$

Taking

$y$ common from all the terms of the given polynomial,

$\Rightarrow y(3y^2+2y+1)\times \cfrac{1}{y}$

$\Rightarrow 3y^2+2y+1$

Hence,

$3y^{3}+2y^{2}+y$ divided by

$y$ is

$3y^2+2y+1$ .

Divide $4p^{2} + 2p + 2$ by $'2p'$ .

Given polynomial

$4P^{2}+2P+2$ divided by

$2p$

$4p^{2}+2p+2\div 2p$

$\Rightarrow 4p^{2}+2p+2\times \cfrac{1}{2p}$

$\Rightarrow 2(2p^{2}+p+1)\times \cfrac{1}{2p}$

$\Rightarrow \cfrac{2p^{2}+p+1}{p}$

Laxmi does not want to disclose the length, breadth and height of a cuboid of her project. She has constructed a polynomial ${ x }^{ 3 }-6{ x }^{ 2 }+11x-6$ by taking the values of length, breadth and height as its zeroes. Can you open the secret [i.e., find the measures of length, breadth and height]?

Solution:

Given polynomial whose zeroes are length, breadth and height respectively is

$x^3-6x^2+11x-6.$

$x^3-6x^2+11x-6=0$

0r,

$x^3-x^2-5x^2+5x+6x-6=0$

or,

$x^2(x-1)-5x(x-1)+6(x-1)=0$

or,

$(x-1)(x^2-5x+6)=0$

or,

$(x-1)(x^2-3x-2x+6)=0$

or,

$(x-1)\left\{x(x-3)-2(x-3)\right\}=0$

or,

$(x-1(x-2)(x-3))=0$

or,

$x=1$ or

$x=2$ or

$x=3$

Therefore, length, breadth and height of cuboid are

$3units,2units$ and

$1unit$ respectively.

Factorise $3{ x }^{ 4 }-10{ x }^{ 3 }+5{ x }^{ 2 }+10x-8$

Factorisation:

$3x^4-10x^3+5x^2+10x-8$

$=3x^4-3x^3-7x^3+7x^2-2x^2+2x+8x-8$

$=3x^3(x-1)-7x^2(x-1)-2x(x-1)+8(x-1)$

$=(x-1)(3x^3-7x^2-2x+8)$

$=(x-1)(3x^3+3x^2-10x^2-10x+8x+8)$

$=(x-1)\left\{3x^2(x+1)-10x(x+1)+8(x+1)\right\}$

$=(x-1)(x+1)(3x^2-10x+8)$

$=(x-1)(x+1)(3x^2-6x-4x+8)$

$=(x-1)(x+1)\left\{3x(x-2)-4(x-2)\right\}$

$=(x-1)(x+1)(x-2)(3x-4)$

Factorize:
$a^{2}x + abx + ac + aby + b^{2}y + bc$

Using common factor method :

$=a^2x+abx+aby+b^2y+ac+bc$

$=ax\left( { a+b } \right) +by\left( { a+b } \right) +c\left( { a+b } \right)$

$=\left( { a+b } \right) \left[ { ax+by+c } \right]$

Hence, the answer is

$\left( { a+b } \right) \left[ { ax+by+c } \right].$

Factorize:
$x^{2} + \dfrac {2}{3}x + \dfrac {1}{9}$

$x^2+\dfrac{2}{3}x+\dfrac{1}{9}$

$=x^2+\dfrac{1}{3}x+\dfrac{1}{3}x+\dfrac{1}{9}$

$=x\left( { x+\dfrac { 1 }{ { 3 } } } \right) +\dfrac { 1 }{ { 3 } }\left( { x+\dfrac { 1 }{ { 3 } } } \right)$

$=\left( { x+\dfrac { 1 }{ { 3 } } } \right) \left( { x+\dfrac { 1 }{ { 3 } } } \right)$

$=\left( { x+\dfrac { 1 }{ { 3 } } } \right)^2.$

Hence, the answer is

$\left( { x+\dfrac { 1 }{ { 3 } } } \right)^2.$

Factorize the following expressions:
$2a^{5}b^{3} - 14a^{2}b^{2} + 4a^{3}b$

Let us first find the HCF of all the terms of the given polynomial

$2a^5b^3-14a^2b^2+4a^3b$ by factoring the terms as follows:

$2a^{ 5 }b^{ 3 }=2\times a\times a\times a\times a\times a\times b\times b\times b\\ 14a^{ 2 }b^{ 2 }=2\times 7\times a\times a\times b\times b\\ 4a^{ 3 }b=2\times 2\times a\times a\times a\times b$

Therefore,

$HCF=2\times a\times a\times b=2a^{ 2 }b$

Now, we factor out the HCF from each term of the polynomial

$2a^5b^3-14a^2b^2+4a^3b$ as shown below:

$2a^{ 5 }b^{ 3 }-14a^{ 2 }b^{ 2 }+4a^{ 3 }b=2a^{ 2 }b(a^{ 3 }b^{ 2 }-7b+2a)$

Hence,

$2a^{ 5 }b^{ 3 }-14a^{ 2 }b^{ 2 }+4a^{ 3 }b=2a^{ 2 }b(a^{ 3 }b^{ 2 }-7b+2a)$ .

Factorise : $2x^{2}+7x+3$ .

$2{x^2} + 7x + 3$

$= 2{x^2} + 6x + x + 3$

$= 2x\left( {x + 3} \right) + 1\left( {x + 3} \right)$

$= \left( {2x + 1} \right)\left( {x + 3} \right)$

Rearrange the like terms:
$7l^{3}m^{2} - 21lm^{2}n + 28lm$

The given terms

$7l^3m^2,21lm^2n$ and

$28lm$ can be rewritten as follows:

$7l^{ 3 }m^{ 2 }=7\times l\times l\times l\times m\times m\\ 21lm^{ 2 }n=7\times 3\times l\times m\times m\times n\\ 28lm=7\times 4\times l\times m$

These terms have

$7,l$ and

$m$ as common factors, therefore,

$7l^{ 3 }m^{ 2 }-21lm^{ 2 }n+28lm\\ =(7\times l\times l\times l\times m\times m)-(7\times 3\times l\times m\times m\times n)+(7\times 4\times l\times m)\\ =7\times l\times m[(l\times l\times m)-(3\times m\times n)+(4)]\quad \quad \quad \quad \quad \quad \quad (Combining\quad like\quad terms)\\ =7lm(l^{ 2 }m-3mn+4)$

Hence,

$7l^{ 3 }m^{ 2 }-21lm^{ 2 }n+28lm=7lm(l^{ 2 }m-3mn+4)$ .

Factorize the following expressions:
$3x^{3} - 5x^{2} + 6x$

Factorize :

$3x^3-5x^2+6x.$

We can take

$x$ common,

$x\left( 3x^2-5x+6 \right)$

Hence, the answer is

$x\left( 3x^2-5x+6 \right).$

Factorize the following:
$ab(x^{2} + 1) + x (a^{2} + b^{2})$

$ab\left( { x }^{ 2 }+1 \right) +x\left( { a }^{ 2 }+{ b }^{ 2 } \right)$

$\Rightarrow abx^2+a^2x+b^2x+ab$

$\Rightarrow ax\left( bx+a \right) +b\left( bx+a \right)$

$\Rightarrow \left( ax+b \right) \left( bx+a \right)$

Hence, the answer is

$\left( ax+b \right) \left( bx+a \right).$

Find the remainder using remainder theorem, when:
$2x^3-4x^2+7x+6$ is divided by $x-2$

According to remainder theorem when

$f(x)$ is divided by

$(x-2)$ , Remainder

$=f(2)$

$f(x)=2x^3-4x^2+7x+6$

$f(2)=2(2)^3-4(2)^2+7(2)+6$

$=16-8+14+6=28$

$\therefore$ Remainder

$=28$

If the polynomials $mx^3-2x^2+25x-26$ and $2x^3-mx+9$ leave the same remainder when they are divided by $(x-2)$ , find the value of m. Also find the remainder.

Let,

$f(x)=mx^3-2x^2+25x-26$

$f(2)=m(2)^3-2(2)^2+25(2)-26$

$=8m-8+50-26$

$=8m+16$

Let,

$p(x)=2x^3-mx+9$

$p(2)=2(2)^3-2m+9$

$=16+9-2m=25-2m$

Given:

$f(x)$ and

$p(x)$ leave the same remainder, when they are divided by

$(x-2)$ ,

$\therefore f(2)=p(2)$

$\implies 8m+16=25-2m$

$\implies 8m+2m=25-16$

$\implies 10m=9$

$\implies m=\dfrac{9}{10}$

Now,

Remainder

$=f(2)=8m+16$

$=8\times \dfrac{9}{10}+16$

$=7.2+16=23.2$

When the polynomial $2x^3-ax^2+9x-8$ is divided by $x-3$ the remainder is Find the value of a.

According to remainder theorem, when

$f(x)$ is divided by

$(x-3)$ ,Remanider

$=f(3)=28$ (given).

$f(x)=2x^3-ax^2+9x-8$

$\implies f(3)=2(3)^3-a(3)^2+9(3)-8$

$\implies 28=54-9a+27-8$

$\implies 9a=73-28=45$

$\implies a=\dfrac{45}{9}=5$

Find the value of m if $x^3-6x^2+mx+60$ leaves the remainder 2 when divided by $(x+2)$ .

According to remainder theorem, when

$f(x)$ is divided by

$(x+2)$ ,Remanider

$=f(-2)=2$ (given).

$f(x)=x^3-6x^2+mx+60$

$f(-2)=(-2)^3-6(-2)^2+(-2)m+60$

$\implies 2=-8-24-2m+60$

$\implies 2m=28-2=26$

$\implies m=\dfrac{26}{2}=13$

Determine whether $(x+1)$ is a factor of the polynomial:
$x^3-14x^2+3x+12$

According to factor theorem, when

$f(x)$ is divided by

$(x+1)$ ,

$f(-1)=0$

$f(x)=x^3-14x^2+3x+12$

$f(-1)=(-1)^3-14(-1)^2+3(-1)+12$

$=-1-14-3+12=-6$

$\because f(-1) \neq 0 , (x+1)$ is not a factor of the polynomial.

If $(x-1)$ divides $mx^3-2x^2+25x-26$ with remainder $0$ , find the value of $m$ .

According to remainder theorem, when

$f(x)$ is divided by

$(x-1)$

Remainder

$=f(1)=0$ (given).

$f(x)=mx^3-2x^2+25x-26$

$f(1)=m(1)^3-2(1)^2+25(1)-26$

$\implies 0=m-2+25-26$

$\implies m-3=0$

$\implies m=3$

Find the remainder using remainder theorem, when:
$x^3-ax^2-5x+2a$ is divided by $x-a$

According to remainder theorem, when

$f(x)$ is divided by

$(x-a)$ , Remainder

$=f(a)$

$f(x)=x^3-ax^2-5x+2a$

$f(a)=a^3-a(a)^2-5(a)+2a$

$=a^3-a^3-5a+2a$

$-3a$

$\therefore$ Remanider

$=-3a$

Determine whether $(x+1)$ is a factor of the polynomial:
$3x^3+8x^2-6x-5$

According to factor theorem, when

$f(x)$ is divided by

$(x+1), f(-1)=0$

$f(x)=3x^3+8x^2-6x-5$

$f(-1)=3(-1)^3+8(-1)^2-6(-1)-5$

$=-3+8+6-5=6$

$\because f(-1) \neq 0, (x+1)$ is not a factor of the given polynomial.

Determine whether $(x+4)$ is a factor of $x^3+3x^2-5x+36$ .

According to factor theorem, when

$f(x)$ is divided by

$(x+4), f(-4)=0$

$f(x)=x^3+3x^2-5x+36$

$f(-4)=(-4)^3+3(-4)^2-5(-4)+36$

$=-64+48+20+36=40$

$\because f(-4) \neq 0, (x+4)$ is not a factor of the given polynomial.

Find the remainder using remainder theorem, when:
$4x^3-3x^2+2x-4$ is divided by $x+3$

According to remainder theorem when

$f(x)$ is divided by

$(x+3)$ , Remainder

$=f(-3)$

$f(x)=4x^3-3x^2+2x-4$

$f(-3)=4(-3)^3-3(-3)^2+2(-3)-4$

$=4\times -27-3\times 9-6-4=-108-27-10=-145$

$\therefore$ Remainder

$=-145$

Determine whether $(2x+1)$ is a factor of $4x^3+4x^2-x-1$ .

According to factor theorem, when

$f(x)$ is divided by

$(2x+1)$ ,

$f(\dfrac{-1}{2})=0$

$f(x)=4x^3+4x^2-x-1$

$f(\dfrac{-1}{2})=4(\dfrac{-1}{2})^3+4(\dfrac{-1}{2})^2-(\dfrac{-1}{2})-1$

$=4 \times \dfrac{-1}{8}+4 \times \dfrac{1}{4}+\dfrac{1}{2}-1$

$=\dfrac{-1}{2}+1+\dfrac{1}{2}-1=0$

$\because f(\dfrac{-1}{2})=0, (2x+1)$ is a factor of the polynomial.

Give possible expressions for the length and breadth of the following rectangles, in which their areas are given:
Area: $35{y}^{2}+13y-12$

$= 35y^{2}+13y-12$

$=35y^{2}+35y-12y-12$

$=35y(y+1)-12(y+1)$

$=(35y-12)(y+1)$

$l=35y-12$ ,

$b=y+1$

$y+1$ ,

$b = 35y-12$

Factorize $x^3 + 3x^2 - x - 3$

Let

$p(x) = x^3 + 3x^2 - x - 3$

p(x) is a cubic polynomial, so it may have three linear factors.

The constant term is -3. The factors of -3 are -1, 1, -3 and 3.

$p(-1) = (-1)^3 + 3 (-1)^2 - (-1) - 3 = -1 + 3 + 1 -3 = 0$

$\therefore (x + 1)$ is a factor of p(x).

$p(1) = (1)^3 + 3(1)^2 - 1 - 3 = 1 + 3 - 1 - 3 = 0$

$\therefore (x - 1)$ is a factor of p(x).

$p(-3) = (-3)^3 + 3(-3)^2 - (-3) - 3 = - 27 + 27 + 3 - 3 = 0$

$\therefore (x + 3)$ is a factor of p(x).

The three factors of p(x) are (x + 1), (x - 1) and (x - 3)

$\therefore x^3 + 3x^2 - x - 3 = (x + 1) (x - 1) (x + 3)$ .

Find the remainder when $x^3-7x^2-x+6$ is divided by $(x+2)$ .

According to remainder theorem, when

$f(x)$ is divided by

$(x+2)$ , Remainder

$=f(-2)$

$f(x)=x^3-7x^2-x+6$

$f(-2)=(-2)^3-7(-2)^2-(-2)+6$

$=-8-7\times 4+2+6$

$=-8-28+8=-28$

$\therefore$ Remainder

$=-28$

Using factor theorem show that $(x- 1)$ is factor of $4x^3-6x^2+9x-7$ .

According to factor theorem, when

$f(x)$ is divided by

$(x-1), f(1)=0$

$f(x)=4x^3-6x^2+9x-7$

$f(1)=4(1)^3-6(1)^2+9(1)-7$

$=4-6+9-7=0$

$\because f(1)=0, (x-1)$ is a factor of the given polynomial

Find the value of a if $2x^3-6x^2+5ax-9$ leaves the remainder 13 when it is divided by x-2.

Let

$p(x)=2x^3-6x^2+5ax-9$
When p(x) is divided by (x -2) the remainder is p(2).
Given that p(2)=13

$\rightarrow 2(2)^3-6(2)^2+5a(2)-9=13$

$2(8)-6(4)+10a-9=13$

$16-24+10a-9=13$

$10a-17=13$

$10a=30$

$\therefore a=3$

Factorize the following
$2x^2 - 15x + 27$

$2x^2 - 15 x + 27$

Coefficient of

$x^2 = 2$ ; constant term

$= 27$

Their product

$= 2 \times 27 = 54$

Coefficient of

$x = -15$

$\therefore$ product

$= 54$ ; sum

$= -15$

$2x^2 - 15x + 27$

$= 2x^2 - 6x - 9x + 27$

$= 2x(x - 3) - 9 (x - 3)$

$= (x - 3)(2x - 9)$

$\therefore 2x^2 - 15x + 27 = (x - 3)(2x - 9)$

Find the remainder when $x^3+ax^2-3x+a$ is divided by x+a.

Let

$p(x)=x^3+ax^2-3x+a$
When p(x) is divided by (x+a) the remainder is p(-a).

$p(-a)=(-a)^3+a(-a)^2-3(-a)+a=-a^3+a^3+4a=4a$

$\therefore$ The remainder is 4a.

Find the quotient and the remainder when $10-4x+3x^2$ is divided by $x-2$ .

Let us first write the terms of each polynomial in descending order ( or ascending order). thus, the given problem becomes

$93x^2-4x+10) \div (x-2)$
(i)

$\frac{3x^2}{x}=3x$
(ii)

$3x(x-2)=3x^2-6x$
(iii)

$\frac{2x}{2}=2$
(iv)

$2(x-2)=2x-4$

$\therefore$ Quotient

$=3x+2$ and Remainders=14

$(i.e. 3x^2-4x+10=(x-2)(3x+2)+14$ and is in form

$Divided =(Divisor \times quotient +Remainder)$

Find the quotient and the remainder $(4x^3+6x^2-23x-15)\div (3+x)$

Write the given polynomials in ascending or decending order.

$(4x^3+6x^2-23x-15)\div (3+x)$
(i)

$\frac{4x^3}{x}=4x^2$
(ii)

$4x^2(x+3)=4x^3+12x^2$
(iii)

$\frac{-6x^2}{x}=6x$
(iv)

$-6x(x+3)=-6x^2-18x$
(v)

$\frac{-5x}{x}=-5$
(vi)

$-5(x+3)=-5x-15$

$\therefore$ Quotient

$=4x^2-6x-5$
Remainder =0

Determine the value of $p$ if $(x+3)$ is a factor of $x^3-3x^2-px+24$ .

According to factor theorem, when

$f(x)$ is divided by

$(x+3), f(-3)=0$

$f(x)=x^3-3x^2-px+24$

$f(-3)=(-3)^3-3(-3)^2-p(-3)+24$

$\implies 0=-27-27+3p+24$

$\implies 0=3p-30$

$\implies 3p=30$

$\implies p=\dfrac{30}{3}=10$

If the polynomials $2x^3+ax^2+4x-12$ and $x^3+x^2-2x+a$ leave the same remainder when divided by (x-3), find the value of a. Also find the remainder.

Let

$p(x)=2x^3+ax^2+4x-12$ ,

$q(x)=x^3+x^2-2x+a$
When p(x) is divided by (x-3) the remainder is p(3). Now,

$p(3)=2(3)^3+a(3)^2+4(3)-12$

$2(27)+a(9)+12-12$

$54+9a$ ....(1)
When q(x) is divided by (x-3) the remainder is q(3). Now,

$q(3)=2(3)^3+(3)^2+2(3)+a$

$(27)+(9)-6+a$

$30+a$ ......(2)
Given that p(3)=q(3). That is,

$54+9a=30+a$ (By (1) and (2))

$9a-a=30-54$

$8a=-24$

$\therefore a=-\frac{24}{8}=-3$
Substituting a=-3 in p(3), we get

$p(3) =54+9(-3) =54 -27=27$

$\therefore$ The remainder is 27.

Find the remainder when $f(x)=12x^3-13x^2-5x+7$ is divided by $(3x+2).$

$f(x)=12x^3-13x^2-5x+7$
When f(x) is divided by (3x+2) the remainder is

$f\left ( -\frac{2}{3} \right )$ . Now,

$f\left ( -\frac{2}{3} \right )=12\left ( -\frac{2}{3} \right )^3-13\left ( -\frac{2}{3} \right )^2-5\left ( -\frac{2}{3} \right )+7$

$=12\left ( -\frac{8}{27} \right )-13\left ( \frac{4}{9} \right )+\frac{10}{3}+7$

$-\frac{32}{9}-\frac{52}{9}+\frac{10}{3}+7=\frac{9}{9}=1$

$\therefore$ The remainder is 1.

Factorize:
$6x^{2}+11x-10$

$6{x^2} + 11x - 10$

$= 6{x^2} + 15x - 4x - 10$

$= 3x\left( {2x + 5} \right) - 2\left( {2x + 5} \right)$

$= \left( {3x - 2} \right)\left( {2x + 5} \right)$

If $a+b+c=0$ , show that $6\left( { a }^{ 5 }+{ b }^{ 5 }+{ c }^{ 5 } \right) =5\left( { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 } \right) \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right)$ .

We have identically
$\left( 1+ax \right) \left( 1+bx \right) \left( 1+cx \right) =1+px+q{ x }^{ 2 }+r{ x }^{ 3 }$ ,
where $p=a+b+c$ , $q=bc+ca+ab$ , $r=abc$ .
Hence, using the condition given,
$\left( 1+ax \right) \left( 1+bx \right) \left( 1+cx \right) =1+q{ x }^{ 2 }+r{ x }^{ 3 }$ .
Taking logarithms and equating the coefficients of ${ x }^{ n }$ , we have $\dfrac { { \left( -1 \right) }^{ n-1 } }{ n } \left( { a }^{ n }+{ b }^{ n }+{ c }^{ n } \right) =$ coefficient of ${ x }^{ n }$ in the expansion of $\log { \left( 1+q{ x }^{ 2 }+r{ x }^{ 3 } \right) } =$ coefficient of ${ x }^{ n }$ in $\left( q{ x }^{ 2 }+r{ x }^{ 3 } \right) -\dfrac { 1 }{ 2 } { \left( q{ x }^{ 2 }+r{ x }^{ 3 } \right) }^{ 2 }+\dfrac { 1 }{ 3 } { \left( q{ x }^{ 2 }+r{ x }^{ 3 } \right) }^{ 3 }-\cdots$
By putting $n=2,3,5$ we obtain
$-\dfrac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 2 } =q$ , $\dfrac { { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 } }{ 3 } =r$ , $\dfrac { { a }^{ 5 }+{ b }^{ 5 }+{ c }^{ 5 } }{ 5 } =-qr$ ;
whence $\dfrac { { a }^{ 5 }+{ b }^{ 5 }+{ c }^{ 5 } }{ 5 } =\dfrac { { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 } }{ 3 } \cdot \dfrac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 2 }$ ,
and the required result at once follows.
If $a=\beta -\gamma ,b=\gamma -\alpha ,c=\alpha -\beta$ , the given condition is satisfied; hence we have identically for all values of $\alpha ,\beta ,\gamma$
$6\left\{ { \left( \beta -\gamma \right) }^{ 5 }+{ \left( \gamma -\alpha \right) }^{ 5 }+{ \left( \alpha -\beta \right) }^{ 5 } \right\} =5\left\{ { \left( \beta -\gamma \right) }^{ 3 }+{ \left( \gamma -\alpha \right) }^{ 3 }+{ \left( \alpha -\beta \right) }^{ 3 } \right\} \left\{ { \left( \beta -\gamma \right) }^{ 2 }+{ \left( \gamma -\alpha \right) }^{ 2 }+{ \left( \alpha -\beta \right) }^{ 2 } \right\}$
that is,
$6\left\{ \left( \beta -\gamma \right)^{ 5 }+{ \left( \gamma -\alpha \right) }^{ 5 }+{ \left( \alpha -\beta \right) }^{ 5 }\right\}=5\left( \beta -\gamma \right) \left( \gamma -\alpha \right) \left( \alpha -\beta \right) \left( { \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 }-\beta \gamma -\gamma \alpha -\alpha \beta \right)$ ;

$6\left ( a^{5}+b^{5}+c^{5} \right )=5\left ( {a^{3}+b^{3}+c^{3}}\right )\left (a^{2}+b^{2}+c^{2} \right )$

Hence proved.

Prove that ${ \left( a+b \right) }^{ 5 }-{ a }^{ 5 }-{ b }^{ 5 }=5ab\left( a+b \right) \left( { a }^{ 2 }+ab+{ b }^{ 2 } \right)$ .

L.H.S

$=(a+b)^5-a^5-b^5$

$=a^5+5ab^4+10a^2b^3+10a^3b^2+5a^4b+b^5-a^5-b^5$

$=5ab^4+10a^2b^3+10a^3b^2+5a^4b$

$=5ab(b^3+2ab^2+2a^{2}b+a^3)$

$=5ab\{(a+b)(a^2-ab+b^2)+2ab(a+b)\}$

$=5ab(a+b)(a^2-ab+b^2+2ab)$

$=5ab(a+b)(a^2+ab+b^2)$

$=$ R.H.S

Alternate Method

Denote the expression on the left by

$E$ ; then

$E$ is a function of

$a$ which vanishes when

$a=0$ ; hence

$a$ is a factor of

$E$ ; similarly

$b$ is a factor of

$E$ . Again

$E$ vanishes when

$a=-b$ , that is

$a+b$ is a factor of

$E$ ; and therefore

$E$ contains

$ab\left( a+b \right)$ as a factor. The remaining factor must be of two dimensions, and, since it is symmetrical with respect to

$a$ and

$b$ , it must be of the form

$A{ a }^{ 2 }+Bab+A{ b }^{ 2 }$ ; thus

${ \left( a+b \right) }^{ 5 }-{ a }^{ 5 }-{ b }^{ 5 }=ab\left( a+b \right) \left( A{ a }^{ 2 }+Bab+A{ b }^{ 2 } \right)$ ,
where

$A$ and

$B$ are independent of

$a$ and

$b$ .
Putting

$a=1$ ,

$b=1$ , we have

$15=2A+B$ ;
putting

$a=2$ ,

$b=-1$ , we have

$15=5A-2B$ ;
Hence

$A=5$ ,

$B=5$ ; and thus the required result at once follows.

Show that ${ \left( x+y \right) }^{ 7 }-{ x }^{ 7 }-{ y }^{ 7 }=7xy\left( x+y \right) { \left( { x }^{ 2 }+xy+{ y }^{ 2 } \right) }^{ 2 }$ .

The expression,

$E$ , on the left vanishes when

$x=0$ ,

$y=0$ ,

$x+y=0$ ; hence it must contain

$xy\left( x+y \right)$ as a factor.
Putting

$x=\omega y$ , we have

$E=\left\{ { \left( 1+\omega \right) }^{ 7 }-{ \omega }^{ 7 }-1 \right\} { y }^{ 7 }=\left\{ { \left( -{ \omega }^{ 2 } \right) }^{ 7 }-{ \omega }^{ 7 }-1 \right\} { y }^{ 7 }$

$=\left( -{ \omega }^{ 2 }-\omega -1 \right) { y }^{ 7 }=0$ ;
hence

$E$ contains

$x-\omega y$ as a factor; and similarly we may show that it contains

$x-{ \omega }^{ 2 }y$ as a factor; that is,

$E$ is divisible by

$\left( x-\omega y \right) \left( x-{ \omega }^{ 2 }y \right)$ , or

${ x }^{ 2 }+xy+{ y }^{ 2 }$ .
Further,

$E$ being of degree seven, and

$xy\left( x+y \right) \left( { x }^{ 2 }+xy+{ y }^{ 2 } \right)$ of five dimensions, the remaining factor must e of the form

$A\left( { x }^{ 2 }+{ y }^{ 2 } \right) +Bxy$ ; thus

${ \left( x+y \right) }^{ 7 }-{ x }^{ 7 }-{ y }^{ 7 }=xy\left( x+y \right) \left( { x }^{ 2 }+xy+{ y }^{ 2 } \right) \left( A{ x }^{ 2 }+Bxy+A{ y }^{ 2 } \right)$ .
Putting

$x=1$ ,

$y=1$ , we have

$21=2A+B$ ;
pitting

$x=2$ ,

$y=-1$ , we have

$21=5A-2B$ ;
whence

$A=7$ ,

$B=7$ ;

$\therefore { \left( x+y \right) }^{ 7 }-{ x }^{ 7 }-{ y }^{ 7 }=7xy\left( x+y \right) { \left( { x }^{ 2 }+xy+{ y }^{ 2 } \right) }^{ 2 }$ .

Factorise : ${y^2} - 4y + 3$

$y^2-4y+3$

$=y^2-y-3y+3$

$=y\left( y-1 \right)-3 \left( y-1 \right)$

$=\left( y-1 \right) \left( y-3 \right)$

Hence, the answer is

$\left( y-1 \right) \left( y-3 \right).$

Factorise: $2{x^2} + 11x - 21$

$2x^2+11x-21$

$\Rightarrow 2x^2+14x-3x-21$

$\Rightarrow 2x\left( { x+7 } \right) -3\left( { x+7 } \right)$

$\Rightarrow \left( { 2x-3 } \right) \left( { x+7 } \right)$

Hence, the answer is

$\left( { 2x-3 } \right) \left( { x+7 } \right).$

The expression $2{ x }^{ 3 }+b{ x }^{ 2 }-cx+d\quad$ leaves the same remainder, when divided by $x+1$ or $x-2$ or $2x-1$ . Find $b$ and $c$ .

$\left(x+1\right)\left(x-2\right)\left(2x-1\right)=2{x}^{3}+b{x}^{2}-cx+d$

$\Rightarrow\,\left({x}^{2}-2x+x-2\right)\left(2x-1\right)=2{x}^{3}+b{x}^{2}-cx+d$

$\Rightarrow\,\left({x}^{2}-x-2\right)\left(2x-1\right)=2{x}^{3}+b{x}^{2}-cx+d$

$\Rightarrow\,2{x}^{3}-{x}^{2}-2{x}^{2}+x-4x+2=2{x}^{3}+b{x}^{2}-cx+d$

$\Rightarrow\,2{x}^{3}-3{x}^{2}-3x+2=2{x}^{3}+b{x}^{2}-cx+d$

Comparing the coefficients, we get

$b=-3,c=3,d=2$

$\therefore\,b=-3,c=3,d=2$

Factorize : $x^2-24x-180$

$x^2-24x-180$

$=x^2-30x+6x-180$

$=x(x-30)+6(x-30)$

$=(x-30)(x+6)$

Factorise : $6 - x - {x^2}$

$\textbf{Step 1: Factorizing}$

$\text{We have }6-x-x^2$

$= 6-3x+2x-x^2$

$= (6-3x)+(2x-x^2)$

$= 3(2-x)+x(2-x)$

$= (3+x)(2-x)$

$\mathbf{\textbf{Hence, }6-x-x^2 \textbf{ can be factorised to }(3+x)(2-x)}$

Factorise : ${x^2} - 11x - 80$

$x^2-11x-80$

$=x^2-16x+5x-80$

$=x\left( x-16 \right) +5\left( x-16 \right)$

$=\left( x-16 \right) \left( x+5 \right)$

Hence, the answer is

$\left( x-16 \right) \left( x+5 \right).$

Simplify :
i) $\cfrac{-14x^8y^5+21x^{10}y-28x^7y^6}{7x^7y^8}$
ii) $\cfrac{15a^4x^8-30a^{7}x^5-45a^6x^6}{20a^{14}x^5}$
iii) $\cfrac{-60x^4a^5-75x^{3}a^6+8x^5a^4}{-20x^8a^4}$

$\cfrac{-14x^8y^5+21x^{10}y-28x^7y^6}{7x^7y^8}$

$=-\cfrac{2x}{y^3}+\cfrac{3x^3}{y^7}-\cfrac{4}{y^2}$

$=\cfrac{1}{y^2}\left(\cfrac{-2x}{y}+\cfrac{3x^3}{y^5}-\small{4}\right)$

ii)

$\cfrac {15a^4x^8-30a^{7}x^5-45a^6x^6}{20a^{14}x^5}$

$=\cfrac {3x^3}{4a^{10}}-\cfrac{6}{4a^7}-\cfrac{9x}{4a^8}$

$=\cfrac {3}{4a^7}\left(\cfrac{x^3}{a^3}-\small{2}-\cfrac{3x}{a}\right)$

iii)

$\cfrac{-60x^4a^5-75x^{3}a^6+8x^5a^4}{-20x^8a^4}$

$=\left(\cfrac{3a}{x^4}+\cfrac{15a^2}{4x^5}-\cfrac{2}{5x^3}\right)$

Factorise : ${z^2} - 32z - 105$

$z^2-32z-105$

$=z^2-35z+3z-105$

$=z\left( z-35 \right) +3\left( z-35 \right)$

$=\left( z+3 \right) \left( z-35 \right)$

Hence, the answer is

$\left( z+3 \right) \left( z-35 \right).$

Factorise : $6{x^2} + 17x + 12$

Factorise :

$6x^2+17x+12$ using common method.

$6x^2+17x+12$

The coefficient of

$x^2$ is

$6$ and constant is

$12.$ The product of two coefficients

$=6\times 12=72$

Now, the coefficient of middle term

$=17$

We know, that the factors of

$72$ such that the sum of these factors is

$17$ are

$8\;\&\;9 \left(8\times 9=72,\; 8+9=17\right)$

So,

$6x^2+8x+9x+12$

$\Rightarrow 2x\left( { 3x+4 } \right) +3\left( { 3x+4 } \right)$

$\Rightarrow \left( { 2x+3 } \right) \left( { 3x+4 } \right).$

Hence, the answer is

$\left( { 2x+3 } \right) \left( { 3x+4 } \right).$

Resolve into factors of $4{x^2} - 25{y^2} + 2x + 5y$

$(2x)^2 - (5y)^2 +2x+5y$

$=(2x-5y)(2x+5y)+1(2x+5y)$

$=(2x+5y)(2x-5y+1)$

Factorise
$x^3 - 2x^2 - x + 2$

$x^3 - 2x^2 - x + 2$

$=x^2(x - 2)- 1(x - 2)$

$=(x^2-1)(x-2)$

$=(x+1)(x-1)(x-2)$

Express in standard form
$(x-a) (x-b)$

$\left( x-a \right) \left( x-b \right)$

$=x^2-\left( a+b \right) x+ab$

Hence, the answer is

$x^2-\left( a+b \right) x+ab.$

Factorise: $6{x^2} + 11x - 10$

$6x^2+11x-10$

$\Rightarrow 6x^2+15x-4x-10$

$\Rightarrow 3x\left( { 2x+5 } \right) -2\left( { 2x+5 } \right)$

$\Rightarrow \left( { 3x-2 } \right) \left( { 2x+5 } \right)$

Hence, the answer is

$\left( { 3x-2 } \right) \left( { 2x+5 } \right).$

Factorise: $18{x^2} + 3x - 10$

$18x^2+3x-10$

$\Rightarrow 18x^2+15x-12x-10$

$\Rightarrow 3x\left( { 6x+5 } \right) -2\left( { 6x+5 } \right)$

$\Rightarrow \left( { 3x-2 } \right) \left( { 6x+5 } \right)$

Hence, the answer is

$\left( { 3x-2 } \right) \left( { 6x+5 } \right).$

Factorize $y^2-( a+ b) y + ab$

$\textbf{Step 1: Open the bracket and simplify.}$

$y^2-(a+b)y + ab=$

$y^2-ay-by+ab$

$= y(y-a)-b(y-a)$

$= (y-a) (y-b)$

$\textbf{Hence, (y-a) (y-b) are the two factors.}$

Factorize:
$x^2+3\sqrt{2}x+4=0$

$x^2+3\sqrt2 x+4=0$

$\implies x^2+2\sqrt2 x +\sqrt2 x+4=0$

$\implies x(x+2\sqrt2)+\sqrt2(x+2\sqrt2)=0$

$\implies (x+\sqrt2)(x+2\sqrt2)=0$

Divide P(x) by g(x)
P(x) $={x^4} - 3{x^2} - 4$
g(x) $= x + 2$

$p\left( x \right) =x^4-3x^2-4$

$g\left( x \right) =x+2$

Since divisor, we can divide the polynomial by synthetic division.

$\therefore$ Quotient

$:x^3-2x^2+x-2.$

Hence, the answer is

$x^3-2x^2+x-2.$

Factorize:
$(i)\ \ \ {x^2} + 9x + 18$
$(ii)\ \ 6{x^2} + 7x - 3$
$(iii)\ 2{x^2} - 7x - 15$
$(iv)\ \ 84 - 2r - 2{r^2}$

$(i).$

$x^2+9x+18 = x^2+3x+6x+18$

$=x{ \left( x+3 \right) }+6{ \left( x+3 \right) }$

$={ \left( x+3 \right) }{ \left( x+6 \right) }$

$(ii).$

$6x^2+7x-3= 6x^2+9x-2x-3$

$=3x{ \left( 2x+3 \right) }-1{ \left( 2x+3 \right) }$

$={ \left( 3x-1 \right) }{ \left( 2x+3 \right) }$

$(iii).$

$2x^2-7x-15=2x^2-10x+3x-15$

$=2x{ \left( x-5 \right) }+3{ \left( x-5 \right) }$

$={ \left( x-5 \right) }{ \left( 2x+3 \right) }$

$(iv).$

$84-2r-2r^2=-\left( 2r^2+2r-84 \right)$

$=-\left(2r^2+14r+12r-84\right)$

$=-\left( 2r\left( r+7 \right) -12\left( r+7 \right) \right)$

$=-\left( r+7 \right) \left( 2r-12 \right)$

If you divide $f\left( x \right) = {x^3} + 3{x^2} - kx - 12$ by $\left( {x - 3} \right)$ , you get remainder $30$ . Find $k$ and also find the quotient.

$f(x)=x^3+3x^2-kx-12$ $

According to remainder theorem

$f(3)=30$

$\implies (3)^2+3(3)^3-k(3)-12=30$

$\implies 27+27-3k-12=30 \implies 3k=54-12-30=12$

$\implies k=4$

$f(x)=x^3+3x^2-4x-12$

$\implies \cfrac{x^3+3x^2-4x-12}{x-3} = x^2+6x+14$

So, Quotient

$=x^2+6x+14$

Simplify $\dfrac {{x}^{2}}{9}-\dfrac {{y}^{2}}{4}$

$\begin{array}{l} \dfrac { { { x^{ 2 } } } }{ 9 } -\dfrac { { { y^{ 2 } } } }{ 4 } \\ ={ \left( { \dfrac { x }{ 3 } } \right) ^{ 2 } }-{ \left( { \dfrac { y }{ 2 } } \right) ^{ 2 } } \\ =\left( { \dfrac { x }{ 3 } -\dfrac { y }{ 2 } } \right) \left( { \dfrac { x }{ 3 } +\dfrac { x }{ 2 } } \right) \, \, \, \left[ { { a^{ 2 } }-{ b^{ 2 } }=\left( { a-b } \right) \left( { a+b } \right) } \right] \end{array}$

Let $p(x) = x^{4}-3x^{2}+2x + 5$ . Find the remainder when $p(x)$ is divided by $(x-1)$ .

We know that the remainder when a polynomial

$p(x)$ is divided by

$(x - a)$ is

$p(a)$ .

Given that,

$p(x) = x^4 -3x^2 +2x +5$

Divisor

$=(x-1)$

So, The Remainder

$= p(1)$

$= (1)^4 - 3(1)^2 + 2(1) + 5$

$= 1 - 3 +2 +5$

$= 5$

factorise
$9x^{2}+12xy$

$9{ x }^{ 2 }+12xy$

$=\left( 3 \right) \left( 3x^2+4xy \right)$

$=\left( 3x \right) \left( 3x+4y \right)$

Divide $({x^2} + 5x + 6 )$ by $( x + 2)$ .

We divide the given polynomial

$x^2+5x+6$ by the monomial

$x+2$ as shown below:

$\dfrac { x^{ 2 }+5x+6 }{ x+2 } \\ =\dfrac { x^{ 2 }+3x+2x+6 }{ x+2 } \\ =\dfrac { x(x+3)+2(x+3) }{ x+2 } \\ =\dfrac { (x+2)(x+3) }{ x+2 } \\ =x+3$

Hence,

$\dfrac { x^{ 2 }+5x+6 }{ x+2 } =x+3$ .

Factorize: $2x^{2}-7x-15$

Simplifying the equation,

$2{x^2} - 7x - 15$

$= 2{x^2} - 10x + 3x - 15$

$= 2x\left( {x - 5} \right) + 3\left( {x - 5} \right)$

$= \left( {2x + 3} \right)\left( {x - 5} \right)$

Factorise: $15x+5$

Simplifying $15x + 5$ .

$15x + 5 = 5\left( {3x + 1} \right)$

Factorize the following: $8{x}^{2}-14x+8x-14$

Simplifying $8{x^2} - 14x + 8x - 14$ .

$8{x^2} - 14x + 8x - 14$

$= 2x\left( {4x - 7} \right) + 2\left( {4x - 7} \right)$

$= \left( {2x + 2} \right)\left( {4x - 7} \right)$

$= 2\left( {x + 1} \right)\left( {4x - 7} \right)$

Factorise ${X}^{2}-7X+10=0$

Simplifying ${X^2} - 7X + 10 = 0$

${X^2} - 5X - 2X + 10 = 0$

$X\left( {X - 5} \right) - 2\left( {X - 5} \right) = 0$

$\left( {X - 5} \right)\left( {X - 2} \right) = 0$

Factorize the following:
$(16-81{x}^{2})$

Simplifying $16 - 81{x^2}$

$16 - 81{x^2}$

$= {\left( 4 \right)^2} - {\left( {9x} \right)^2}$

$= \left( {4 + 9x} \right)\left( {4 - 9x} \right)$

Find the remainder when ${x^2}\, - \,a{x^2}\, + \,8x\, + \,a$ is divided by $(x - a)$

Here the concept of reminder theorem can be used.

this theorem states that the reminder when f(x) is divided by (x-a) is f(a).

let

$f(x)$ =

$x^2$ -

$ax^2$ +

$8x$ +

$a$

We can find f(a) by substituting '

$a$ ' in the replace of

$x$ in the above equation.

$f(a)$ =

$a^2$ -

$a \times a^2$ +

$8a$ +

$a$

$f(a)$ =

$a^2$ -

$a^3$ +

$9a$

Hence the reminder is

$a^2$ -

$a^3$ +

$9a$

Divide $p(x) = 2x^3 - 11x^2 + 19x - 10$ by g(x) = 2x -Find the quotient and remainder.

Given,

$p(x)=2x^3-11x^2+19x-10$

$g(x)=2x-5$

$\underline {\ \ \ x^2-3x+2\ \ \ \ \quad \quad \quad \quad }$

$2x-5)2x^3-11x^2+19x-10$

$2x^3-5x^2$

$(-)$

$(+)$

____________________

$-6x^2+19x$

$-6x^2+15x$

$(+)$

$(-)$

____________________

$+4x-10$

$(-)$

$(+)$

_________________

$0$

$\Rightarrow 2x^3-11x^2+19x-10=(2x-5)(x^2-3x+2)$

$\therefore$ Quotient

$=x^2-3x+2$ and remainder

$=0$ .

Factorize the following: $3x-{3}^{2}$

Simplifying $3x - {3^2}$

$=3x - {3^2}$

Taking 3 common from both the terms:

$= 3\left( {x - 3} \right)$

Factorize:
$3x^{2}-14x+18$

$3{x^2} - 14x + 18$

Factor with discriminant method,

$x = \dfrac{{14 \pm \sqrt {{{\left( { - 14} \right)}^2} - 4 \times 3 \times 18} }}{{2 \times 3}}$

$x = \dfrac{{14 \pm \sqrt {20} i}}{6}$

$= \left( {x - \left( {\dfrac{{14 - \sqrt {20} i}}{6}} \right)} \right)\left( {x - \left( {\dfrac{{14 + \sqrt {20} i}}{6}} \right)} \right)$

Factorize $y^{2}-10y+25$

Simplifying ${y^2} - 10y + 25$ .

${y^2} - 10y + 25$

$= {y^2} - 5y - 5y + 25$

$= y\left( {y - 5} \right) - 5\left( {y - 5} \right)$

$= \left( {y - 5} \right)\left( {y - 5} \right)$

Factorise $(i) x^{2}+3\sqrt{3}x+6$
$(ii) 6x^{2}+\sqrt{5}x-10$

(i)

$x^2+3\sqrt 3x+6$

This can be written as

$\Rightarrow x^2+\sqrt 3x+2\sqrt 3x+6$

$\Rightarrow x(x+\sqrt 3)+2\sqrt 3(x+\sqrt 3)$

$\Rightarrow (x+\sqrt 3)(x+2\sqrt 3)$

Therefore,

$\Rightarrow x^2+3\sqrt3x+6= (x+\sqrt 3)(x+2\sqrt 3)$

(ii)

$6x^2+\sqrt 5x-10$

This can be written as

$\Rightarrow 6(x^2+\dfrac{\sqrt 5}{6}x-\dfrac{5}{3})$

$\Rightarrow 6(x^2-\dfrac{\sqrt 5}{2}x+\dfrac{2\sqrt 5}{3}x-\dfrac{5}{3})$

$\Rightarrow 6[x(x-\dfrac{\sqrt 5}{2})+\dfrac{2\sqrt 5}{3}(x-\dfrac{\sqrt 5}{2})]$

$\Rightarrow 6(x+\dfrac{2\sqrt 5}{3})(x-\dfrac{\sqrt 5}{2})$

Therefore,

$\Rightarrow 6x^2+\sqrt5x-10= 6(x+\dfrac{2\sqrt 5}{3})(x-\dfrac{\sqrt 5}{2})$

Factorise : $x^{2}+3\sqrt{3}x+6$

$x^2+3\sqrt 3x+6$

This can be written as

$\Rightarrow x^2+\sqrt 3x+2\sqrt 3x+6$

$\Rightarrow x(x+\sqrt 3)+2\sqrt 3(x+\sqrt 3)$

$\Rightarrow (x+\sqrt 3)(x+2\sqrt 3)$

Therefore,

$\Rightarrow x^2+3\sqrt3x+6= (x+\sqrt 3)(x+2\sqrt 3)$

Find the remainder when the polynomial $4y^3 - 3y^2 - 5y + 1$ is divided by $2y + 3$ .

$2y+3$

$)4y^3-3y^2-5y+1($

$2y^2-\dfrac{9}{2}y+\dfrac{17}{4}$

$4y^3+6y^2$

$(-)$

----------------------------------------

$-9y^2-5y$

$-9y^2-\dfrac{27}{2}y$

$(-)$

----------------------------------------

$\dfrac{17}{2}y+1$

$\dfrac{17}{2}y+\dfrac{51}{4}$

$(-)$

---------------------------------------------

$-\dfrac{47}{4}$

$\dfrac{4y^3-3y^2-5y+1}{2y+3} = 2y^2-\dfrac{9}{2}y+\dfrac{17}{4} - \dfrac {47}{4(2y+3)}$

Express in the product of two polynomial $x^4+y^4+x^2y^2$ .

Let us add and subtract

$x^2y^2$ in the given expression

$x^4+y^4+x^2y^2$ and then solve as shown below:

$(x^{ 4 }+y^{ 4 }+x^{ 2 }y^{ 2 }+x^{ 2 }y^{ 2 })-x^{ 2 }y^{ 2 }\\ =[(x^{ 2 })^{ 2 }+(y^{ 2 })^{ 2 }+2x^{ 2 }y^{ 2 }]-x^{ 2 }y^{ 2 }\\ =[(x^{ 2 }+y^{ 2 })^{ 2 }]-x^{ 2 }y^{ 2 }\quad \quad \quad \quad \left( \because \quad (a+b)^{ 2 }=a^{ 2 }+b^{ 2 }+2ab \right) \\ =(x^{ 2 }+y^{ 2 })^{ 2 }-(xy)^{ 2 }\\ =(x^{ 2 }+y^{ 2 }-xy)(x^{ 2 }+y^{ 2 }+xy)\quad \quad (\because \quad a^{ 2 }-b^{ 2 }=(a-b)(a+b))$

Hence,

$x^{ 4 }+y^{ 4 }+x^{ 2 }y^{ 2 }=(x^{ 2 }+y^{ 2 }-xy)(x^{ 2 }+y^{ 2 }+xy)$ .

Factorize:
$(1){ \quad x }^{ 3 }-{ 2x }^{ 2 }-x+2\\ (2)\quad { x }^{ 3 }-{ 3x }^{ 2 }-9x-5\\ (3)\quad { x }^{ 3 }+{ 13x }^{ 2 }+32x+20\\ (4)\quad { y }^{ 3 }+{ y }^{ 2 }-y-1$

(1)

$x^{3}-2x^{2}-x+2$

$=x^{2}(x-2)-1(x-2)$

$=(x-2)(x^{2}-1)$

$=(x-2)(x-1)(x+1)$

(2)

$x^{3}-3x^{2}-9x-5$

$=x^{3}+(x^{2}-4x^{2})-(4x+5x)-5$

$=(x^{3}+x^{2})-(4x^{2}+4x)-(5x+5)$

$=x^{2}(x+1)-4x(x+1)-5(x+1)$

$=(x^{2}-4x-5)(x+1)-5(x+1)$

$=(x^{2}-4x-5)(x+1)$

$=(x^{2}-5x+x-5)(x+1)$

$=(x-5)(x+1)^{2}$

(3)

$x^{3}+13x^{2}+32x+20$

$=(x^{3}+x^{2})+(12x^{2}+12x)+(20x+20)$

$=(x^{2}+12x+20)(x+1)$

$=(x+1)(x+2)(x+10)$

(4)

$y^{3}+y^{2}-y-1=y^{2}(y+1)-1(y+1)$

$=(y^{2}-1)(y+1)$

$=(y-1)(y+1)^{2}$

Find the remainder when $x^4+4x^3-5x^2-6x+7$ is divisible by $x-3$ .

We divide

$x^4+4x^3-5x^2-6x+7$ by

$(x-3)$ as shown in the above image:

From the division, we observe that the quotient is

$x^3+7x^2+16x+42$ and the remainder is

$133$ .

Hence, the remainder is

$133$ .

1231014_1074148_ans_ec2bb7f3465e48e9b87658d8ef85c084.png

Factorise $x^2-3x+2$ .

In the given polynomial

$x^2-3x+2$ ,

The first term is

$x^2$ and its coefficient is

$1$ .

The middle term is

$-3x$ and its coefficient is

$-3$ .

The last term is a constant term

$2$ .

Multiply the coefficient of the first term by the constant

$2\times 1=2$ .

We now find the factor of

$2$ whose sum equals the coefficient of the middle term, which is

$-3$ and then factorize the polynomial

$x^2-3x+2$ as shown below:

$x^{ 2 }-3x+2\\ =x^{ 2 }-x-2x+2\\ =x(x-1)-2(x-1)\\ =(x-1)(x-2)$

Hence,

$x^{ 2 }-3x+2=(x-1)(x-2)$ .

Divide :
${2x}^{3}-{5x}^{2}-{3x}\ by (x-1)$ ?

$\begin{array}{c}2{x^3} - 5{x^2} - 3x = x\left( {2{x^2} - 5x - 3} \right)\\ = x\left( {2{x^2} - 6x + x - 3} \right)\\ = x\left( {2x\left( {x - 3} \right) + 1\left( {x - 3} \right)} \right)\\ = x\left( {2x + 1} \right)\left( {x - 3} \right)\end{array}$

So,

$2{x^3} - 5{x^2} - 3x$ by x-1 is ,

$\frac{{x\left( {2x + 1} \right)\left( {x - 3} \right)}}{{x - 1}}$

Factorise the following:
$27a^{3} - 8b^{3} - 54a^{2}b + 36ab^{2}$

$27a^{3} - 8b^{3} - 54a^{2}b + 36ab^{2}$

$= (3a)^{3} - (2b)^{3} - 3(3a)(2b) [3a - 2b]$

$= (3a - 2b)^{3}$

$[\because (a - b)^{3} = a^{3} - b^{3} - 3ab (a - b)]$

Find the remainder when $x^4+4x^3-5x^2-6x+7$ is divided by $x+1$ .

We have,

$P(x)=x^{4}+4x^{3}-5x^{2}-6x+7$

here Divisor is

$x+1$

$\qquad \qquad { x }^{ 3 }+{ 3x }^{ 2 }-8x+1\\ (x+1)\sqrt { { x }^{ 4 }+{ 4x }^{ 3 }-{ 5x }^{ 2 }-6x+7\\ { x }^{ 3 }+{ x }^{ 3 } } \\ \qquad \quad \quad -\quad \quad -\\ \qquad \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \qquad \qquad \qquad { 3x }^{ 3 }-{ 5x }^{ 2 }\\ \qquad \qquad \qquad { 3x }^{ 3 }+{ 3x }^{ 2 }\\ \qquad \qquad \qquad -\quad \quad -\\ \qquad \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \qquad \qquad \qquad \qquad -{ 8x }^{ 2 }-6x\\ \qquad \qquad \qquad \qquad -{ 8x }^{ 2 }-8x\\ \qquad \qquad \qquad \qquad +\quad \quad +\\ \qquad \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \quad \qquad \qquad \qquad \quad 2x+7\\ \qquad \qquad \qquad \qquad \qquad \quad 2x+2\\ \qquad \qquad \qquad \qquad \qquad -\quad -\\ \qquad \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 5$

We get

$5$ as remainder.

Find the remainder when $x^4+4x^3-5x^2-6x+7$ is divided by $x-2$ .

We divide

$x^4+4x^3-5x^2-6x+7$ by

$(x-2)$ as shown in the above image:

From the division, we observe that the quotient is

$x^3+6x^2+7x+8$ and the remainder is

$23$ .

Hence, the remainder is

$23$ .

1231017_1074154_ans_c9883a455f0f4673bf98eaec94d990e2.png

Find the remainder when $x^3-ax^2+6x-a$ is divided by $x-a$ .

We divide

$x^3-ax^2+6x-a$ by

$(x-a)$ as shown in the above image:

From the division, we observe that the quotient is

$x^2+6$ and the remainder is

$5a$ .

Hence, the remainder is

$5a$ .

1232059_1075288_ans_93b8c98ba38b48d3ba54bb7b427e5567.jpg

Find the remainder when $x^4+4x^3-5x^2-6x+7$ is divided by $x+2$ .

We divide

$x^4+4x^3-5x^2-6x+7$ by

$(x+2)$ as shown in the above image:

From the division, we observe that the quotient is

$x^3+2x^2-7x+8$ and the remainder is

$-9$ .

Hence, the remainder is

$-9$ .

1231020_1074157_ans_757c0fe2ba854ec1abd12436f597f7ce.png

Find the remainder when $x^4+4x^3-5x^2-6x+7$ is divided by $x-1$ .

We divide

$x^4+4x^3-5x^2-6x+7$ by

$(x-1)$ as shown in the above image:

From the division, we observe that the quotient is

$x^3+5x^2-6$ and the remainder is

$1$ .

Hence, the remainder is

$1$ .

1231023_1074159_ans_7286b11cc4c74e3c8999c9b07be39f9d.png

Solve:
$x^2+9x+18$ .

In the given polynomial

$x^2+9x+18$ ,

The first term is

$x^2$ and its coefficient is

$1$ .

The middle term is

$9x$ and its coefficient is

$9$ .

The last term is a constant term

$18$ .

Multiply the coefficient of the first term by the constant

$1\times 18=18$ .

We now find the factor of

$18$ whose sum equals the coefficient of the middle term, which is

$9$ and then factorize the polynomial

$x^2+9x+18$ as shown below:

$x^{ 2 }+9x+18\\ =x^{ 2 }+3x+6x+18\\ =x(x+3)+6(x+3)\\ =(x+3)(x+6)$

Hence,

$x^{ 2 }+9x+18=(x+3)(x+6)$ .

Find the remainder:
$\left( {{x^3} - 4{x^2} + 7x - 2} \right) \div \left( {x - 2} \right)$ .

$(x^3 -4x^2+7cx-2)\div (x-2)$

$[\because \ x^3=x.x^2]$

$[\because \ -2x^2=x.(-2x)]$

$[\because \ 3x^x.(3)]$

1420330_1078364_ans_874d970176274dcdb254d86889594463.png

Evaluate:
$\dfrac{1}{2}x^2-3x+4$ .

$\dfrac { 1 }{ 2 } { x }^{ 2 }-3x+4$

$=\dfrac { 1 }{ 2 } { x }^{ 2 }-2x-x+4$

$=1/2x\left( x-4 \right) -1\left( x-4 \right)$

$=\left( x-4 \right) \left( 1/2x-1 \right)$

Factorise $a^3-36a$ .

$\begin{array}{l} a\left( { { a^{ 2 } }-36 } \right) \\ a\left( { a-b } \right) \left( { a+b } \right) \end{array}$

Factorise
$12x +15$

$=12x+15$

$=3×4×x+3×5$

$=3(4x+5)$

$3x^2-x-4$ .

In the given polynomial

$3x^2-x-4$ ,

The first term is

$3x^2$ and its coefficient is

$3$ .

The middle term is

$-x$ and its coefficient is

$-1$ .

The last term is a constant term

$-4$ .

Multiply the coefficient of the first term by the constant

$3\times -4=-12$ .

We now find the factor of

$-12$ whose sum equals the coefficient of the middle term, which is

$-1$ and then factorize the polynomial

$3x^2-x-4$ as shown below:

$3x^{ 2 }-x-4\\ =3x^{ 2 }+3x-4x-4\\ =3x(x+1)-4(x+1)\\ =(3x-4)(x+1)$

Hence,

$3x^{ 2 }-x-4=(3x-4)(x+1)$ .

Solve:
$5{x^2} - 22x - 15 = 0$

By using middle term split method,

$5{ x }^{ 2 }-22x-15=0$

$\Rightarrow 5{ x }^{ 2 }-25x+3x-15=0$

$\Rightarrow 5x(x -5)+3(x-5)=0$

$\Rightarrow (x-5)(5x+3)=0$

Either

$x-5=0 \Rightarrow x = 5$ or

$5x+3=0 \Rightarrow x=-\dfrac{3}{5}$

Factorize - ${x^2} + \dfrac{1}{{{x^2}}} - 2x - \dfrac{2}{x} + 2$

We have,

${{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2x-\dfrac{2}{x}+2$

${{x}^{2}}+\dfrac{1}{{{x}^{2}}}+2-2\left( x+\dfrac{1}{x} \right)$

${{\left( x+\dfrac{1}{x} \right)}^{2}}-2\left( x+\dfrac{1}{x} \right)$

$\left( x+\dfrac{1}{x} \right)\left( x+\dfrac{1}{x}-2 \right)$

Hence, this is the answer.

Give possible expressions for the length and breadth of the following rectangles, in which their areas are given:
Area: $25{a}^{2}-35a+12$

$=25a^2-35a+12$

$=25a^2-25a-12a+12$

$=25a(a-1)-12(a-1)$

$=(25a-12)(a-1)$

Area

$=l.b =$ length

$\times$ breath

$l=25a-12,b=a-1$

$l=a-1,b=25a-12$

$4x^2+4\sqrt{3}x+3=0$ .

In the given quadratic polynomial

$4x^{ 2 }+4\sqrt { 3 } x+3=0$ ,

The first term is

$4x^2$ and its coefficient is

$4$ .

The middle term is

$4\sqrt { 3 } x$ and its coefficient is

$4\sqrt { 3 }$ .

The last term is a constant term

$3$ .

Multiply the coefficient of the first term by the constant

$4\times 3=12$ .

We now find the factor of

$12$ whose sum equals the coefficient of the middle term, which is

$4\sqrt { 3 }$ and then factorize the quadratic polynomial

$4x^{ 2 }+4\sqrt { 3 } x+3=0$ as shown below:

$4x^{ 2 }+4\sqrt { 3 } x+3=0\\ \Rightarrow 4x^{ 2 }+2\sqrt { 3 } x+2\sqrt { 3 } x+3=0\\ \Rightarrow 2x(2x+\sqrt { 3 } )+\sqrt { 3 } (2x+\sqrt { 3 } )=0\\ \Rightarrow (2x+\sqrt { 3 } )(2x+\sqrt { 3 } )=0\\ \Rightarrow (2x+\sqrt { 3 } )=0,\quad (2x+\sqrt { 3 } )=0\\ \Rightarrow 2x=-\sqrt { 3 } ,\quad 2x=-\sqrt { 3 } \\ \Rightarrow x=-\dfrac { \sqrt { 3 } }{ 2 } ,\quad x=-\dfrac { \sqrt { 3 } }{ 2 }$

Hence,

$x=-\dfrac { \sqrt { 3 } }{ 2 } ,-\dfrac { \sqrt { 3 } }{ 2 }$ .

Factorise:
$a^2 + 12a+32$

In the given polynomial

$a^2+12a+32$ ,

The first term is

$a^2$ and its coefficient is

$1$ .

The middle term is

$12a$ and its coefficient is

$12$ .

The last term is a constant term

$32$ .

Multiply the coefficient of the first term by the constant

$1\times 32=32$ .

We now find the factor of

$32$ whose sum equals the coefficient of the middle term, which is

$12$ and then factorize the polynomial

$a^2+12a+32$ as shown below:

$a^{ 2 }+12a+32\\ =a^{ 2 }+4a+8a+32\\ =a(a+4)+8(a+4)\\ =(a+4)(a+8)$

Hence,

$a^{ 2 }+12a+32=(a+4)(a+8)$ .

Factorise:
$z^2+13z+40$

In the given polynomial

$z^2+13z+40$ ,

The first term is

$z^2$ and its coefficient is

$1$ .

The middle term is

$13z$ and its coefficient is

$13$ .

The last term is a constant term

$40$ .

Multiply the coefficient of the first term by the constant

$1\times 40=40$ .

We now find the factor of

$40$ whose sum equals the coefficient of the middle term, which is

$13$ and then factorize the polynomial

$z^2+13z+40$ as shown below:

$z^{ 2 }+13z+40\\ =z^{ 2 }+8z+5z+40\\ =z(z+8)+5(z+8)\\ =(z+5)(z+8)$

Hence,

$z^{ 2 }+13z+40=(z+5)(z+8)$ .

Factorise:
$s^2+12s+27$

In the given polynomial

$s^2+12s+27$ ,

The first term is

$s^2$ and its coefficient is

$1$ .

The middle term is

$12s$ and its coefficient is

$12$ .

The last term is a constant term

$27$ .

Multiply the coefficient of the first term by the constant

$1\times 27=27$ .

We now find the factor of

$27$ whose sum equals the coefficient of the middle term, which is

$12$ and then factorize the polynomial

$s^2+12s+27$ as shown below:

$s^{ 2 }+12s+27\\ =s^{ 2 }+9s+3s+27\\ =s(s+9)+3(s+9)\\ =(s+3)(s+9)$

Hence,

$s^{ 2 }+12s+27=(s+3)(s+9)$ .

Factorize $1+2p + p^2$

${p}^{2}+2p+1$

${p}^{2}+p+p+1$

$p\left(p+1\right)+1\left(p+1\right)$

$\left(p+1\right) \left(p+1\right)={ \left(p+1\right)}^{2}$

Factorise
$a^2+4b^2-4ab-4c^2$ .

${ a }^{ 2 }+{ (2b) }^{ 2 }-2(2b)(a)-{ (2c) }^{ 2 }$

using

${ x }^{ 2 }+{ y }^{ 2 }-2xy={ (x-y) }^{ 2 }$ we get ,

${ (a-2b) }^{ 2 }-{ (2c) }^{ 2 }$

using

${ x }^{ 2 }-{ y }^{ 2 }=(x+y)(x-y)$ , we get

$(a-2b+2c).(a-2b-2c)$

Factories the given algebraic expression by taking out the common terms. Also, write the common term
$8ab+14a^{2}b-2a^{3}-12ab^{2}$

We first factorize each term of the given expression

$8ab+14a^2b−2a^3−12ab^2$ as shown below:

$8ab=2\times 2\times 2\times a\times b\\ 14a^{ 2 }b=2\times 7\times a\times a\times b\\ 2a^{ 3 }=2\times a\times a\times a\\ 12ab^{ 2 }=2\times 2\times 3\times a\times b\times b$

The common factors are

$2$ and

$a$ , therefore, the HCF is

$2\times a=2a$ .

Now, by taking out the common term(HCF), we have:

$8ab+14a^{ 2 }b−2a^{ 3 }−12ab^{ 2 }=2a(4b+7ab-2a^{ 2 }-6b^{ 2 })$

Hence, the common term is

$2a$ .

Factorise each of the following:
$9x^2-6x+1$

In the given polynomial

$9x^2-6x+1$ ,

The first term is

$9x^2$ and its coefficient is

$9$ .

The middle term is

$-6x$ and its coefficient is

$-6$ .

The last term is a constant term

$1$ .

Multiplication of the coefficient of the first term by the constant is 9$$.

We now find the factor of

$9$ whose sum equals the coefficient of the middle term, which is

$-6$ and then factorize the polynomial

$9x^2-6x+1$ as shown below:

$9x^{ 2 }-6x+1\\ =9x^{ 2 }-3x-3x+1\\ =3x(3x-1)-1(3x-1)\\ =(3x-1)(3x-1)\\ =(3x-1)^{ 2 }$

Hence,

$9x^{ 2 }-6x+1=(3x-1)^{ 2 }$ .

Factorise:
$p^2 + p - 132$

In the given polynomial

$p^2+p-132$ ,

The first term is

$p^2$ and its coefficient is

$1$ .

The middle term is

$p$ and its coefficient is

$1$ .

The last term is a constant term

$-132$ .

Multiply the coefficient of the first term by the constant

$1\times -132=-132$ .

We now find the factor of

$-132$ whose sum equals the coefficient of the middle term, which is

$1$ and then factorize the polynomial

$p^2+p-132$ as shown below:

$p^{ 2 }+p-132\\ =p^{ 2 }+12p-11p-132\\ =p(p+12)-11(p+12)\\ =(p-11)(p+12)$

Hence,

$p^{ 2 }+p-132=(p-11)(p+12)$ .

$\frac{{10{x^2} + 15x + 63}}{{5{x^2} - 25x + 12}} = \frac{{2x + 3}}{{x - 5}}$

$\begin{array}{l} \frac { { 10{ x^{ 2 } }+15x+63 } }{ { 5{ x^{ 2 } }-25x+12 } } =\frac { { 2x+3 } }{ { \left( { x-5 } \right) } } \\ \Rightarrow 10{ x^{ 3 } }-50{ x^{ 2 } }+15{ x^{ 2 } }-75x+63x-315=\left( { 10{ x^{ 3 } }-50{ x^{ 2 } }+24x+15{ x^{ 2 } }-75x+36 } \right) \\ \Rightarrow 10{ x^{ 3 } }-10{ x^{ 3 } }-50{ x^{ 2 } }+50{ x^{ 2 } }+15{ x^{ 2 } }-15{ x^{ 2 } }-75x+75x+63x-24x=351 \\ \Rightarrow 39x=351 \\ \therefore x=9 \end{array}$

Factorise:
$a^2+5a-104$

In the given polynomial

$a^2+5a-104$ ,

The first term is

$a^2$ and its coefficient is

$1$ .

The middle term is

$5a$ and its coefficient is

$5$ .

The last term is a constant term

$-104$ .

Multiply the coefficient of the first term by the constant

$1\times -104=-104$ .

We now find the factor of

$-104$ whose sum equals the coefficient of the middle term, which is

$5$ and then factorize the polynomial

$a^2+5a-104$ as shown below:

$a^{ 2 }+5a-104\\ =a^{ 2 }+13a-8a-104\\ =a(a+13)-8(a+13)\\ =(a-8)(a+13)$

Hence,

$a^{ 2 }+5a-104=(a-8)(a+13)$ .

Factorize $2x^2+x-4=0$ .

$\begin{matrix} 2{ x^{ 2 } }+x-4=0\, \, factorise \\ \to It\, \, is\, \, not\, \, factor\, \, by\, \, common\, \, factorisation. \\ \end{matrix}$

Find the common factor of the given term.
$3a, 21ab$ .

$3a,\, \, 21ab$

$3a=3\times a$

$21ab=3\times 7\times a\times b$

common factor is

$=3a\,$

Solve the factorisation method: $x + \frac{1}{x} = 11\frac{1}{{11}}$

$x+\dfrac{1}{x}=11\dfrac{1}{11}$

$\Rightarrow \dfrac{x^{2}+1}{x}=\dfrac{(11\times 11+1)}{11}$

$\Rightarrow 11x^{2}+11=122x$

$\Rightarrow 11x^{2}-122x+11=0$

$\Rightarrow 11x^{2}-121x-x+11=0$

$\Rightarrow 11x(x-11)-(x-11)=0$

$\Rightarrow (11x-1)(x-11)=0$

$\Rightarrow 11x-1=0$ or

$x-11=0$

$\Rightarrow x=\dfrac{1}{11}$ or

$x=11$

Enter 1 if it is true else 0.
$(25m^{4}-15m^{3}+10m+8)\div5m^{3}$ $=5m-3+\dfrac {2}{m^2}+\dfrac {8}{5m^3}$

$\dfrac{25{m}^{4}-15{m}^{3}+10m+8}{5{m}^{3}}$

$=\dfrac{25{m}^{4}}{5{m}^{3}}-\dfrac{15{m}^{3}}{5{m}^{3}}+\dfrac{10m}{5{m}^{3}}+\dfrac{8}{5{m}^{3}}$

$=5m-3+\dfrac{2}{{m}^{2}}+\dfrac{8}{5{m}^{3}}$

Find the common factor of the given term : $8x, 24$

$8x,\, \, 24 \\ \Rightarrow 8x=2\times 2\times 2\times x \\ \Rightarrow 24=2\times 2\times 2\times 3$

Common factor is

$=2\times 2\times 2=8\, \,$

Factorize: $3 a ^ { 2 } b - 12 a ^ { 2 } - 9 b + 36$

Let us factorize the given expression

$3a^2-12a^2-9b+36$ as shown below:

$3a^{ 2 }b-12a^{ 2 }-9b+36\\ =3a^{ 2 }(b-4)-9(b-4)\\ =(b-4)(3a^{ 2 }-9)\\ =(b-4)3(a^{ 2 }-3)\\ =3(b-4)(a^{ 2 }-3)\\ =3(b-4)(a-3)(a+3)\quad \quad \quad \quad (\because \quad x^{ 2 }-y^{ 2 }=(x-y)(x+y))$

Hence,

$3a^{ 2 }b-12a^{ 2 }-9b+36=3(b-4)(a-3)(a+3)$ .

Factors : $\left(5x-\dfrac{1}{x}\right)^2 + 5 \left(5x- \dfrac{1}{x}\right)+6$

Let

$(5x-\dfrac1x)=k$

So,

$k^2+5k+6=k^2+3k+2k+6=k(k+3)+2(k+3)=(k+2)(k+3)\\=(5x-\dfrac1x+2)(5x-\dfrac1x+3)$

Divide the polynomial $3x^{4}-4x^{3}-3x-1$ by $x-1$ . B using long division method only. Also check whether $x+1$ is a factor or not.

$3x^3-x^2-x-4$

__________________

$x-1$

$) 3x^4-4x^3+0x^2-3x-1$

$3x^4-3x^3$

__________________

$-x^3+0x^2-3x$

$-x^3+x^2$

________________

$-x^2-3x-1$

$-x^2+x$

_______________

$-4x-1$

$-4x+4$

_______________

$-5$

$\therefore \dfrac{3x^4-4x^3-3x-1}{x-1}=3x^3-x^2-x-4$

Remainder

$=-5$

Now to check whether

$x+1$ is factor or not,

$x+1=0$

$x=-1$

Put

$x=-1$ in

$3x^4-4x^3-3x-1$

$3(-1)^4-4(-1)^3-3(-1)-1$

$=3+4+3-1$

$=9\neq 0$

$\therefore$ By Remainder theorem, we conclude

$(x+1)$ is not a factor.

1069164_1159958_ans_e1e1895c86a045afad35d8464fe50d00.png

If $x+a$ is a common factor of $f\left( x \right) = {x^2} + x - 6$ and $g\left( x \right) = {x^2} + 3x - 18$ Then find the value of a

$f\left(x\right)={x}^{2}+x-6$

$={x}^{2}+3x-2x-6$

$=x\left(x+3\right)-2\left(x+3\right)$

$=\left(x+3\right)\left(x-2\right)$

$g\left(x\right)={x}^{2}+3x-18$

$={x}^{2}+6x-3x-18$

$=x\left(x+6\right)-3\left(x+6\right)$

$=\left(x-3\right)\left(x+6\right)$

There is no Common factor between

$\left(x+3\right)\left(x-2\right)$ and

$\left(x-3\right)\left(x+6\right)$

Hence value of

$a$ is unknown.

Divide ${4x}^{3}+{20x}^{2}+33x+18\ by \left (x+2\right)$

1118050_1163955_ans_0ccba4765f8a4d309166df77592cd0c5.jpeg

Factorize using identities
$t^{4}-625$

${t}^{4}-256$

$={t}^{4}-{16}^{2}$

$=\left({t}^{2}-16\right)\left({t}^{2}+16\right)$

$=\left({t}^{2}-{4}^{2}\right)\left({t}^{2}+16\right)$

$=\left(t-4\right)\left(t+4\right)\left({t}^{2}+16\right)$

Find the remainder when $f(x) = 2{x^3} - 6{x^2} + 4x - 2$ is divided by $g(x)=2x - 1$

When

$f(x)$ is divisible by

$(x-a)$ then the remainder is

$f(a)$

We have

$g(x)=2x-1=2\left(x-\dfrac{1}{2}\right)$

Therefore

$a=\dfrac{1}{2}$

$f\left(\dfrac{1}{2}\right)=2\left(\dfrac{1}{2}\right)^3-6\left(\dfrac{1}{2}\right)^2+4\left(\dfrac{1}{2}\right)-2$

$=2\cdot \dfrac{1}{8}-6\cdot \dfrac{1}{4}+4\cdot \dfrac{1}{2}-2$

$=\dfrac{1}{4}-\dfrac{3}{2}+2-2$

$=\dfrac{1}{4}-\dfrac{3}{2}$

$f\left(\dfrac{1}{2}\right)=-\dfrac{5}{4}$

Thus the remainder is

$-\dfrac{5}{4}$

Simplify:
$\left(1+\dfrac{x^2-y^2-z^2}{2yz}\right)\left(\dfrac{1}{x+y-z}+\dfrac{1}{z+x-y}\right)$

$\bullet$ Now,

$\left (1+\dfrac {x^2+y^2+z^2}{2yz}\right) \left (\dfrac {1}{x+y+z}+\dfrac {1}{z+x-y}\right)$

$=\left (\dfrac {x^2-(y^2-2yz+z^2)}{2yz}\right) \left (\dfrac {z+x-y+x+y-z}{(x+y-z)(z+x-y)}\right)$

$=\dfrac {x^2-(y-z)^2}{2yz}, \dfrac {2x}{(x+y-z)(z+x-y)}$

$=\dfrac {(x-y+z)(x+y-z)}{2yz}, \dfrac {2x}{(x+y-z)(z+x-y)}$

$=\left (\dfrac {2x}{2yz}\right)=\left (\dfrac {x}{yz}\right)$

Divide. Write the quotient and the remainder.
$(5x^{3}-3x^{2}) \div x^{2}$

1146929_1214694_ans_ed5674a074a74c9d9a4091c6aae79b56.jpg

Divide : $(6x^{5}-4x^{4}+8x^{3}+2x^{3})\div 2x^{2}$

$(6x^5-4x^4+8x^3+2x^3)+2x^2$

$=(6x^5-4x^4+10x^3)\div 2x^2$

Taking

$x^2$ common

$\dfrac{x^2(6x^3-4x+0)}{2x^2}=\dfrac{6x^3-4x+10}{2}$

$=3x^3-2x+5$ .

factorise :
$12(x^2 + 7x )^2 - 8(x^2 + 7x)(2x - 1) -15 (2x - 1)^2$

$12(x^{2}+7x)^{2}-8(x^{2}+7x)(2x-1)-15(2x-1)^{2}$

$= 12(x^{2}+7x)^{2}-18(x^{2}+7x)(2x-1)+10(x^{2}+7x)(2x-1)$

$-15(2x-1)^{2}$

$= 6(x^{2}+7x)[2(x^{2}+7x)-3(2x-1)]+$

$5(2x-1)[2(x^{2}+7x)+3(2x-1)]$

$= [2(x^{2}+7x)-3(2x-1)][6(x^{2}+7x)+5(2x-1)]$

$= [2x^{2}+14x-6x+3][6x^{2}+42x+10x-5]$

$= (2x^{2}+8x+3)(6x^{2}+52x-5)$

$\therefore 12(x^{2}+7x^{2})^{2}-8(x^{2}+7x)(2x-1)-15(2x-1)^{2}$

$= [2(x^{2}+7x)-3(2x-1)][6(x^{2}+7x)+5(2x-1)]$

$= (2x^{2}+8x+3)(6x^{2}+52x-5)$

1176067_1203561_ans_d13d9313be394c378e3034319d7d034b.jpg

Factorise
$2 a ^ { 2 } + 10 a - 28 = 0$

$\begin{matrix} 2{ a^{ 2 } }+10a-28=0 \\ 2{ a^{ 2 } }+14a-3a-28=0 \\ 2a\left( { a+7 } \right) -4\left( { a+7 } \right) =0 \\ \left( { 2a-4 } \right) \left( { a+7 } \right) =0 \\ \left( { 2a-4 } \right) \left( { a+7 } \right) =0 \\ 2a-4=0 \\ a=2\, \, \, or\, \, a=7\, \, \, Ans. \\ \end{matrix}$

Factorise : $x^{4}+5x^{3}+5x^{2}-5x-6.$

1115646_1201953_ans_d68413c5bf5b48ba889af926891d9a0e.jpg

Solve : $(2x^4+3x^3+4x-2x^2)\div (x+3)$

1148427_1186784_ans_4c43ebc730bf456784201b2d4632bbbc.jpg

Solve: $6x-3y$

Consider the given expression,

$\Rightarrow 6x-3y$

$\Rightarrow 3\left( 2x-y \right)$

Hence, this is the answer.

Simplify:
${a^2} + 4{b^2} - 4a -8b +4ab$

Now,

$a^2+4b^2-4a-8b+4ab$

$=(a^2+4ab+4b^2)-4(a+2b)$

$=(a+2b)^2-4(a+2b)$

$=(a+2b)(a+2b-4)$ .

Simplify:
$a{x^2}y - bxyz - a{x^2}z + bx{y^2}$

Given,

$ax^{2}y-bxyz- ax^{2} z+ bxy^{2}$

$=xy (ax+ by) - xz (by+ ax)$

$= (ax+by) (xy-xz)$

$= (ax+by)x(y-z)$

Factorize the given polynomial
$x^{3}-3x^{2}-9x-5$

Given

$x^3-3x^2-9x-5$

Let

$x=1$

So,

$x^3-3x^2-9x-5$

$=(1)^3-3(1)^2-9(1)-5$

$=1-3-9-5$

$=-16\neq 0$

So,

$1$ is not a factor

Let

$x=-1$

So,

$x^3-3x^2-9x-5$

$=(-1)^3-3(-1)^2-9(-1)-5$

$=-1-3+9-5$

$=0$

So,

$-1$ is a factor

Divide

$x^3-3x^2-9x-5$ by

$(x+1)$

(Refer image)

$\therefore x^2-4x-5=x^2-5x+x-5$

$=x(x-5)+1(x-5)$

$=(x-5)(x+1)$

So, factors are

$(x+1)(x-5)(x+1)$ .

1381355_1211123_ans_2dde85e2aa7643238b72596424ef2b17.PNG

Factories ${p}^{2}-36p+99$

$p^2-36p+99$

$p^2-33p-3p-99$

$p(p-33)-3(p-33)$

$(p-33)(p-3)$

$p-33=0$

$p-3=0$

$p=33$

$p=3$ .

Factorise the following :
$\begin{array} { l } { (i)6 a + 6 b } \\ { (ii)a x + b x } \\ { (iii)3 x ^ { 2 } - 6 a ^ { 6 } } \end{array}$
$\begin{array} { l } {(iv) 9 x ^ { 2 } + 3 x } \\ { (v)12 x ^ { 3 } y - 4 x y ^ { 2 } } \end{array}$
$\begin{array} { l } {(vi) \dfrac { 1 } { 2 } x + \dfrac { 1 } { 2 } } \\ (vii){ c d m + c d t } \\ (viii){ 36 a ^ { 2 } b ^ { 3 } - 18 a ^ { 3 } b ^ { 2 } } \\(ix) { 25 m ^ { 2 } n ^ { 3 } - 5 m n } \\(x) { 3 a y + 3 a z } \end{array}$
$\begin{array} { l } { (xi)185 a + 185 b } \\(xii) { 28 x - 14 y } \\ (xiii){ a x - a y } \\ (xiv){ 12 y ^ { 3 } + 6 a ^ { 3 } } \\ (xv){ 3 x + 9 y } \end{array}$

(i)

$6a+6b=6(a+b)$

(ii)

$ax+bx=(a+b)x$

(iii)

$3{x}^{2}-6{a}^{6}=3({x}^{2}-2{a}^{6})=3(x-\sqrt 2{a}^{3})(x+\sqrt{2}{a}^{3})$

(iv)

$9{x}^{2}+3x=3x(3x+1)$

(v)

$12{x}^{3}y-4x{y}^{2}=4xy(3{x}^{2}-y)$

(vi)

$\cfrac{1}{2}x+\cfrac{1}{2}=\cfrac{1}{2}(x+1)$

(vii)

$cdm+cdt=cd(m+t)$

(viii)

$36{a}^{2}{b}^{3}-18{a}^{3}{b}^{2}=18{a}^{2}{b}^{2}(2b-a)$

(ix)

$25{m}^{2}{n}^{3}-5mn=5mn(5m{n}^{2}-1)$

(x)

$3ay+3az=3a(y+z)$

(xi)

$185a+185b=185(a+b)$

(xii)

$28x-14y=14(2x-y)$

(xiii)

$ax-ay=a(x-y)$

xiv)

$12{y}^{3}+6{a}^{3}=6(2{y}^{3}+{a}^{3})$

(xv)

$3x+9y=3(x+3y)$

Divide $4{x^2} - 8x + 3$ by $2x - 1$ .

We can do this by long division method.

$\cfrac{4x^{2}}{2x}=2x$

ii)

$\cfrac{-6x}{2x}=-3$

$\therefore 4x^{2}-8x+3=(2x-1)(2x-3)$

1185470_1220362_ans_8c03905c5e294e58a3b01e27d5f27e2a.png

What is the remainder if ${ p }^{ 11 }+{ p }^{ 9 }+{ p }^{ 7 }+{ p }^{ 5 }+{ p }^{ 3 }+{ p }^{ 2 }$ divided by p + 1

Let

$P^{11}+P^7+P^3+P^2=f(P)$

Using remainder theorem, get

Remainder

$=f(P)=f(-1)=(-1)^{11}+(-1)^9+(-1)^7+(-1)^5+(-1)^3+(-1)^2=-4$

$(5p^2-25p+20)\div (p-1)$ .

1299215_1225931_ans_89e79bf8c71f42e89d0db0b5033e6d7b.jpg

Divide the given polynomial by the given monomial:
(i) $\left( 5 x ^ { 2 } - 6 x \right) \div 3 x$
(ii) $8$ $\left( x ^ { 3 } y ^ { 2 } z ^ { 2 } + x ^ { 2 } y ^ { 3 } z ^ { 2 } + x ^ { 2 } y ^ { 2 } z ^ { 3 } \right) \div 4 x ^ { 2 } y ^ { 2 } z ^ { 2 }$

1193571_1223883_ans_8c17d9953d054e1ca5d94f16017bd511.jpeg

Solve $\dfrac{x^2+5x+6}{x+3}$ .

$\dfrac{x^2+5x+6}{x+3}$

$=\dfrac{x^2+3x+2x+6}{x+3}$

$=\dfrac{(x+3)(x+2)}{x+3}$

$=x+2$

Factories : $am^2+bm^2+bn^2+an^2$

Consider the given expression,

$\Rightarrow a{{m}^{2}}+b{{m}^{2}}+b{{n}^{2}}+a{{n}^{2}}$

$\Rightarrow m\left( a+b \right)+{{n}^{2}}\left( b+a \right)$

$\Rightarrow \left( a+b \right)\left( m+n \right)$

Hence, this is the answer.

Solve:
$x^{2}-2x-8=0$

$x^2-2x-8=0$

$\Rightarrow x^2-4x+2x-8=0$

$\Rightarrow x(x-4)+2(x-4)=0$

$\Rightarrow (x-4)(x+2)=0$

$\Rightarrow x=4, \ x=-2$

Factorise
$x^{2}+19x-150$

$x^2+19x-150=x^2+25x-6x-150$

$x^2+19x-150=x(x+25)-6(x+25)$

$x^2+19x-150=(x+25)(x-6)$

Show that $\frac { x ^ { 2 } + 2 x y + y ^ { 2 } - a ^ { 2 } + 2 a b - b ^ { 2 } } { ( x + y - a + b ) } = ( x + y + a - b )$

$\dfrac{x^2 + 2 xy + y^2 - (a^2 + 2ab + b^2)}{(x + y - a + b)}$

$\dfrac{(x + y)^2 - (a - b)^2}{x + y - (a + b)} = \dfrac{(x + y - (a - )) (x + y + (a - b))}{[x + y - (a - b)]}$

$[ \because a^2 - b^2 = (a + b) (a - b)]$

$= x + y + a - b$ Hence, proved

$(3x+2x^2+4x^3)\div (x-4)$ .

$\Rightarrow \cfrac{(3x+2x^{2}+4x^{3})}{x-4}=4x^{2}+18x+75+\cfrac{300}{x-4}$
1380627_1240098_ans_20ccd1f8d46b47959591a6016280edb2.png

Factorize the expression and divide them as directed.
$4yz(z^2+6z-16)\div 2y(z+8)$ .

1302540_1242750_ans_2e43f484199f4e958cc06f308a4b0a0a.jpeg

Solve for r:
$41=400$ $[\big[\frac{100+r}{100}\big]^{2}-1]$

$41=400\left[\dfrac{10000+r^2+200r-100}{100}\right]$

$41=4[r^2+200r+9900]$

$41=4r^2+800r+39600$

$4r^2+800r+39559=0$

$4r^2+442r+358r+39559=0$

$2r(2r+221)+179(2r+221)=0$

$(2r+221)(2r+179)=0$

$\therefore r=\dfrac{-221}{2}, r=\dfrac{-179}{2}$ .

1127992_1244741_ans_e2e3f88fd24c49dbbb142723b3f61fda.jpg

Factorise
$4 u ^ { 2 } + 8 u$

$4{u}^{2}+8u$

$=4u(u+2)$

Factorise:
$12{p^5} + 16{p^4} - 20{p^3}$

1334611_1265924_ans_4b2bece38f6c469f84cc948ecbe50362.PNG

Factorize: $x ^ { 2 } - m ^ { 2 } + 6 m n - 9 n ^ { 2 }$

$x^2-(m^2-6mn+9n^2)\\=x^2-[m^2-2(m)(3n)+(3n)^2]\\=x^2-(m-3n)^2\\=[x-(m-3n)][x+(m-3n)]\\=(x-m+3n)(x+m-3n)$

Factorise:
${p^3} - 3pq + p{q^2}$

1335835_1265893_ans_3db0186d86d04f6e82ce535517c0e69b.PNG

Prove that ${n} ^ { 7 } - 7{ n} ^ { 5 } + 14 {n} ^ { 3 } - 8 n$ is divisible by $840$ for all $n \in N$

$n^7-7n^5+14n^3-8^n$

$=n(n-1)(n+1)(n^4-6n^2+8)$

$=n(n-1)(n+1)(n^2-2)(n^2-4)$

$=(n-2)(n-1)(n)(n+1)(n+2)(n^2-2)$

This expression is a product of

$5$ consecutive terms. Hence the given expression is divisible by

$5!=120$

We need further to prove that the given expression is divisible by

$7$ .

Let

$n=7a+k$ where

$k=0, 1, 2, 3, 4, 5, 6$

$n=7a+0$

$\implies n$ is divisible by

$7$

$n=7a+1$

$\implies n-1$ is divisible by

$7$

$n=7a+2$

$\implies n-2$ is divisible by

$7$

$n=7a+5$

$\implies n+2$ is divisible by

$7$

$n=7a+6$

$\implies n+1$ is divisible by

$7$

$n=7a+3$

$n^2-2=(7a+3)^2-2$

$=7p+q-2=7q$

$n=7a+4$

$n^2-2=(7a+4)^2-2$

$=7r+16-2=7r$

Hence, the given expression is divisible by

$120\times 7=840$ .

Factorise :
$6x^{2}+7x-3$

In order to factorise

$6x^2 + 7x - 3$ , we have to find two numbers p and q such that

$p + q = 7$ and

$pq = -18$ .

Clearly,

$9 + (-2) = 7$ and

$9 \times (-2) = -18$ .

So, we write the middle term

$7x$ as

$9x + (-2x ),$ i.e.,

$9x -2x$ .

$\therefore 6x^2 + 7x - 3 = 6x^2 + 9x - 2x - 3$

$= 3x (2x + 3) - 1 (2x + 3)$

$= (2x + 3)(3x - 1)$

Show that (2x+1) is a factor of ${ 4x }^{ 3 }+{ 12x }^{ 2 }+11x+3$ . Hence, factories ${ 4x }^{ 3 }+{ 12x }^{ 2 }+11x+3$ .

1867727_1270301_ans_15c0e3d5f0274861ba8d05ade681073d.jpg

The polynomial $6{x^4} + 8{x^3} - 5{x^2} + ax + b$ is exactly divisible by polynomial $2x-5$ then find the value of $2b-a$

1335858_1266279_ans_1556e951f09f46b5b98b13c5b8769140.PNG

Factorise completely by removing a monomial factor
$7{x^3} - 5{y^2}$

1335837_1265900_ans_c93d58c3f2664d57a508ebb5371a2e8b.PNG

Factorise: $1+2ab-\left( { a }^{ 2 }{ +b }^{ 2 } \right)$

1304729_1269240_ans_abb423da52f44bad87a7f3f95d43dc77.JPG

Factorise:
$2a{b^2} - 6bc + 8abc$

$2ab^2-6bc+8abc$

$=2b(ab-3c+4bc)$

factorise
$9{y^2} + 15ya$

1353302_1270598_ans_2f260dd416a145a8bd8f3a2be645c7b8.PNG

If $x^{3}+8y^{3}+24xy=64$ then $x+2y=$

1353322_1282893_ans_d0737bd49da640a484f55df70774312b.PNG

Factorise completely: $2{x^3} + {x^2} - 2x - 1$

given,

$2x^{3}+x^{2}-2x-1$

$=x^{2}(2x+1)-1(2x+1)$

$=(x^{2}-1)(2x+1)$

$=(x+1)(x-1)(2x+1)$

Factorise the following expressions.
$\left( i \right)\,\,\,\,{p^2} + 6p + 8\,\,\,\,\left( {ii} \right)\,\,\,{q^2} - 10q + 21\,\,\,\,\left( {iii} \right)\,\,\,{p^2} + 6p - 16$

$(i)$

$\begin{aligned}{}{p^2} + 6p + 8& = {p^2} + 4p + 2p + 8\\ &= p\left( {p + 4} \right) + 2\left( {p + 4} \right)\\& = \left( {p + 2} \right)\left( {p + 4} \right)\end{aligned}$

$(ii)$

$\begin{aligned}{}{q^2} -10q +21& = {q^2} -7q- 3q +21\\ &= q\left( {q -7} \right) - 3\left( {q -7} \right)\\& = \left( {q - 3} \right)\left( {q -7} \right)\end{aligned}$

$(iii)$

$\begin{aligned}{}{p^2} + 6p -16& = {p^2} + 8p - 2p -16\\ &= p\left( {p + 8} \right) - 2\left( {p + 8} \right)\\& = \left( {p - 2} \right)\left( {p + 8} \right)\end{aligned}$

factorize
$a^{2}-5b+ab-25$

$\begin{array}{l} { a^{ 2 } }-5b+ab-25 \\ \Rightarrow { a^{ 2 } }-{ 5^{ 2 } }+ab-5b \\ \Rightarrow { a^{ 2 } }-{ 5^{ 2 } }+b\left( { a-5 } \right) \\ \Rightarrow \left( { a-5 } \right) \left( { a+5 } \right) +b\left( { a-5 } \right) \\ \Rightarrow \left( { a-5 } \right) \left[ { a+5+b } \right] \\ \Rightarrow \left( { a-5 } \right) \left( { a+b+5 } \right) \end{array}$

Factorise the following
(i) $p^2 +6p+8$

$p^2+6p+8$ [By splitting the middle term]

$\Rightarrow p^2+2p+4p+8$

$\Rightarrow p(p+2)+4(p+2)$

$\Rightarrow (p+4)(p+2)$ .

Factorise
$49a^{2}b^{4}-4a^{2}b^{6}$

$\begin{array}{l} 49{ a^{ 2 } }{ b^{ 4 } }-4{ a^{ 2 } }{ b^{ 6 } } \\ { \left( { 7a{ b^{ 2 } } } \right) ^{ 2 } }-{ \left( { 4a{ b^{ 3 } } } \right) ^{ 2 } }\, \, \left[ { { a^{ 2 } }-{ b^{ 2 } }=\left( { a-b } \right) \left( { a+b } \right) } \right] \\ \left( { 7a{ b^{ 2 } }-4a{ b^{ 3 } } } \right) \left( { 7a{ b^{ 2 } }+4a{ b^{ 3 } } } \right) \end{array}$

Find, in each case, the remainder when: $x^{4}-3x^{2}+2x+1$ is divided by $x-1$ .

From remainder theorem, we know that when a polynomial

$f (x)$ is divided by

$(x – a)$ , then the remainder is

$f(a).$

Given,

$f(x) = x^4 – 3x^2 + 2x + 1$ is divided by

$x – 1$

So, remainder

$= f(1) = (1)^4 – 3(1)^2 + 2(1) + 1 = 1 – 3 + 2 + 1 = 1$

Factorise using
$12x+75x^{5}-60x^{8}$

$\begin{array}{l} 12x+75{ x^{ 5 } }-60{ x^{ 8 } } \\ \Rightarrow 3x\left( { 4+25{ x^{ 4 } }-20{ x^{ 7 } } } \right) \\ 3x\left\{ { 4+5{ x^{ 4 } }\left( { 5-4{ x^{ 3 } } } \right) } \right\} \end{array}$

Factorise
(i) $a^4 -b^4$ (ii) $p^4 -81$ (iii) $x^4 -(y+z)^4$ (iv) $x^4 -(x-z)^4$ (v) $a^4 -2a^2 b^2 +b^4$

$i){ a }^{ 4 }-{ b }^{ 4 }=({ a }^{ 2 }-{ b }^{ 2 })({ a }^{ 2 }+{ b }^{ 2 })=(a+b)(a-b)({ a }^{ 2 }+{ b }^{ 2 })\\ =(a+b)({ a }^{ 2 }+{ b }^{ 2 })(a-b)\\ii){ p }^{ 4 }-81\\ =(p^{ 2 }-9)({ p }^{ 2 }+9)\\ =(p+3)(p-3)({ p }^{ 2 }+9)\\iii){ x }^{ 4 }-{ (y+z) }^{ 4 }=({ x }^{ 2 }-{ (y+z) }^{ 2 })({ x }^{ 2 }+{ (y+z) }^{ 2 })\\ =(x+y+z)(x-y-z)({ x }^{ 2 }+{ (y+z) }^{ 2 })\\ iv){ x }^{ 4 }-{ (x-z) }^{ 4 }=({ x }^{ 2 }-{ (x-z) }^{ 2 })({ x }^{ 2 }+{ (x-z) }^{ 2 })\\ =(x+x-z)(x-x+z)({ x }^{ 2 }+{ (x-z) }^{ 2 })\\ =(2x-z)(z)({ x }^{ 2 }+{ (x-z) }^{ 2 })\\v){ a }^{ 4 }-2{ a }^{ 2 }{ b }^{ 2 }+{ b }^{ 4 }={ a }^{ 4 }-{ a }^{ 2 }{ b }^{ 2 }-{ a }^{ 2 }{ b }^{ 2 }+{ b }^{ 4 }\\ ={ a }^{ 2 }({ a }^{ 2 }-{ b }^{ 2 })-{ b }^{ 2 }({ a }^{ 2 }-{ b }^{ 2 })\\ ={ ({ a }^{ 2 }-{ b }^{ 2 }) }^{ 2 }\\ ={ (a+b) }^{ 2 }{ (a-b) }^{ 2 }$

Factorise :
$x^{2}+5x+1$

$x^{2}+5x+1=0$

$=x^{2}+5x+1-1=0-1$

$=1-1+5x+x^{2}=0-1$

$=5x+x^{2}=0-1$

$=x^{2}+5x=-1$

coeff of x is 5. Take

$\frac{5}{2}$ , square it & add it on both sides

$=5x+6.25+x^{2}=-1+6.25$

$=6.25+5x+x^{2}=5.25$

$=(x+2.5)(x+2.5)= 5.25$

$\therefore x+2.5=2.29$

$x=-0.208$

$x+2.5=-2.29$

$x=-4.79$

1186831_1292447_ans_b4d478d5d6f94825991cc68258a2bedc.jpg

$\left( {{x^2} + 7x + 12} \right)/\left( {x + 3} \right)$

$\begin{array}{l} \dfrac { { { x^{ 2 } }+7x+12 } }{ { x+3 } } \\ \dfrac { { { x^{ 2 } }+4x+3x+12 } }{ { x+3 } } \\ \dfrac { { x\left( { x+4 } \right) +3\left( { x+4 } \right) } }{ { \left( { x+3 } \right) } } \\ \dfrac { { \left( { x+3 } \right) \left( { x+4 } \right) } }{ { \left( { x+3 } \right) } } \\ =x+4 \end{array}$

Solve the following.
$\dfrac{{x - 1}}{{x - 2}} + \dfrac{{x - 3}}{{x - 4}} = 3\dfrac{1}{3}(x \ne 2,4)$

$\frac{(x-1)(x-4)+(x-3)(x-2)}{(x-2)(x-4)} = \frac{10}{2}$

$3[(x^{2}-5x+4)+(x^{2}-5x+6)] = 10[x^{2}-6x+8]$

$3[2x^{2}-10x+10] = 10x^{2}-60x+80$

$6x^{2}-30x+30=10x^{2}-60x+80$

$4x^{2}-30x+50=0.$

$2x^{2}-15x+25=0.$

$2x^{2}-10x-5x+25=0.$

$\Rightarrow 2x(x-5)-5(x-5)=0$

$(2x-5)(x-5)=0$

$(2x-5)=0$ or

$x-5=0$

$x=5/2$ or

$x=5$

1212361_1298478_ans_49956c3427ff49c7ba59faa31e3fd34c.JPG

Factorise
$18x^{2}+48x+32$

$18{x^2} + 48x + 32 = 18{x^2} + 24x + 24x + 32$

$= 6x\left( {3x + 4} \right) + 8\left( {3x + 4} \right)$

$= \left( {6x + 8} \right)\left( {3x + 4} \right)$

Solve: $(7x^4 - 21x^2 + 15) \div 7x^2$ .

1181567_1292949_ans_8f4267053909499ab5889f454196cd95.jpg

Divides and write the quotient and the remainder.
$\left( { 6x }^{ 5 }-{ 4x }^{ 4 }+{ 8x }^{ 3 }+{ 2x }^{ 2 } \right) \div { 2x }^{ 2 }$

$=\dfrac{6{x}^{5}-4{x}^{4}+8{x}^{3}+2{x}^{2}}{2{x}^{2}}$

$=\dfrac{2{x}^{2}\left(3{x}^{3}-2{x}^{2}+4x+1\right)}{2{x}^{2}}$

$=3{x}^{3}-2{x}^{2}+4x+1$

$\therefore$ Quotient

$=3{x}^{3}-2{x}^{2}+4x+1$

and Remainder

$=0$

Divide $6{x}^{3}-{x}^{2}-10x-2$ by $2x-3$

img

1194073_1367941_ans_a3fcf33657a34cd8939543172a12d26e.JPG

If ${x}^{3}+a{x}^{2}+bx+6$ is divisible by $\left(x-2\right)$ and leaves remainder $3$ when divided by $\left(x-3\right)$ then find the value of $a$ and $b$ .

Let the given polynomial be

$p\left(x\right) ={x}^{3}+a{x}^{2}+ bx + 6$

Given

$p\left(x\right)$ is divisible by

$\left(x – 2\right)$ , hence

$p\left(2\right)= 0$

Put

$x = 2$ in

$p\left(x\right)$

$p(2) = {2}^{3}+ a{\left(2\right)}^{2}+ b(2) + 6$

$\Rightarrow 0 = 8 + 4a + 2b + 6$

$\Rightarrow 4a + 2b = – 14$

$\Rightarrow 2a + b = – 7$ .........

$(1)$

It is also given that when

$p\left(x\right)$ is divided by

$\left(x – 3\right)$ the remainder is

$3$

That

$p\left(3\right)= 3$

Put

$x = 3$ in

$p\left(x\right)$

$p\left(3\right) = {3}^{3}+ a{\left(3\right)}^{2}+ b(3) + 6$

$\Rightarrow 3 = 27 + 9a + 3b + 6$

$\Rightarrow 9a + 3b = – 30$

$\Rightarrow 3a + b = – 10$ .....

$(2)$

Subtract

$(2)$ from

$(1)$

$\Rightarrow 3a+b-2a-b= -10+7$

$\therefore a = – 3$

Substitute

$a = – 3$ in

$(1)$ , we get

$2\left(– 3\right) + b = –7$

$\therefore b = –1$

Divide $x^{2}+3x+2$ by $x+1$

$x+2$

___________

$x+1)x^2+3x+2$

$x^2+x$

___________

$2x+2$

___________

$x$

___________

Quotient

$=x+2$ .

1185909_1349745_ans_555d44c90019460d847f3d0d2f2f90ab.jpg

Simplify : $\dfrac { a ^ { 3 } - 27 } { 5 a ^ { 2 } - 16 a + 3 } \div \dfrac { a ^ { 2 } + 3 a + 9 } { 25 a ^ { 2 } - 1 }$

$\begin{array}{l} \dfrac { { { a^{ 3 } }-27 } }{ { 5{ a^{ 2 } }-16a+3 } } \div \dfrac { { { a^{ 2 } }+3a+9 } }{ { 25{ a^{ 2 } }-1 } } \\ =\dfrac { { \left( { a-3 } \right) \left( { { a^{ 2 } }+9+3a } \right) } }{ { \left( { 5{ a^{ 2 } }-15a-a+3 } \right) } } \times \dfrac { { \left( { 5a-1 } \right) \left( { 5a+1 } \right) } }{ { \left( { { a^{ 2 } }+3a+9 } \right) } } \\ =\dfrac { { \left( { a-3 } \right) \times \left( { 5a-1 } \right) \times \left( { 5a+1 } \right) } }{ { \left( { 5a-1 } \right) \left( { a-3 } \right) } } \\ =5a+1 \end{array}$

$a^{3}-12a-16$

Let

$f{\left( a \right)} = {a}^{3} - 12a - 16$

$f{\left( -2 \right)} = {\left( -2 \right)}^{3} - 12 \left( -2 \right) - 16 = 0$

Thus,

$\left( a + 2 \right)$ is a factor of

${a}^{3} - 12a - 16$ .

Now dividing

$f{\left( a \right)}$ by

$\left( a + 2 \right)$ , we get

$\left( {a}^{3} - 12a - 16 \right) \div \left( a + 2 \right) = {a}^{2} - 2a - 8$

Now,

${a}^{2} - 2a - 8$

$= {a}^{2} - 4a + 2a - 8$

$= \left( a - 4 \right) \left( a + 2 \right)$

Therefore,

${a}^{3} - 12a - 16 = \left( a + 2 \right) \left( a - 4 \right) \left( a + 2 \right)$

$\Rightarrow {a}^{3} - 12a - 16 = {\left( a + 2 \right)}^{2} \left( a - 4 \right)$

Divide and write the quotient and the remainder
$(21x^{4}-14x^{2}+7x)\div 7x^{3}$

1172941_1338693_ans_afcb9ead1dd6475a9b8752c91e90ef2f.jpg

factorize
$4x+8$

Here, 4 is the common factor

Now,

$4(x+4)$ are the factor.

Solve:
$2a(3x+5y)-5b(3x+5y)$

$2a (3x + 5y) - 5b (3x + 5y)$

$\Rightarrow (3x + 5y) (2a - 5b)$

1185853_1348760_ans_2c154516fb2342909b288275afa3ab30.jpg

Divide the given polynomial by the given monomial
8 $\left( x ^ { 3 } y ^ { 2 } z ^ { 2 } + x ^ { 2 } y ^ { 3 } z ^ { 2 } + x ^ { 2 } y ^ { 2 } z ^ { 3 } \right) \div 4 x ^ { 2 } y ^ { 2 } z ^ { 2 }$

given,

$8(x^3y^2z^2+x^2y^3z^2+x^2y^2z^3)\div 4x^2y^2z^2$

$=8x^2y^2z^2(x+y+z)\div 4x^2y^2z^2$

$=\dfrac{8x^2y^2z^2}{4x^2y^2z^2}(x+y+z)$

$=2(x+y+z)$ .

Factorize:
$3+2a-a^{2}$

$3 + 2a - {a}^{2}$

$= 3 + 3a - a - {a}^{2}$

$= 3 \left( 1 + a \right) - a \left( 1 + a \right)$

$= \left( 3 - a \right) \left( 1 + a \right)$

Therefore

$3 + 2a - {a}^{2} = \left( 3 - a \right) \left( 1 + a \right)$

Solve the following
$(6x^{4}+4x^{2}+9x+5) \div (2x+3)$

$\left( 6 {x}^{4} + 4 {x}^{2} + 9x + 5 \right) \div \left( 2x + 3 \right)$

Quotient

$= \left( 3 {x}^{3} - \cfrac{9}{2} {x}^{2} + \cfrac{35}{4} x - \cfrac{69}{8} \right)$

Remainder

$= \cfrac{247}{8}$

1289188_1337743_ans_c56a883a5b824769a1f583f8ada2c045.png

Divide and write the equation $\left( {{y^2}+10y + 24} \right) \div \left( {y + 4} \right)$

We have,

$\left( {{y^2}+10y + 24} \right) \div \left( {y + 4} \right)$

$\Rightarrow \dfrac{\left( {{y^2}+10y + 24} \right)}{\left( {y + 4} \right)}$

$\Rightarrow \dfrac{\left( y+6 \right) \left( y+4 \right) }{\left( {y + 4} \right)}$

$\Rightarrow y+6$

Hence, this is the answer.

Factorise: ${x}^{4}+{x}^{2}+1$

$={x}^{4}+2\times {x}^{2}+1-{x}^{2}$

$={\left({x}^{2}+1\right)}^{2}-{x}^{2}$

$=\left({x}^{2}+1+x\right)\left({x}^{2}+1-x\right)$

$=\left({x}^{2}+x+1\right)\left({x}^{2}-x+1\right)$

Factorize -
${ ax }^{ 2 }+{ bx }^{ 2 }-{ ay }^{ 2 }-{ by }^{ 2 }$

$a{x}^{2}+b{x}^{2}-a{y}^{2}-b{y}^{2}$

$=\left(a+b\right){x}^{2}-\left(a+b\right){y}^{2}$

$=\left(a+b\right)\left({x}^{2}-{y}^{2}\right)$

$=\left(a+b\right)\left(x+y\right)\left(x-y\right)$

Find the factorss of $x^{4}+3x^{3}-7x^{2}-27x-18$

$x = a$ is a zero, then

$x – a$ is a factor

$f\left(x\right)=x^4+3x^3−7x^2−27x−18$

$f\left(-1\right)=1-3−7+27−18=0$

$\left(x+1\right)$ is a factor of

$f\left(x\right)$

The other factors are calculated by using synthetic division.

$\left(x+1\right)\left(x^3+2x^2-9x-18\right)$

$=\left(x+1\right)\left(x^2\left(x+2\right)-9\left(x+2\right)\right)$

$=\left(x+1\right)\left(x+2\right)\left({x}^{2}-9\right)$

$=\left(x+1\right)\left(x+2\right)\left(x-3\right)\left(x+3\right)$

Factorise:
$(x^{3}-2x^{2}-5x+6)$

Let

$f\left(x\right)={x}^{3}-2{x}^{2}-5x+6$

$\Rightarrow f\left(1\right)={1}^{3}-2{1}^{2}-5\times 1+6=1-2-5+6=0$

Using synthetic division, we have

${x}^{2}-x-6=0$ is the other factor

$\left(x-1\right)\left({x}^{2}-x-6\right)=0$

$=\left(x-1\right)\left({x}^{2}-3x+2x-6\right)=0$

$=\left(x-1\right)\left(x\left(x-3\right)+2\left(x-3\right)\right)=0$

$=\left(x-1\right)\left(x+2\right)\left(x-3\right)$ are the factors of

$f\left(x\right)$

Divide $15{x}^{3}-20{x}^{2}+13x-12$ by $3x-6$

img

1188892_1367917_ans_c4d60b68fd4d4b0cabf911ce45e827a6.jpg

Verify whether $2y-5$ is a factor of $4{y}^{4}-10{y}^{3}-10{x}^{2}+30y-15$

1194072_1367932_ans_448f493289a7423093ef489dc66da470.PNG

Write the factors of
$15xy-6x+5y-2$

$=15xy-6x+5y-2$

$=3x\left(5y-2\right)+1\left(5y-2\right)$

$=\left(3x+1\right)\left(5y-2\right)$

Work out the following divisions
(i) $\left(11x-121\right)\div 11$
(ii) $\left(15x-25\right)\div \left(3x-5\right)$
(iii) $10y\left(9y+21\right)\div 2\left(3y+7\right)$
(iv) $9{p}^{2}{q}^{2}(3z-12)\div 27pq\left(z-4\right)$

$(i) (11x-121) \div 11$

$=\dfrac{11(x-11)}{11} = x-11$

$(ii) (15x-25) \div (3x-5)$

$=\dfrac{5(3x-5)}{3x-5} = 5$

$(iii) 10y(9y+21) \div 2(3y+7)$

$=\dfrac{5y(9y+21)}{3y+7} = 5y \times 3 = 15y$

$(iv) 9p^{2}q^{2}(3z-12) \div 27pq(z-4)$

$=\dfrac{pq(3z-12)}{3(z-4)}$

$=pq$

Find the value of $a$ and $b$ , if the zeros of the polynomial. $x^{3}-3x^{2}+x+1$ are $a-b,\ a,\ a+b$

1302719_1370313_ans_ac72039910b24ae1bdbfbb2c7f94a66c.jpg

$({ y }^{ 2 }+10y+24)\div (y+4)$

$({ y }^{ 2 }+10y+24)\div (y+4)$ =

$\dfrac{y^{2}+10y+24}{y+4}$

$\dfrac{y^{2}+4y+6y+24}{y+4}$

$\dfrac{(y+6)(y+4)}{y+4}$

$y+6$

Factorise.
$6a^{2}+7a-5$

$6{a}^{2}+7a-5$

$=6{a}^{2}+10a-3a-5$

$=2a\left(3a+5\right)-1\left(3a+5\right)$

$=\left(2a-1\right)\left(3a+5\right)$

Hence the factors are

$\left(2a-1\right)\left(3a+5\right)$

Factorise ${ \left( x-2y \right) }^{ 2 }+7\left( x-2y \right) +12$

Let

$\alpha=x-2y$ then

$={\alpha}^{2}+7\alpha+12$

$={\alpha}^{2}+4\alpha+3\alpha+12$

$=\left(\alpha+4\right)\left(\alpha+3\right)$

$=\left(x-2y+4\right)\left(x-2y+3\right)$ where

$\alpha=x-2y$

Divide $-12a^{3}b+18a^{2b^{2}}-24ab^{3}$ by $-6ab$

1194617_1374769_ans_2fefbaa8b98147369208245cda0519a6.JPG

Factorise.
$6x^{2}-11xy-10y^{2}$

$6{x}^{2}-11xy-10{y}^{2}$

$=6{x}^{2}-15xy+4xy-10{y}^{2}$

$=3x\left(2x-5y\right)-2y\left(2x-5y\right)$

$=\left(2x-5y\right)\left(3x-2y\right)$

Hence the factors are

$\left(2x-5y\right)\left(3x-2y\right)$

Factorize :
$8(a+1)^{2}+2(a+1)(b+2)-15(b+2)^{2}$

$8(a+1)^{2}+2(a+1)(b+2)-15(b+2)^{2}$

Let

$a+1=x$ and

$b+2=y$ . Then,

$8x^{2}+2xy-15y^{2}$

$=8x^{2}+12xy-10xy-15y^{2}$

$=4x(2x+3y)-5y(2x+3y)$

$=(2x+3y)(4x-5y)$

$=(2a+3b+8)(4a-5b-6)$

Factorise.
$3x^{2}+11xy+6y^{2}$

$3{x}^{2}+11xy+6{y}^{2}$

$=3{x}^{2}+9xy+2xy+6{y}^{2}$

$=3x\left(x+3y\right)+2y\left(x+3y\right)$

$=\left(3x+2y\right)\left(x+3y\right)$

Hence the factors are

$\left(3x+2y\right)\left(x+3y\right)$

Factorize the following:
$20a^{12}b^{2}-15a^{8}b^{4}$

$20{a}^{12}{b}^{2}-15{a}^{8}{b}^{4}$

$=5{a}^{8}{b}^{2}\left(4{a}^{4}-3{b}^{2}\right)$

$=5{a}^{8}{b}^{2}\left({\left(\sqrt{2}a\right)}^{4}-{\left(\sqrt{3}b\right)}^{2}\right)$

$=5{a}^{8}{b}^{2}\left({\left(\sqrt{2}a\right)}^{2}+{\left(\sqrt{3}b\right)}^{2}\right)\left({\left(\sqrt{2}a\right)}^{2}-{\left(\sqrt{3}b\right)}^{2}\right)$

$=5{a}^{8}{b}^{2}\left({\left(\sqrt{2}a\right)}^{2}+{\left(\sqrt{3}b\right)}^{2}\right)\left(\sqrt{2}a-\sqrt{3}b\right)\left(\sqrt{2}a+\sqrt{3}b\right)$

Simplify: $30a^{3}b^{3}c^{3}+45abc$

$30{a}^{3}{b}^{3}{c}^{3}+45abc$

$=15abc\left(2{a}^{2}{b}^{2}{c}^{2}+3\right)$

Factorise: $4xy - x + 12y -3$

$4xy-x+12y-3$

$=x\left(4y-1\right)+3\left(4y-1\right)$

$=\left(4y-1\right)\left(x+3\right)$

Write the common factors of : ${ 4a }^{ 2 }b$ and $3ab$

Factors of

$4{a}^{2}b=2\times 2\times a\times a\times b$

Factors of

$3ab=3\times a\times b$

Common factor

$=a\times b=ab$

Factorize: ${x}^{2}-15x+56$

Given,

${x}^{2}-15x+56$

$={x}^{2}-7x-8x+56$

$=x\left(x-7\right)-8\left(x-7\right)$

$=\left(x-7\right)\left(x-8\right)$

Resolve ${x}^{8}+{x}^{4}+1$ in to factors.

${x}^{8}+{x}^{4}+1$

$={\left({x}^{4}\right)}^{2}+2{x}^{4}+1-{x}^{4}$

$={\left({x}^{4}+1\right)}^{2}-{\left({x}^{2}\right)}^{2}$

$=\left({x}^{4}+1+{x}^{2}\right)\left({x}^{4}+1-{x}^{2}\right)$

$=\left({x}^{4}+{x}^{2}+1\right)\left({x}^{4}-{x}^{2}+1\right)$

$=\left({\left({x}^{2}\right)}^{2}+2{x}^{2}+1-{x}^{2}\right)\left({\left({x}^{2}\right)}^{2}-2{x}^{2}+1+{x}^{2}\right)$

$=\left({\left({x}^{2}+1\right)}^{2}-{x}^{2}\right)\left({\left({x}^{2}-1\right)}^{2}+{x}^{2}\right)$

$=\left({x}^{2}+x+1\right)\left({x}^{2}-x+1\right)\left({\left({x}^{2}-1\right)}^{2}+{x}^{2}\right)$

Solve
$\left(x^ {4}+x^ {3}+3x^ {2}+3x+12\right)\div \left(x^ {2}+2\right)$

Remainder

$=x+10$

1336488_1385299_ans_d02dcdbbcbe840cfad8c8d9104dffee7.PNG

Factorize:
$pqr-{ p }^{ 2 }q+{ pq }^{ 2 }r$

$pqr-{p}^{2}q+p{q}^{2}r$

$=pq\left(r-p+qr\right)$

Solve:
$x^{3}-x^{2}-14x+14$

${x}^{3}-{x}^{2}-14x+14$

$={x}^{2}\left(x-1\right)-14\left(x-1\right)$

$=\left(x-1\right)\left({x}^{2}-14\right)$

$=\left(x-1\right)\left({x}^{2}-{\left(\sqrt{14}\right)}^{2}\right)$

$=\left(x-1\right)\left(x-\sqrt{14}\right)\left(x+\sqrt{14}\right)$

Factorize:
$4x^2-12xy+9y^2+2x-3y$

Now,

$4x^2-12xy+9y^2+2x-3y$

$=(2x-3y)^2+(2x-3y)$

$=(2x-3y)(2x-3y+1)$ .

write the quotient and the remainder.
$\left( { 2x }^{ 4 }+{ 3x }^{ 3 }+4x-2\right) \div \left( x+2 \right)$

Remainder

$=2$

Quotient

$=2{x}^{3}-{x}^{2}+2x$

1327202_1414059_ans_2fe6c7fba82a4ad1b8e6b881b349cc51.PNG

Divide $(-x^6+2x^4+4x^3+2x^2)$ by $(2\sqrt2 x^2)$

$(-x^6+2x^4+4x^3+2x^2)\div$

$(2\sqrt2 x^2)=\dfrac{-x^6+2 x^4+4 x^3+2 x^2}{2\sqrt{2}x^2}=\dfrac{x^2(-x^4+2 x^2+4 x+2)}{2 \sqrt{2} x^2}$

$=\dfrac{1}{2\sqrt{2}}(-x^4+2 x^2+4 x+2)$

Factorize:
$x^2-7x+10$ .

Now,

$x^2-7x+10$

$=x^2-5x-2x+10$

$=(x-2)(x-5)$ .

Solve the following
$8x^{3}-6x^{2}+x=0$

Now,

$8x^{3}-6x^{2}+x=0$

or,

$x(8x^2-6x+1)=0$

or,

$x(8x^2-4x-2x+1)=0$

or,

$x\{4x(2x-1)-1(2x-1)\}=0$

or,

$x(4x-1)(2x-1)=0$

or,

$x=0,\dfrac{1}{4},\dfrac{1}{2}$ .

Factorise.
$x^{ 2 } + xy + 8x + 8y$

$x^2+x{y}+8 x+8 y$

$=x(x+y)+8(x+y)$

$=(x+8)(x+y)$

Divide $(5{p^2} - 25p + 20) \div (p-1)$

$(5p^2-25p+20)\div 5(p-4)$

$\Rightarrow 5p^2-25p+20=5(p^2-5p+4)$

$=5[p^2-4p-p+4]$

$=5[p(p-1)-4(p-1)]$

$=5[(p-1)(p-4)]$

$\therefore$

$(5p^2-25p+20)\div (p-1)=\dfrac{5(p-1)(p-4)}{(p-1)}$

$=5(p-4)$ .

$\therefore (5p^2-25p+20)\div (p-1)=5(p-4)$ .

1196717_1394116_ans_824b363c1bac4cf7bdbc6dde910ea286.jpg

Solve :
$15x + 5$

$15x + 5$

Simplifying we get

$15x+5=5(3x+1)$

Factorise :-
${a^4} - 2{a^2}{b^2} + {b^4}$

$a^4-2 a^2 b^2+b^4=a^4-a^2 b^2-a^2 b^2+b^4=a^2(a^2-b^2)- b^2(a^2-b^2)=(a^2-b^2)^2$

As we know that

$x^2-y^2=(x+y)(x-y)$

$\implies a^4-2 a^2 b^2+b^4=(a^2-b^2)^2=(a-b)^2(a+b)^2$

Factorise: $\sqrt{3} x^{2}+4x-7\sqrt{3}$

The given expression is

$\sqrt{3}{x}^{2}+4x-7\sqrt{3}$

We are asked to factorise the given expression.

We can factorise the given expression by rewriting it and taking out common terms.

$\sqrt{3}{x}^{2}+4x-7\sqrt{3}$

The expression can be written as:

$\sqrt{3}{x}^{2}-3x+7x-7\sqrt{3}$

$\Rightarrow \sqrt{3}{x}^{2}-\sqrt3 \times \sqrt3 x+7x-7\sqrt{3}$

Taking out

$\sqrt3 x$ common from the first two terms and

$7$ from the last two.

$\Rightarrow \sqrt{3}x\left(x-\sqrt{3}\right)+7\left(x-\sqrt{3}\right)$

$\Rightarrow \left(\sqrt{3}x+7\right)\left(x-\sqrt{3}\right)$

Thus,

$\left(\sqrt{3}x+7\right)\ and \ \left(x-\sqrt{3}\right)$ are the factors of the given expression.

Hence,

$\sqrt{3}{x}^{2}+4x-7\sqrt{3}=\left(\sqrt{3}x+7\right)\left(x-\sqrt{3}\right)$

Factorise: $x^2+2x+1$

Given

$x^{2}+2x+1$

$\Rightarrow x^{2}+x+x+1$

$\Rightarrow x (x+1)+1(x+1)$

$\Rightarrow (x+1)(x+1)$

Simplify :-
$\dfrac{8({x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3})}{4{x^2}{y^2}{z^2}}$

$\dfrac{8(x^3y^2z^2+x^2y^3z^2+x^2y^2z^3)}{4x^2y^2z^2}$

$=\dfrac{8x^3y^2z^2+8x^2y^3z^2+8x^2y^2z^3}{4x^2y^2z^2}$

$=\dfrac{8x^2y^2z^2(x+y+z)}{4x^2y^2z^2}$

$=2(x+y+z)$ .

Factorise : $a^2x^2 +(ax^2 +1)x +a$

Given :

$a^2x^2 +(ax^2 +1)x +a$

By multiplying the terms

$a^2x^2 +(ax^2 +1)x +a$

$=a^2x^2 +ax^3 +x+a$

By taking the common terms out

$a^2x^2 +(ax^2 +1)x +a$

$= ax^2 (a+x) +1(x+a)$

So we get,

$a^2x^2 +(ax^2 +1)x +a$

$=(a+x)(ax^2 +1)$

Solve:
$49(x^4-5x^3-24x^2)\div 11x(x-8)$

${x}^{4} - 5{x}^{3} - 24 {x}^{2}$

$= {x}^{2} \left( {x}^{2} - 5x - 24 \right)$

$= {x}^{2} \left[ {x}^{2} - 8x + 3x - 24 \right]$

$= {x}^{2} \left( x - 8 \right) \left( x + 3 \right)$

Therefore,

$49 \left( {x}^{4} - 5 {x}^{3} - 24 {x}^{2} \right) \div 11x \left( x - 8 \right)$

$= \cfrac{49 \left( {x}^{4} - 5{x}^{3} - 24{x}^{2} \right)}{11x \left( x - 8 \right)}$

$= \cfrac{49 {x}^{2} \left( x - 8 \right) \left( x + 3 \right)}{11x \left( x - 8 \right)}$

$= \cfrac{49}{11} x \left( x + 3 \right)$

Hence

$49 \left( {x}^{4} - 5 {x}^{3} - 24 {x}^{2} \right) \div 11x \left( x - 8 \right) = \cfrac{49}{11} x \left( x + 3 \right)$

Solve:
$(m^2-14m-32)\div (m+2)$

${m}^{2} - 14m - 32$

$= {m}^{2} - 16m + 2m - 32$

$= \left( m - 16 \right) \left( m + 2 \right)$

Therefore,

$\left( {m}^{2} - 14m - 32 \right) \div \left( m + 2 \right)$

$= \cfrac{\left( m - 1 6 \right) \left( m + 2 \right)}{\left( m + 2 \right)}$

$= \left( m - 16 \right)$

Simplify:
$z - 7 + 7xy - xyz$

Given:

$z-7+7xy-xyz$

$=z-xyz-7+7xy$

$= z(1-xy)-7(1-xy)$

$= (1-xy)(z-7)$

Factorise the following expressions.
$\left(1\right){p}^{2}+6p+8$
$\left(2\right){q}^{2}-10q+21$
$\left(3\right){p}^{2}+6p-16$

$1) p^{2}+6p+8$

$=p^{2}+4p+2p+8$

$=p(p+4)+2(p+4)$

$=(p+4)(p+2)$

$2)q^{2}-10q+21$

$=q^{2}3q-7q+21$

$=q(q-3)-7(q-3)$

$=(q-3)(q-7)$

$3)p^{2}+6p-16$

$=p^{2}-2p+8p-16$

$=p(p-2)+8(p-2)$

$=(p+8)(p-2)$

Simplify:
$15xy-6x+5y-2$

Given expression is

$15xy-6x+5y-2$ .

Simplifying the given expression by combining the like terms or taking out common parts:

$= (15xy+5y)-(6x+2)$

$= 5y(3x+1)-2(3x+1)$

$=( 5y-2)(3x+1)$

$= (3x+1)(5y-2)$

Hence,

$15xy-6x+5y-2=(3x+1)(5y-2)$ .

Factorize:
$x^2-100x+99$

Now,

$x^2-100x+99$

$=x^2-99x-x+99$

$=(x-99)(x-1)$ .

Factorise the expression
$am^{2}+bm^{2}+bn^{2}+an^{2}$

1306769_1457760_ans_fd3576e8e2514d339ec14fd82376ad85.jpeg

Factorise:
$z-7+7xy-xyz$

$z - 7 + 7xy - xyz$

$= z - xyz - 7 + 7xy$

$= z \left( 1 - xy \right) - 7 \left( 1 - xy \right)$

$= \left( 1 - xy \right) \left( z - 7 \right)$

Hence

$z - 7 + 7xy - xyz = \left( 1 - xy \right) \left( z - 7 \right)$ .

Factorize: ${ q }^{ 2 }-10 q+21.$

1318647_1490386_ans_5a852de0d51b48a5b332fac6c6b372c3.JPG

Factorise:
$6xy(a^2+b^2)+8yz(a^2+b^2)-10xz(a^2+b^2)$

$6xy \left( {a}^{2} + {b}^{2} \right) + 8yz \left( {a}^{2} + {b}^{2} \right) - 10xz v$

$= \left( {a}^{2} + {b}^{2} \right) \left( 6xy + 8yx - 10xz \right)$

$= 2 \left( {a}^{2} + {b}^{2} \right) \left( 3xy + 4 yz - 5 xz \right)$

Factorise:
$6xy(a^2+b^2)+8yz(a^2+b^2)-10xz(a^2+b^2)$

$6xy \left( {a}^{2} + {b}^{2} \right) + 8yz \left( {a}^{2} + {b}^{2} \right) - 10xz \left( {a}^{2} + {b}^{2} \right)$

$= \left( {a}^{2} + {b}^{2} \right) \left( 6xy + 8yz - 10xz \right)$

$= 2 \left( {a}^{2} + {b}^{2} \right) \left( 3xy + 4yz - 5xz \right)$

Factorise:
$36a^{2}+12abc-15b^{2}c^{2}$

$36 {a}^{2} + 12abc - 15 {b}^{2} {c}^{2}$

$= 3 \left( 12 {a}^{2} + 4 abc - 5 {b}^{2} {c}^{2} \right)$

$= 3 \left( 12 {a}^{2} + 10abc - 6 abc - 5 {b}^{2} {c}^{2} \right)$

$= 3 \left( 2a \left( 6a + 5bc \right) - bc \left( 6a - 5bc \right) \right)$

$= 3 \left( 2a - bc \right) \left( 6a + 5bc \right)$

Divide 48a $^{3}$ by 6a

1318686_1490209_ans_2f0ed174750149d7a09968bc0b2345ec.JPG

Factorize: $(p - x) (p + x) + y (p - x) = 0$

1323535_1491306_ans_b69fa613584642708fa22c18ff93ebae.JPG

Simplify:
$[4{ y }^{ 3 }+5{ y }^{ 2 }+6y]\div 2y$

Given:

$\dfrac{4{y}^{3}+5{y}^{2}+6y}{2y}$

$=\dfrac{4{y}^{3}}{2y}+\dfrac{5{y}^{2}}{2y}+\dfrac{6y}{2y}$

$=2{y}^{2}+\dfrac{5y}{2}+3$

Simplify:
$x^2-33x+90$

Now,

$x^2-33x+90$

$=x^2-30x-3x+90$

$=(x-30)(x-3)$ .

Factorize $y^2 + 10y + 24$

1894545_1473447_ans_923fe8a9e5c8441bbf70e8aeca29b1e6.jpg

Divide as directed.
5 $( 2 x + 1 ) ( 3 x + 5 ) \div ( 2 x + 1 )$

1326365_1494319_ans_447df0bea48043d6aa997407ed85c51b.JPG

Factorise the expression ${ 2a }^{ 3 }-{ 3a }^{ 2 }b+{ 5ab }^{ 2 }-ab$

Given:

$2a^{3}-3a^{2}b+5ab^{2}-ab$

$=a(2a^{2}-3ab+5b^{2}-b)$

Factorise $x^3-2x^2-x+2$

1311064_1535009_ans_22a361e1b29646c4b6db6bcdd6681caa.jpeg

Look at several examples of rational numbers in the form $\dfrac{p}{q}(q\neq 0)$ , where $p$ and $q$ are integers with no common factors other than $1$ and having terminating decimal representations. Can you guess what property $q$ must satisfy?

Prime factorization of

$q$ must have only powers of

$2$ or

$5$ or both.

Consider the following rational numbers:

$(i)\dfrac{7}{8}=\dfrac{7}{2^{3}}=\dfrac{7\times 5^{3}}{2^{3}\times 5^{3}}=\dfrac{875}{(2\times 5)^{3}}=\dfrac{875}{10^{3}}=\dfrac{875}{1000}=0.875$

$(ii) \dfrac{3}{40}=\dfrac{3}{2^{3}\times 5}=\dfrac{3\times 5^{2}}{2^{3}\times 5^{3}}=\dfrac{75}{(2\times 5)^{3}}=\dfrac{75}{1000}=0.075$

$(iii)\dfrac{4}{25}=\dfrac{4\times 2^{2}}{5^{2}\times 2^{2}}=\dfrac{4\times 4}{(5\times 2)^{2}}=\dfrac{16}{100}=0.16$

We observe that the denominators of all the above rational numbers are of the form

$2^{m}\times 5^{n}$ i.e., the prime factorizations of denominators have only powers of

$2$ or powers of

$5$ or both.

Factorise : $ax^{3}y^{2} + bx^{2}y^{3} + cx^{2}y^{2}z$ .

$ax^3y^2+bx^2y^3+cx^2y^2z$

Minimum power of

$x$ and

$y$ present in each term is 2. Hence

$x^2y^2$ can be taken common.

$ax^3y^2+bx^2y^3+cx^2y^2z=x^2y^2(ax+by+cz)$

Divide :
$x + 2x^{2} + 3x^{4} - x^{5}$ by $2x$ .

$\dfrac{x+2x^2+3x^4-x^5}{2x}$

$\Rightarrow \dfrac{x}{2x}+\dfrac{2x^2}{2x}+\dfrac{3x^4}{2x}-\dfrac{x^5}{2x}$

$\Rightarrow \dfrac{1}{2}+x+\dfrac{3}{2}x^3-\dfrac{1}{2}x^4$

Factorize:
$8{x}^{3}+27{y}^{3}+36{x}^{2}y+54x{y}^{2}$

1650964_1669428_ans_df813793b78f4c7eadce4e664522f0d5.jpg

Divide :
$9x^{2} y - 6xy + 12xy^{2}$ by $-\dfrac {3}{2} xy$ .

1507708_1672529_ans_75305cf7be9744e48497a11e04f54164.jpg

Divide :
$3x^{3}y^{2} + 2x^{2}y + 15xy$ by $3xy$ .

1507710_1672530_ans_1190e19e7a124ebfa351252b9d4ce0bc.jpg

Divide :
$5z^{3} - 6z^{2} + 7z$ by $2z$ .

$\dfrac{5z^3-6z^2+7z}{2z}$

$=\dfrac{5z^3}{2z}-\dfrac{6z^2}{2z}+\dfrac{7z}{2z}$

$=\dfrac{5}{2}z^2-\dfrac{6}{2}z+\dfrac{7}{2}$

$=\dfrac{5}{2}z^2-\dfrac{2\times 3}{2}z+\dfrac{7}{2}$

$=\dfrac{5}{2}z^2-3z+\dfrac{7}{2}$

Divide :
$y^{4} - 3y^{3} + \dfrac {1}{2} y^{2}$ by $3y$ .

$\dfrac{y^4-3y^3+\dfrac{1}{2}y^2}{3y}$

$=\dfrac{y^4}{3y}$

$-\dfrac{3y^3}{3y}$

$+\dfrac{1}{2\times 3y}y^2$

$=\dfrac{1}{3}y^3-y^2+\dfrac{1}{6}y$

Divide :
$-4a^{3} + 4a^{2} + a$ by $2a$ .

$\dfrac{-4a^3+4a^2+a}{2a}$

$\Rightarrow \dfrac{-4a^3}{2a}+\dfrac{4a^2}{2a}+\dfrac{a}{2a}$

$\Rightarrow \dfrac{-2\times 2a\times a^2}{2a}+\dfrac{2\times 2a\times a}{2a}+\dfrac{a}{2a}$

$\Rightarrow -2a^2+2a+\dfrac{1}{2}$

$2\sqrt 2{a}^{3}+3\sqrt 3{b}^{3}+{c}^{3}-3\sqrt 6abc$

1652202_1669523_ans_4abee9e24c0543c78d63beb9b5cac089.jpeg

Divide :
$4z^{3} + 6z^{2} - z$ by $-\dfrac {1}{2}z$ .

To divide

$(4z^{3} + 6z^{2} - z)$ by

$-\dfrac {1}{2}z$

Now

$\dfrac{4z^{3} + 6z^{2} - z}{-\dfrac {1}{2}z}$

$=-\dfrac{2(4z^{3} + 6z^{2} - z)}{z}$

$=-\dfrac{2z(4z^{2} + 6z - 1)}{z}$

$=-2(4z^{2} + 6z - 1)$

Hence, the answer is

$=-2(4z^{2} + 6z - 1)$

Divide :
$\sqrt {3}a^{4} + 2\sqrt {3} a^{3} + 3a^{2} - 6a$ by $3a$ .

Given expressions are

$\sqrt{3} a^4 + 2\sqrt{3} a^3 + 3a^2 - 6a$ and

$3a$ .

On dividing

$\sqrt{3} a^4 + 2\sqrt{3} a^3 + 3a^2 - 6a$ by

$3a$ , we get:

$\dfrac{\sqrt{3} a^4 + 2\sqrt{3} a^3 + 3a^2 - 6a}{3a}$

$=\dfrac{\sqrt{3} a^4}{3a} + \dfrac{2\sqrt{3} a^3}{3a} + \dfrac{3a^2}{3a} - \dfrac{6a}{3a}$

$=\dfrac{\sqrt{3} a^4}{\sqrt{3} \times \sqrt{3} a} + \dfrac{2 \sqrt{3} a^3}{\sqrt{3} \times \sqrt{3}a} + \dfrac{3a^2}{3a} - \dfrac{6a}{3a}$

$=\dfrac{1}{\sqrt{3}} a^3 + \dfrac{2}{\sqrt{3}} a^2 + a - 2$

Hence,

$\dfrac{\sqrt{3} a^4 + 2\sqrt{3} a^3 + 3a^2 - 6a}{3a}=\dfrac{1}{\sqrt{3}} a^3 + \dfrac{2}{\sqrt{3}} a^2 + a - 2$ .

Factorize the following.
$a^4b-3a^2b^2-6ab^3$ .

1682590_1673434_ans_6cd9410b98c5458fbe7a7680b91d4595.jpg

Factorize the following.
$x^4y^2-x^2y^4-x^4y^4$ .

1682601_1673437_ans_609c288551984d13a8081b35ffc0f225.jpg

Factorize the following.
$20x^3-40x^2+80x$ .

$\textbf{Step-1: Take greatest common factor common }$

$\text{Given polynomial :}$

$20x^3−40x^2+80x$

$\Rightarrow 20x(x^2-2x+4)$

$\textbf{Hence , 20x and }\boldsymbol{x^2 - 2x + 4}\textbf{ is the factors }$

Factorize the following.
$10m^3n^2+15m^4n-20m^2n^3$ .

1682453_1673429_ans_5fabf1fc58d1450a8b9ebdada4efb00b.jpg

Factorize the following.
$2l^2mn-3lm^2n+4lmn^2$ .

1682596_1673436_ans_e1a4e711cd71456a8ea56f94ca482462.jpg

Factorize the following.
$72x^6y^7-96x^7y^6$ .

1682372_1673424_ans_73175d5718934290999d9b32cbc8b3d1.jpg

Factorize the following expression.
$3x^3y-243xy^3$ .

$\textbf{Step 1: Factorize the given expression by using known identities.}$

$\text{Given expression,}$

$3x^3y-243xy^3$

$=3xy(x^2-81y^2)$

$=3xy[x^2-(9y)^2]$

$=3xy(x-9y)(x+9y)$

$[\because\boldsymbol{x^2-y^2=(x-y)(x+y)}]$

$\textbf{Hence, }\boldsymbol{3x^3y-243xy^3=3xy(x-9y)(x+9y).}$

Factorize the following expression.
$16(2x-1)^2-25y^2$ .

$\textbf{Step 1: Factorize the given expression by using known identities.}$

$\text{Given expression,}$

$16(2x-1)^2-25y^2$

$=4^2(2x-1)^2-5^2y^2$

$=[4(2x-1)]^2-[5y]^2$

$=[4(2x-1)-5y][4(2x-1)+5y]$

$[\because\boldsymbol{a^2-b^2=(a-b)(a+b)}]$

$=(8x-5y-4)(8x+5y-4)$

$\textbf{Hence, }\boldsymbol{16(2x-1)^2-25y^2=(8x-5y-4)(8x+5y-4).}$

Divide the polynomial $(9x^2 + 12x + 10)$ by $(3x + 2)$ and write the quotient and the remainder.

Divide

$(9x^2+12x+10)$ by

$(3x+2)$

Refer image.

So Quotient is

$3x+2$ and Remainder is

$6$ .

1501124_1700912_ans_31d3e45869a34ba38cb8b0704898c6f4.png

Factorise : $9x^2 + 12 xy$

Given :

$9x^ 2 + 12 xy$

Consider

$9x^ 2 + 12 xy$

Now by taking

$3x$ as common,

$9x^ 2 + 12 xy$

$= 3x(3x +4y)$

Factorise : $18x^2 y -24xyz$

Given :

$18x^2 y -24xyz$

Considering

$18x^2 y -24xyz$

Now by taking

$6xy$ as common in the equation
we get,

$18x^2 y -24xyz$

$= 6xy ( 3x -4z)$

Factorise : $a^3 +a -3a^2 -3$

Given :

$a^3 +a -3a^2 -3$

We can further write it as

$a^3 +a -3a^2 -3$

$= a(a^2 +1)-3(a^2 +1)$

So we get,

$a^3 +a -3a^2 -3$

$=(a-3)(a^2 +1 )$

Factorise : $2x+4y-8xy-1$

Given :

$2x+4y-8xy-1$

We can further write it as

$2x+4y-8xy-1$

$= 2x-1 -8xy +4y$

By taking common terms out

$2x+4y-8xy-1$

$=(2x-1)-4y(2x-1)$

So we get,

$2x+4y-8xy-1$

$=(2x-1)(1-4y)$

Factorise : $x^2 +y-xy -x$

Given :

$x^2 +y -xy -x$

By taking

$x$ as common in the first and

$1$ as common in the second term

$x^2 +y -xy -x$

$= x(x-y)-1(x-y)$

So we get,

$x^2 +y -xy -x$

$= (x-1)(x-y)$

Factorise : $abx^2+a^2x+b^2x+ab$

Given :

$abx^2+a^2x+b^2x+ab$

We can further write it as

$abx^2+a^2x+b^2x+ab$

$=ax(bx +a) +b(bx+a)$

So we get,

$abx^2+a^2x+b^2x+ab$

$=(ax+b)(bx+a)$

Factorise : $2a^2 + bc-2ab- ac$

Given :

$2a^2 + bc-2ab- ac$

We can further write it as

$2a^2 + bc-2ab- ac$

$=2a^2 -2ab -ac+ bc$

By taking the common terms out

$2a^2 + bc-2ab- ac$

$= 2a(a-b) -c(a-b)$

So we get,

$2a^2 + bc-2ab- ac$

$= (a-b) (2a-c)$

Factorise : $ab(x^2+y^2) -xy(a^2+b^2)$

Given :

$ab (x^2 +y^2) - xy (a^2 +b^2 )$

By multiplying the terms

$ab (x^2 +y^2) - xy (a^2 +b^2 )$

$= abx^2 +aby^2 -a^2xy -b^2 xy$

We can further write it as

$ab (x^2 +y^2) - xy (a^2 +b^2 )$

$= abx^2 -a^2xy +aby^2 -b^2xy$

By taking out the common terms out

$ab (x^2 +y^2) - xy (a^2 +b^2 )$

$=ax(bx-ay) +by(ay-bx)$

By taking the negative sign out

$ab (x^2 +y^2) - xy (a^2 +b^2 )$

$=ax(bx-ay) -by(bx -ay)$

So we get ,

$ab (x^2 +y^2) - xy (a^2 +b^2 )$

$= ( ax-by)(bx-ay)$

Factorise : $a^2 +ab(b+1)+b^3$

Given :

$a^2 +ab(b+1)+b^3$

By multiplying the terms

$a^2 +ab(b+1)+b^3$

$=a^2 +ab^2 +ab+b^3$

We can further write it as

$a^2 +ab(b+1)+b^3$

$= a^2 + ab+ab^2+ b^3$

By taking the common terms out

$a^2 +ab(b+1)+b^3$

$= a(a+b) +b^2 (a+b)$

So we get,

$a^2 +ab(b+1)+b^3$

$= (a+b^2)(a+b)$

Factorise : $20x^2 -45$

Given :

$20x^2 -45$

By taking

$5$ as common

$20x^2 -45$

$= 5(4x^2 -9)$

We can write it as

$20x^2 -45$

$= 5((2x)^2 -3^2 )$

Based on the equation

$a^2 -b^2 =(a+b)(a-b)$

We get,

$20x^2 -45$

$= 5(2x+3)(2x-3)$

Factorise : $a(a+b-c)-bc$

Given :

$a(a+b-c)-bc$

We can further write it as

$a(a+b-c)-bc$

$=a^2 +ab -ac-bc$

By taking the common terms out

$a(a+b-c)-bc$

$= a(a+b) -c(a+b)$

So we get,

$a(a+b-c)-bc$

$=(a+b)(a-c)$

Factorise : $ab(x^2+ 1) +x(a^2 +b^2 )$

1563622_1714984_ans_b3970a533e9341cdb5e919c384fba9e4.png

Factorise : $a^2 -b^2 -a -b$

$\textbf{Step 1 : Simplify the form}$

$\text{Consider } a^2 - b^2 - a - b\space$

$=(a - b)(a + b) - a - b$

$\quad \quad \quad \mathbf{[(x^2-y^2) = (x-y)(x+y)]}$

$=(a + b ){(a - b - 1 )}$

$\quad \textbf{[Take (a-b) common from both the terms]}$

$\textbf{Hence , }\boldsymbol{a^2 - b^2 - a - b} \textbf{ is equal to } \boldsymbol{(a + b)(a - b - 1)}$

Factorise : $9a^2 +6a+ 1 -36b^2$

Given :

$9a^2 +6a+ 1-36b^2$

We can further write its as

$9a^2 +6a+ 1-36b^2$

$= (9a^2 +6a +1 )-36b^2$

So we get

$9a^2 +6a+ 1-36b^2$

$= [(3a)^2 +2 \times 3 \times 1 +1^2 ]-(6b)^2$

$9a^2 +6a+ 1-36b^2$

$= (3a+1)^2 -(6b)^2$

Based on the equation

$a^2 -b^2= (a+b)(a-b)$

We get,

$9a^2 +6a+ 1-36b^2$

$=(3a+1+6b)(3a+1-6b)$

Factorise : $4a^2 -4b^2 +4a +1$

Given :

$4a^2 -ab^2 +4a +1$

We can write it as

$4a^2 -ab^2 +4a +1$

$=(4a^2 +4a +1 )-4b^2$

We get,

$4a^2 -ab^2 +4a +1$

$= [(2a)^2 +2 \times 2a \times 1 +1^2]-(2b)^2$

On further calculation

$4a^2 -ab^2 +4a +1$

$=(2a +1)^2 -(2b)^2$

Based on the equation

$a^2 -b^2 = (a+b)(a-b)$

Using the formula

$4a^2 -ab^2 +4a +1$

$=(2a +1+ 2b)(2a+1-2b)$

So we get

$4a^2 -ab^2 +4a +1$

$= ( 2a+ 2b+1)(2a-2b +1 )$

Factorise : $a^2 -b^2 -4ac +4c^2$

Given :

$a^2 -b^2-4ac +4c^2$

We can write it as

$a^2 -b^2-4ac +4c^2$

$= a^2 -4ac +4c^2 -b^2$

So we get

$a^2 -b^2-4ac +4c^2$

$=a^2 -2 \times a \times 2x + 2c^2 -b^2$

$a^2 -b^2-4ac +4c^2$

$= (a-2c)^2 -b^2$

Based on the equation

$a^2 -b^2= (a+b)(a-b)$

We get,

$a^2 -b^2-4ac +4c^2$

$=(a-2c +b)(a-2c-b)$

Factorise : $5x^2 -16x -21$

Given :

$5x^2 -16x-21$

We can further write it as

$5x^2 -16x-21$

$=5x^2 +5x -21x -21$

by taking out the common terms

$5x^2 -16x-21$

$= 5x(x+1)-21(x+1)$

So we get,

$5x^2 -16x-21$

$=(5x-21)(x+1)$

Factorise : $18x^2 + 3x -10$

Given :

$18x^2 +3x -10$

We can further write it as:

$18x^2 +3x -10$

$= 18x^2 -12 x+ 15x -10$

By taking out the common terms:

$18x^2 +3x -10$

$= 6x(3x-2) +5(3x-2)$

So we get,

$18x^2 +3x -10$

$=(6x+5)(3x-2)$

Factorise : $25x^2 -10x+ 1-36y^2$

Given :

$25x^2 -10 x+ 1-36y^2$

We can further write the given question as

$25x^2 -10 x+ 1-36y^2$

$=(25x^2 -10 x+ 1)-36y^2$

So we get

$25x^2 -10 x+ 1-36y^2$

$=[(5x)^2 -2 \times 5x \times 1 +1^2]-(6y)^2$

$25x^2 -10 x+ 1-36y^2$

$=(5x -1)^2 -(6y)^2$

Based on the equation

$a^2 -b^2 =(a+b)(a-b)$

We get,

$25x^2 -10 x+ 1-36y^2$

$=(5x-1 +6y)(5x-1-6y)$

Factorise : $a^2 +2ab +b^2 -9c^2$

Given :

$a^2 +2ab +b^2 -9c^2$

We can write the given question as

$a^2 +2ab +b^2 -9c^2$

$=(a +b)^2 -(3c)^2$

Based on the equation

$a^2 -b^2 =(a+b)(a-b)$

We get,

$a^2 +2ab +b^2 -9c^2$

$=( a+ b+ 3c)(a+b-3c)$

Factorise : $x^3 - 5x^2 -x+5$

Given :

$x^3 -5x^2 -x+5$

By taking common terms

$x^3 -5x^2 -x+5$

$=x^2 (x-5)-1(x-5)$

So we get

$x^3 -5x^2 -x+5$

$=(x-5)(x^2 -1)$

Based on the equation

$a^2 -b^2= (a+b)(a-b)$

We get,

$x^3 -5x^2 -x+5$

$=(x-5)(x+1)(x-1)$

Factorise : $a^2 -b^2 +2bc - c^2$

Given :

$a^2-b^2+2bc-c^2$

We can write it as

$a^2-b^2+2bc-c^2$

$= a^2 -(b^2 -2bc +c^2 )$

We know that,

$(a-b)^2=a^2-2ab+b^2$

$a^2-(b^2-2bc+c^2)$

$= a^2 -(b-c)^2$

Based on the identity,

$a^2 -b^2 = (a+b)(a-b)$

We get,

$a^2-(b-c)^2$

$= [a + (b-c)[a- (b-c)]$

$\implies a^2-(b-c)^2$

$=( a+b-c)(a-b+c)$

$\therefore$

$a^2-b^2+2bc-c^2=(a+b-c)(a-b+c)$

Factorise : $x^2 +y^2 -z^2 -2xy$

Given :

$x^2 +y^2 -z^2 -2xy$

We can write it as

$x^2 +y^2 -z^2 -2xy$

$=(x^2 +y^2 -2xy)-z^2$

So we get

$x^2 +y^2 -z^2 -2xy$

$= ( x-y)^2 -z^2$

Based on the equation

$a^2 -b^2 =(a+b)(a-b)$

We get ,

$x^2 +y^2 -z^2 -2xy$

$=(x-y+z)(x-y-z)$

Divide:
$8x^{2}y^{2}-6xy^{2}+10 x^{2}y^{3}$ by $2xy$

$8x^2y^2-6xy^2+10x^2y^3$ by

$2xy$

On dividing polynomial by a monomial we have divide every variables of polynomial.

$\Rightarrow$

$\dfrac{8x^2y^2-6xy^2+10x^2y^3}{2xy}$

$=\dfrac{8x^2y^2}{2xy}-\dfrac{6xy^2}{2xy}+\dfrac{10x^2y^3}{2xy}$

$=\dfrac{8}{2}x^{2-1}y^{2-1}-\dfrac{6}{2}x^{1-1}y^{2-1}+\dfrac{10}{2}x^{2-1}y^{3-1}$

$=4xy-3y+5xy^2$

Factorise: $36x^3y –60x^2y^3z$

Given:

$36x^3y - 60x^2y^3z$

By taking

$12x^2y$ as a common factor for the above equation we get,

$36x^3y - 60x^2y^3z = 12x^2y(3x - 5y^2z)$

Factorise : $x^4y^4 - xy$

Given expression is

$x^4y^4-xy$

We can take out

$xy$ as common.

Hence, the expression becomes:

$xy \left(x^3y^3-1\right)$

$\Rightarrow xy \left[(xy)^3-1^3\right]$

We know that,

$a^3-b^3=(a-b)(a^2+b^2+ab)$

Hence, the expression becomes:

$xy\left[(xy-1)((xy)^2+1^2+1(xy))\right]$

$\Rightarrow xy\left[(xy-1)(x^2y^2+1+xy)\right]$

$\Rightarrow xy(xy-1)(x^2y^2+xy+1)$

Hence,

$x^4y^4-xy=xy(xy-1)(x^2y^2+xy+1)$ .

Divide $12x^{4}+8x^{3}-6x^{2}$ by $-2x^{2}$

While dividing a polynomial by a monomial, we have to divide every variables of polynomial.

$\Rightarrow$

$\dfrac{12x^4+8x^3-6x^2}{-2x^2}=\dfrac{12x^4}{-2x^2}+\dfrac{8x^3}{-2x^2}-\dfrac{6x^2}{-2x^2}$

$=-\dfrac{12}{2}x^{4-2}-\dfrac{8}{2}x^{3-2}+\dfrac{6}{2}x^{2-2}$

$=-6x^2-4x+3$

Factorise : $16x^4 +54 x$

Given :

$16x^4+54x$

We can write the given question as,

$16x^4+54x=2x(8x^3+27)$

According to the equation,

$a^3+b^3=(a+b)(a^2-ab+b^2)$

So we get,

$2x(8x^3+27)$

$=2x((2x)^3+3^3)$

Using the equation,

$=2x((2x)+3)((2x)^2-(2x)(3)+(3)^2)$

$16x^4+54x$

$= 2x(2x+3)(4x^2-6x+9)$

Factorise : $x - 8xy^3$

Given :

$x-8xy^3$

We can write the given question as,

$x-8xy^3=x(1-8y^3)$

According to the equation,

$a^3-b^3=(a-b)(a^2+ab+b^2)$

So we get,

$x(1-8y^3)$

$=x(1^3-(2y)^3)$

using the equation,

$=x[(1-2y)(1^2+2y+(2y)^2)]$

$x-8xy^3$

$=x(1-2y)(1+2y+4y^2)$

Divide:
$9x^{2}y-6xy+12xy^{2}$ by $-3xy$

${\textbf{Step - 1: Write the equivalent expression by diving each term of polynomial }}$

${\textbf{by the monomial}}{\text{.}}$

${\text{Dividing each term by the monomial we get, }}\\\Rightarrow\dfrac{{{\text{9}}{{\text{x}}^{\text{2}}}{\text{y}}}}{{{\text{ - 3xy}}}}{\text{ - }}\dfrac{{{\text{6xy}}}}{{{\text{ - 3xy}}}}{\text{ + }}\dfrac{{{\text{12x}}{{\text{y}}^{\text{2}}}}}{{{\text{ - 3xy}}}}$

${\text{= ( - 3}}{{\text{x}}^{{\text{2 - 1}}}}{{\text{y}}^{{\text{1 - 1}}}}{\text{ + 2}}{{\text{x}}^{{\text{1 - 1}}}}{{\text{y}}^{{\text{1 - 1}}}}{\text{ - 4}}{{\text{x}}^{{\text{1 - 1}}}}{{\text{y}}^{{\text{2 - 1}}}}{\text{)}}{\text{.}}$

${\text{= - 3x + 2 - 4y}}$

${\textbf{Hence, the required answer is }}{\mathbf{ - 3x + 2 - 4y}}{\text{.}}$

Find the values of a and b so that the polynomial $(x^4+ax^3-7x^2+8x+b)$ is exactly divisible by $(x+2)$ as well as $(x+3)$ .

We are given that polynomial

$(x^4+ax^3-7x^2+8x+b)$ is exactly divisible by

$(x+2)$ as well as

$(x+3)$ .

The two divisors in the question are

$(x + 2)$ and

$(x + 3)$ . As it is given that the given polynomial is divisible by both these divisors, that means;

$x + 2 = 0$ and

$x + 3 = 0$

So,

$x = -2$ and

$x = -3$ will make the remainder zero when these values of x are substituted in the given polynomial.

$f(x) = x^{4}+ax^{3} -7x^{2} +8x+b$

So,

$f(-2)$ and

$f(-3)$ will be equal to zero.

$f(-2) = (-2)^{4}+a(-2)^{3} -7(-2)^{2} +8(-2)+b =0$

$f(-3) = (-3)^{4}+a(-3)^{3} -7(-3)^{2} +8(-3)+b =0$

$f(-2) = 16-8a -28 -16+b =0$

$\Rightarrow -8a+b-28=0\quad \quad ....(1)$

$f(-3) = 81-27a -63 -24+b =0$

$\Rightarrow -27a+b -6 =0 \quad \quad ....(2)$

Now, subtracting

$(2)$ from

$(1)$ :

$-8a+b -28 =0$

$-27a+b -6 =0$

+ - + -

$\overline{\ \ 19a - \ \ 22 \ \ =\ \ 0}$

$\Rightarrow a = \dfrac{22}{19}$

Putting the value of

$a$ in

$(1)$ , we get;

$-8a+b -28 =0$

$-8\left(\dfrac{22}{19}\right) +b -28 =0$

$\Rightarrow b = 28 + \dfrac{176}{19}$

$\Rightarrow b = \dfrac{708}{19}$

Hence, the value of

$a$ is

$\dfrac{22}{19}$ and the value of

$b$ is

$\dfrac{708}{19}$ .

Find the value of 'a' for which the polynomial $(x^4-x^3-11x^2-x+a)$ is divisible by $(x+3)$ .

$P(x) = x^4-x^3-11x-x+a$

$= x^4-x^3-12x + a$

$P(-3) = 81+27+36+a =0$

So,

$a = -144.$

Factorise : $1029 - 3x^3$

1673382_1715284_ans_973915ff2da440409cdb2e60fd174cf3.jpg

Factorise:
$16a^2 – 24ab$

Given:

$16a^2$

$– 24ab$

By taking

$8a$ as a common factor for the above equation we get,

$16a^2$

$– 24ab = 8a (2a – 3b)$

Factorise :
$14x^3+21x^4y-28x^2y^2$

let’s take HCF of above equationBy taking

$7x^2$ as a common factor for the above equation we get,

$14x^3+21x^4y-28x^2y^2 = 7x^2(2x + 3x^2y- 4y^2)$

Factorise

$9x^3 – 6x^2 + 12x$

let’s take HCF of above equationBy taking

$3x$ as a common factor for the above equation we get,

$9x^3 -6x^2 + 12x = 3x(3x^2 - 2x + 4)$

Factorise

$6a(a-2b) + 5b(a-2b)$

By taking

$a-2b$ as a common factor for the above equation we get,

$6a (a-2b) + 5b (a-2b) = (a-2b) (6a + 5b)$

Factorise

$10x^3 -15x^2$

let’s take HCF of above equationBy taking

$5x^2$ as a common factor for the above equation we get,

$10x^3 - 15x^2 = 5x^2(2x-3)$

Factorise

$2m(1-n) + 3(1-n)$

By taking

$(1-n)$ as a common factor for the above equation we get

$\Rightarrow 2m (1-n) + 3(1-n) = (1-n) (2m + 3)$

The sum of first $n$ natural numbers is given by the expression $\dfrac{n^{2}}{2} + \dfrac{n}{2}$ . Factorise this expression.

Given,

Sum of first

$n$ natural numbers

$=\dfrac{n^{2}}{2} + \dfrac{n}{2}$

Factorizing the given expression, we get,

$\dfrac{n^{2}}{2} + \dfrac{n}{2}$

$= \dfrac{1}{2}(n^{2}+n)$

$= \dfrac{1}{2} n(n +1)$

$\therefore \dfrac{n^{2}}{2} + \dfrac{n}{2}= \dfrac{1}{2} n(n +1)$

Perform the following division:
$(3pqr-6p^{2}q^{2}r^{2}) \div 3pq$

The dividion is carried out as follows :

$\dfrac{3pqr-6p^{2}q^{2}r^{2}}{3pq}$

$\\=\dfrac{3pqr}{3pq} -\dfrac{6p^{2}q^{2}r^{2}}{3pq}$

$=\dfrac {3\cdot p\cdot q\cdot r}{3\cdot p\cdot q}-\dfrac{2\cdot3\cdot p\cdot p\cdot q\cdot r\cdot r}{3\cdot p\cdot q}$

Cancelling out the common factors in numerator and denominator,

$=r-2\cdot p\cdot q\cdot r\cdot r$

$=r-2pqr^2$

Write the greatest common factor of the following terms $-18a^{2}$ , $108a$

We have

$-18a^2=(-1)\times 18\times a\times a$

$108a=18\times 6\times a$

$\therefore$ Greatest common factor is

$18a$

Write the greatest common factor of the following terms $3x^{2}y$ , $18xy^{2}$ , $-6xy$

We have

$3x^2y=3\times x\times x\times y$

$18xy^2=3\times 6\times x \times y\times y$

$-6xy=(-1)\times 6\times x \times y$

$\therefore$ Greatest common factor is

$3xy$

The common factor method of factorisation for a polynomial is based on __________ property.

1974136_1792827_ans_8a98965c4b0b45f48214d05f730d17ed.jpg

The factorisation of 2x + 4y is __________ .

We have,

$2x + 4y$

Factoring

$2$ out, we get

$= 2(x+2y)$

Hence

$2x + 4y= 2(x+2y)$

Factorise the following polynomials :
(i) $25 abc^2 - 15a^2b^2c$
(ii) $x^2yz + xy^2z + xyz^2$

$(i) 25abc^2-15a^2b^2c=5abc(5c-3ab)$

$(ii) x^2yz+xy^2z+xyz^2=xyz(x+y+z)$

Factorise the following polynomials :
(i) $8xy^3 + 12x^2 y^2$
(ii) $15ax^3 - 9ax^2$

$(i) 8xy^3 + 12x^2y^2=4xy^2(2y+3x)$

$(ii) 15 ax^3-9ax^2=3ax^2(5x-3)$

Factorise the following polynomials :
$10a( 2p +q)^3 - 15b ( 2p +q)^2 + 35 ( 2p + q)$

$10a(2p+q)^3-15b(2p+q)^2+35(2p+q)$

$5[2a(2p+q)]^3-3b(2p+q)^2+7(2p+q)]$

$5(2p+q)[2a(2p+q)^2-3b(2p+q)+7 ]$

Factorise the following polynomials :
(i) $21 py^2 - 56 py$
(ii) $4x^3 - 6x^2$

$(i) 21py^2-56py=7py(3y-8)$

$(ii) 4x^3-6x^2=2x^2(2x-3)$

Factorise the following polynomials :
(i) $8x^3 - 6x^2 + 10 x$
(ii) $14mn + 22m -62 p$

$(i) 8x^3-6x^2+10x=2x(4x^2-3x+5)$

(Take

$2x$ common from each term)

$(ii) 14mn +22 m-62p=2(7mn+11m-31p)$

(Take

$2$ common from each term)

Factorise the following polynomials :
(i) $6(x+2y)^3 + 8 ( x+ 2 y)^2$
(ii) $14 (a -3b)^3 - 21 p(a -3b)$

$(i) \ 6(x+2y)^3+8(x+2y)^2$

Taking

$(x+2y)^2$ common, we get:

$(x+2y)^2[6(x+2y)+8]$

$= (x+2y)^2[6x+12y+8]$

$=(x+2y)^2(2)(3x+6y+4)$

$=2(x+2y)^2(3x+6y+4)$

Hence,

$6(x+2y)^3+8(x+2y)^2 =2(x+2y)^2(3x+6y+4)$

$(ii) \ 14(a-3b)^3-21p(a-3b)$

Taking

$7$ common, we get:

$7[2(a-3b)^3-3p(a-3b)]$

$= 7(a-3b)\{2(a-3b)^2-3p\}$

Hence,

$14(a-3b)^3-21p(a-3b) = 7(a-3b)\{2(a-3b)^2-3p\}$

Factorise the following polynomials:
(i) $18p^2q^2 -24pq^2 +30p^2q$
(ii) $27a^3b^3 - 18a^2 b^3 + 75 a^3 b^2$

$(i)$ Factor out

$6pq$ from the given expression,

$18{p^2}{q^2} - 24p{q^2} + 30{p^2}q = 6pq(3pq - 4q + 5p)$

$(ii)$ Factor out

$3 a^2b^2$ from the given expression.

$27a^3b^3-18a^2b^3+75a^3b^2= 3a^2b^2(9ab-6b+25a)$

Find the quotient and the remainder when $P(x)=3 x^{3}+x^{2}+2 x+5$ is divided by
$g(x)=x^{2}+2 x+1 \qquad\qquad$

The quotient and remainder when

$P(x)=3 x^{3}+x^{2}+2 x+5 \quad$ is divided by

$g(x)=x^{2}+2 x+1$ are

$3 x-5$ and

$9 x+10$ respectively.
1679202_1803908_ans_f7d0011ac1884947b3d124d9d6a43d64.png

Factorise the expressions and divide them as directed:
$(x^{2}-22x+117) \div (x-13)$

Factorising the dividend,

$x^2-22x+117$

$=x^2-13x-9x+117\qquad$ [Splitting the middle term]

$=x(x-13)-9(x-13)$

$=(x-13)(x-9)$

$\therefore (x^{2}-22x+117) \div (x-13)$

$=\dfrac{x^{2}-22x+117}{x-13}$

$=\dfrac{(x-13)(x-9)}{x-13}$

Cancelling out the common factor,

$=x-9$

Factorise the following polynomials :
(i) $15a( 2p - 3p) - 10b ( 2 p - 3q)$
(ii) $3a(x^2 +y^2) + 6b ( x^2 +y^2 )$

$(i) 15a(2p-3q)-10b(2p-3q)$

$(2p-3q)(15a-10b)$

$(2p-3q)(5)(3a-2b)$

$5(2p-3q)(3a-2b)$

$(ii) 3a(x^2+y^2)+6b(x^2+y^2)$

$(x^2+y^2)(3a+6b)$

$(x^2+y^2)(3)(a+2b)$

$3(x^2+y^2)(a+2b)$

Divide:
$9x^4 - 8x^3 - 12x + 3$ by $3x$

$\frac{9x^4 - - 8x^3 - 12x + 3}{3x}$
=

$\frac{9x^4}{3x} - \frac{8x^3}{3x} - \frac{12x}{3x} + \frac{3}{3x}$
=

$3x^3 - \frac{8}{3}x^2 - 4 + \frac{1}{x}$

$2 \pi r^2 - 4 \pi r$

$2 \pi r^2-4 \pi r$
Take out common in both terms,
Then,

$2 \pi r (r-2)$
Therefore,HCF of

$2\pi r^2 \quad and \quad 4 \pi r \quad is \quad 2 \pi r..$

$\text{Factorise } x^3 +x +2$

${\textbf{Step -1: Simplification}}$

${\text{As we know that }}{{\text{x}}^3} + {1} = (x + 1)({x^2} - x + 1)$

${\text{Given, }}{{\text{x}}^3} + 1 + x + 1$

$= (x + 1)({x^2} - x + 1) + x + 1$

$= (x + 1)({x^2} - x + 1) + (x + 1)$

$= (x + 1)({x^2} - x + 1 + 1)$

$= (x + 1)({x^2} - x + 2)$

${\textbf{Hence, the answer is }}\mathbf{(x + 1)({x^2} - x + 2)}$

$32x^4 -500 x$

$32x^4 -500 x$
Take out common in all terms we get,

$4x(8x^3-125)$
Above terms can be written as,

$4x((2x)^3-5^3)$
We know that,

$a^3-b^3=(a-b)(a^2+ab+b^2)$
Where,

$a=2x,b=5$
=

$4x(2x-5)((2x)^2+(2x \times 5)+5^2)$
=

$4x(2x-5)(4x^2+10x+25)$

$(x^6 / 343) +( 343 / x^6 )$

$(x^6 / 343) +( 343 / x^6 )$
Above terms can be written as,

$(x^2/7)^3+(7/x^2)^3$
We know that,

$a^3+b^3=(a+b)(a^2-ab+b^2)$
Where,

$a=(x^2/7),b=(7/x^2)$
Then,

$(x^2/7)^3+(7/x^2)^3=[(x^2/7)+(7/x^2)][(x^2/7)^2-((x^2/7)\times (7/x^2))+(7/x^2)^2]$
=

$[(x^2/7)+(7/x^2)][(x^4/49)-1+(49/x^4)]$

$x^2 +x^5$

$x^2 +x^5$
Take out common in all terms we get,

$x^2(1+x^3)$

$x^2(1^3+x^3)$
We know that,

$a^3+b^3=(a+b)(a^2-ab+b^2)$
Where,

$a=1,b=x$
=

$x^2[(1+x)(1^2-(1\times x)+x^2)]$
=

$x^2(1+x)(1-x+x^2)$

$a^3 - a -120$

$a^3 - a -120$
Above terms can be written as,

$a^3-a-125+5$
Rearranging the above terms,we get

$a^3-125-a+5$

$(a^3-125)-(a-5)$

$(a^3-5^3)-(a-5)$
We know that,

$a^3-b^3=(a-b)(a^2+ab+b^2)$

$[(a-5)(a^2+5a+5^2)]-(a-b)$

$(a-5)(a^2+5a+25)-(a-5)$

$(a-5)(a^2+5a+25-1)$

$(a-5)(a^2+5a+24)$

$a^2(b+c) - ( b+c)^3$

$a^2(b+c) - ( b+c)^3$
Take out common in all terms,

$(b+c)(a^2 - (b+c)^2)$
We know that

$a^2 -b^2 = (a+b)(a- b)$

$( b+c)(a+b+c) (a-b-c)$

$x^4 +5x^2 + 9$

$x^4 + 5x^2 + 9$

$x^4 + 6x^2 – x^2 + 9$

$(x^4 + 6x^2 + 9) – x^2$

$((x^2)2 + (2 \times x^2 \times 3) + 3^2)$
We know that,

$(a + b)^2 = a^2 + 2ab + b^2$ ,

$((x^2)^2 + (2 \times x^2 \times 3) + 3^2)$
So,

$(x^2 + 3)^2 – x^2$
We know that,

$a^2 – b^2 = (a + b) (a – b)$

$(x^2 + 3 + x) (x^2 + 3 – x)$

$2x^4 -32$

$2x^4 -32$
Take out common in all terms,

$2(x^4 -16)$
Above terms can be written as,

$2((x^2)^2 - 4^2 )$
We know that

$a^2 -b^2 =(a +b)(a-b)$

$2(x^2 + 4)(x^2 -4)$

$2(x^2 +4)(x^2 -2^2)$

$2(x^2 +4)(x+2)(x-2)$

$x^4 - 1/ x^4$

$x^4 - 1/x^4 =(x^2 +1/x^2)(x^2 -1/x^2)$

$= ( x^2 +1/x^2)(x+1/x)(x-1/x)$
From

$(i), (ii)$ and

$(iii)$ substitute the values we get,

$= ( 97 /36) \times ( 13 / 6) \times ( 5/ 6)$

$= 6305 / 1296$

$a^4 +b^4 - 7a^2b^2$

$a^4 +b^4- 7a^2b^2$
Above terms can be written as

$a^4 +b^4 +2a^2 b62 -9a^2 b^2$
we know that

$(a+b)^2 =a^2 +2ab +b^2$

$[ (a^2)^2 + (b^2)^2 +( 2 \times a^2 \times b^2)] - (3ab)^2$

$( a^2 +b^2)^2 -(3ab)^2$
we know that

$(a+b)^2 =a^2 +2ab +b^2$

$[(a^2)^2 +(b^2)^2 + ( 2 \times a^2 \times b^2 ) ] -(3ab)^2$

$(a^2 +b^2)^2 - (3ab)^2$
We know that

$a^2 -b^2 = ( a+b)(a-b)$

$(a^2 +b^2 +3ab)(a^2 +b^2 -3ab)$

Divide:
$9 x^3 - 6x^2\,\, by\,\, 3x$

We have to divide

$9x^3 - 6x^2\,\, by\,\, 3x$

$9x^3 - 6x^2 \div 3x = \dfrac{(9 \times x^3 - 6 \times x^2) }{ (3 \times x)}$

Separating the terms, we get

$=\dfrac{ (9 \times x^3) }{ (3 \times x)} - \dfrac{(6 \times x^2) }{ (3 \times x)}$

We get,

$= 3 \times x^{3-1} - 2 \times x^{2-1}$

$= 3x^2 - 2x$

Hence,

$9x^3 - 6x^2 \div 3x = 3x^2 - 2x$

$5a^4 - 5a^3 +30a^2 -30 a$

$5a^4 - 5a^3 +30a^2 -30 a$
Take out common in both terms,

$5a(a^3-a^2+6a-6)$

$5a[a^2(a-1)+6(a-1)]$

$5a(a-1)(a^2+6)$

$15(2x-3)^3 - 10 (2x-3 )$

$15(2x-3)^3 - 10 (2x-3 )$
Take out common in both terms,
Then,

$5(2x-3)[3(2x-3)^2-2]$

Factorize the following
$a^{12}x^{4} - a^{4} x^{12}$

$[(a^6)^2 (x^2)^2 - (a^2)^2 (x^6)^2]$

$= [(a^6x^2)^2 - (a^2 x^6)^2]$

$= (a^6 x^2 + a^2x^6)(a^6 x^2 - a^2 x^6)$

$= a^2x^2 (a^4 + x^4) [(a^3)^2 x^2 - a^2(x^3)^2]$

$= a^2x^2 (a^4 + x^4) [(a^3x)^2 - (ax^3)^2]$

$= a^2x^2 (a^4 + x^4)[(a^3x + ax^3) (a^3x - ax^3)]$

$=a^2x^2 (a^4 + x^4) [ax(a^2 + x^2). ax(a^2 - x^2)]$

$= a^2x^2 (a^4 + x^4)[a^2x^2 (a^2 + x^2)(a^2 - x^2)]$

$= a^4 x^4 (a^4 + x^4)(a^2 + x^2)(a + x)(a - x)$

Divide:
$15x^3y^2 + 25x^2y^3 - 36x^4y^4 \,\,by\,\, 5x^2y^2$

We have to devide

$15x^3y^2 + 25x^2y^3 - 36x^4y^4 \,\,by\,\, 5x^2y^2$

$15x^3y^2 + 25x^2y^3 - 36x^4y^4 \div 5x^2y^2 \\= \dfrac{(15x^3y^2 + 25x^2y^3 - 36x^4y^4) }{ (5x^2y^2)}$

$= \dfrac{(15 \times x^3 \times y^3)} {(5 \times x^2 \times y^2)} +\dfrac{ (25 \times x^2 \times y^3) }{ (5 \times x^2 \times y^2)} - \dfrac{(36 \times x^4 \times y^4) }{ (5 \times x^2 \times y^2)}$

On further calculation, we get

$= 3 \times x^{3-2} \times y^{2-2} + 5 \times x^{2-2} \times y^{3-2} -\dfrac{ (36 \times x^{4-2} \times y^{4-2})} 5$

$= 3 \times x^1 \times y^0 + 5 \times x^0 \times y^1 - \dfrac{(36 \times x^2 \times y^2) } 5$

$= 3x + 5y - \dfrac{(36x^2y^2) }5$

Hence,

$15x^3y^2 + 25x^2y^3 - 36x^4y^4 \div 5x^2y^2 = 3x + 5y - \dfrac{(36x^2y^2) }5$

$a^3 - ( 1/a^3) -2a +2/a$

$a^{12}x^4 - a^4 x^{12}$

$a^{12}x^4 - a^4 x^{12}$
Take out common in both terms,

$a^4x^4(a^8-x^8)$

$a^4x^4((a^4)^2-(x^4)^2)$
We know that,

$(a^2-b^2)=(a+b)(a-b)$

$a^4x^4(a^4+x^4)(a^4-x^4)$

$a^4x^4(a^4+x^4)((a^2)^2-(x^2)^2)$

$a^4x^4(a^4+x^4)(a^2+x^2)(a^2-x^2)$

$a^4x^4(a^4+x^2)(a+x)(a-x)$

$x^3 -(8 /x)$

Above terms can be written as,

$(1/x)(x^3-8)$

$(1/x)[(x)^3-(2)^3]$

We know that,

$a^3-b^3=(a-b)(a^2+ab+b^2)$

So,

$a=x,b=2$

$(1/x)(x-2)(x^2+2x+4)$

Carry out the following divisions:
$(x^3 + 2x^2 + 3x) \div 2x$

$\begin{aligned}{}\left( {{x^3} + 2{x^2} + 3x} \right) \div 2x &= \frac{{{x^3} + 2{x^2} + 3x}}{{2x}}\\ &= \frac{{{x^3}}}{{2x}} + \frac{{2{x^2}}}{{2x}} + \frac{{3x}}{{2x}}\\ &= \frac{1}{2}{x^2} + x + \frac{3}{2}\end{aligned}$

$9x^3y +41x^2y^2 +20xy^3$

$9x^3y +41x^2y^2 +20xy^3$
Take out common in all terms,

$xy(9x^2+41xy+y^2)$
Above terms can be written as,

$xy(9x^2+36xy+5xy+20y^2)$

$xy[9x(x+4y)+5y(x+4y)]$

$xy(x+4y)(9x+5y)$

Divide:
$6m^2 - 16 m^3 + 10 m^4\,\, by\,\, - 2m$

We have to divide

$6 m^2 - 16m^3 + 10m^4 \,\,by\,\, - 2m$

$6m^2 - 16m^3 + 10 m^4 \div - 2m\\ = \dfrac{(6 \times m^2- 16 \times m^3 + 10 \times m^4) }{ - 2 \times m}\\$

Separating the terms, we get

$= \dfrac{(6 \times m^2)}{ (- 2 \times m)} - \dfrac{(16 \times m^3)}{ (- 2 \times m)} +\dfrac{ (10 \times m^4) }{ (- 2 \times m)}$

$= - 3 \times m^{2-1} + 8 \times m^{3-1} - 5 \times m^{4-1}$

$= - 3 \times m + 8 \times m^2 - 5 \times m^3$

$= - 3m + 8m^2 - 5m^3$

Hence,

$6m^2 - 16m^3 + 10m^4 \div - 2m = - 3m + 8m^2 - 5m^3$

Solve:
$17a^6b^8-34a^4b^6+51a^2b^4$

$17a^6b^8-34a^4b^6+51a^2b^4$

Simplifying we get

$17a^2b^6(a^4b^4-2a^2b^2+3)$

Solve:
$4a^2-8ab$

Simplifying we get

$4a^2-8ab=4a(a-2b)$

Divide:
$36a^3x^5 - 24a^4x^4 + 18a^5x^3\,\, by\,\, - 6a^3x^3$

We have to divide

$36a^3x^5 - 24a^4x^4 + 18a^5x^3 \,\,by\,\, - 6a^3x^3$

$36a^3x^5 - 24a^4x^4 + 18a^5x^3 \div (- 6a^3x^3)\\ =\dfrac{ (36a^3x^5 - 24a^4x^4 + 18a^5x^3)}{ - 6a^3x^3}\\= \dfrac{(36.a^3.x^5) }{ (-6.a^3.x^3)} - \dfrac{(24.a^4.x^4) }{ (-6.a^3.x^3)} + \dfrac{(18.a^5.x^3) }{ (-6.a^3.x^3)}$

We get,

$= - 6.x^{5-3} + 4.a^{4-3}.x^{4-3} - 3.a^{5-3}$

$= -6x^2 + 4ax - 3a^2$

Hence,

$36a^3x^5 24a^4x^4 + 18a^5x^3 \div (- 6a^3x^3) = - 6x^2 + 4ax - 3a^2$

Simplify:
$a^3b-a^2b^2-b^3$

Given:

$a^3b-a^2b^2-b^3$

Taking

$b$ as a common factor we get,

$a^3b-a^2b^2-b^3$

$=b(a^3-a^2b-b^2)$

Solve:
$3x^2+6x^3$

Simplifying we get

$3x^2+6x^3=3x^2(1+2x)$

Solve:
$a^3-a^2+a$

Simplifying we get

$a^3-a^2+a=a(a^2-a+1)$

Solve:
$3x^5y-27x^4y^2+12x^3y^3$

Simplifying we get

$3x^5y-27x^4y^2+12x^3y^3=3x^3y(x^2-9xy+4y^2)$

Solve:
$6x^2y+9xy^2+4y^3$

Simplifying we get

$6x^2y+9xy^2+4y^3=y(6x^2+9xy+4y^2)$

Solve:
$2x^3b^2-4x^5b^4$

Simplifying we get

$2x^3b^2-4x^5b^4=2x^3b^2(1-2x^2b^2)$

Solve:
$15x^4y^3-20x^3y$

Simplifying we get

$15x^4y^3-20x^3y=5x^3y(3xy^2-4)$

Solve:
$12abc-6a^2b^2c^2+3a^3b^3c^3$

Simplifying we get

$12abc-6a^2b^2c^2+3a^3b^3c^3=3abc(4-2abc+a^2b^2c^2)$

Factorise by taking out the common factors :
$ab(a^2 + b^2- c^2) - bc(c^2- a^2- b^2) + ca(a^2 + b^2- c^2)$

$ab(a^2 + b^2- c^2) - bc(c^2- a^2- b^2) + ca(a^2 + b^2- c^2)$

$=ab(a^2 + b^2- c^2) + bc(-c^2+ a^2+ b^2) + ca(a^2 + b^2- c^2)$

$=(a^2+b^2-c^2)[ab+bc+ca]$

Factorise by taking out the common factors:
$2x(a - b) + 3y(5a - 5b) + 4z(2b - 2a)$

$2x(a - b) + 3y(5a - 5b) + 4z(2b - 2a)$

$=2x(a - b) + 15y(a - b) + 8z(b - a)$

$=(a-b)[2x+15y-8z]$

Factorise :
$3x^7y-81x^4y^4$

$3x^7y-81x^4y^4=3xy(x^6-27x^3y^3)$

$3x^7y-81x^4y^4=3xy((x^2)^3-(3xy)^3)$

$3x^7y-81x^4y^4=3xy(x^2-3xy)[x^4+3x^3y+9x^2y^2]$

$3x^7y-81x^4y^4=3xy(x (x-3y)x^2[x^2+3xy+9y^2])$

$3x^7y-81x^4y^4=3x^4y((x-3y)[x^2+3xy+9y^2])$

$3x^7y-81x^4y^4=3x^4y(x-3y)[x^2+3xy+9y^2]$

Find the prime factorisation of the following numbers:
${ 85 }^{ 3 }-{ 68 }^{ 3 }+{ 5 }^{ 3 }-{ 22 }^{ 3 }$

Let us consider

$68-5+22=85$ , therefore, the given problem can be rewritten as:

$[(68)+(-5)+(22)]^3-(68)^3-(-5)^3-(22)^3$

We know the identity:

$(a+b+c)^3-a^{ 3 }-b^{ 3 }-c^3=3(a+b)(b+c)(c+a)$

Using the above identity taking

$a=68$ ,

$b=-5$ and

$c=22$ , the equation

$[(68)+(-5)+(22)]^3-(68)^3-(-5)^3-(22)^3$ can be factorised as follows:

$[(68)+(-5)+(22)]^{ 3 }-(68)^{ 3 }-(-5)^{ 3 }-(22)^{ 3 }=3\left( 68-5 \right) \left( -5+22 \right) \left( 22+68 \right) =3\times 63\times 17\times 90$

$=2\times (3\times 3\times 3\times 3\times 3)\times 5\times 7\times 17=2\times 3^{ 5 }\times 5\times 7\times 17$

Hence,

$85^3-68^3+5^3-22^3=2\times 3^{ 5 }\times 5\times 7\times 17$

Find the prime factorisation of the following numbers:
${ 100 }^{ 3 }-{ 49 }^{ 3 }+{ 10 }^{ 3 }-{ 61 }^{ 3 }$

Let us consider

$49-10+61=100$ , therefore, the given problem can be rewritten as:

$[(49)+(-10)+(61)]^3-(49)^3-(-10)^3-(61)^3$

We know the identity:

$(a+b+c)^3-a^{ 3 }-b^{ 3 }-c^3=3(a+b)(b+c)(c+a)$

Using the above identity taking

$a=49$ ,

$b=-10$ and

$c=61$ , the equation

$[(49)+(-10)+(61)]^3-(49)^3-(-10)^3-(61)^3$ can be factorised as follows:

$[(49)+(-10)+(61)]^{ 3 }-(49)^{ 3 }-(-10)^{ 3 }-(61)^{ 3 }=3\left( 49-10 \right) \left( -10+61 \right) \left( 61+49 \right) =3\times 39\times 51\times 110$

$=2\times (3\times 3\times 3)\times 5\times 11\times 13\times 17=2\times 3^{ 3 }\times 5\times 11\times 13\times 17$

Hence,

$100^3-49^3+10^3-61^3=2\times 3^{ 3 }\times 5\times 11\times 13\times 17$

Find the prime factoristion of the following numbers:
${ 30 }^{ 3 }-{ 12 }^{ 3 }-{ 10 }^{ 3 }-{ 8 }^{ 3 }$

Let us consider

$12+10+8=30$ , therefore, the given problem can be rewritten as:

$[(12)+(10)+(8)]^3-(12)^3-(10)^3-(8)^3$

We know the identity:

$(a+b+c)^3-a^{ 3 }-b^{ 3 }-c^3=3(a+b)(b+c)(c+a)$

Using the above identity taking

$a=12$ ,

$b=10$ and

$c=8$ , the equation

$[(12)+(10)+(8)]^3-(12)^3-(10)^3-(8)^3$ can be factorised as follows:

$[(12)+(10)+(8)]^{ 3 }-(12)^{ 3 }-(10)^{ 3 }-(8)^{ 3 }=3\left( 12+10 \right) \left( 10+8 \right) \left( 8+12 \right) =3\times 22\times 18\times 20$

$=(2\times 2\times 2\times 2)\times (3\times 3\times 3)\times 5\times 11=2^{ 4 }\times 3^{ 3 }\times 5\times 11$

Hence,

$30^3-12^3-10^3-8^3=2^{ 4 }\times 3^{ 3 }\times 5\times 11$

Factorise: $3{ x }^{ 2 }+6x+6$

Since in the equation

$3x^2+6x+6$ ,

$3$ is the common term, therefore, can be factorised as follows:

$3x^2+6x+6=3(x^2+2x+2)$

Factorise: $2p(x+y)-3q(x+y)$

Since in the expression

$2p(x+y)-3q(x+y)$ ,

$(x+y)$ is the common term, therefore, can be factorised as follows:

$2p(x+y)-3q(x+y)=(x+y)(2p-3q)$

Factorise: $18{ x }^{ 2 }y=24xyz$

Since in the expression

$18x^2y-24xyz$ ,

$6xy$ is the common term, therefore, can be factorised as follows:

$18x^2y-24xyz=6xy(3x-4z)$

Factorise: $5{ x }^{ 2 }-20xy$

Since in the expression

$5x^2-20xy$ ,

$5x$ is the common term, therefore, can be factorized as follows:

$5x^2-20xy=5x(x-4y)$

Factorize: $4{ (a+b) }^{ 2 }-6(a+b)$

Since in the expression

$4(a+b)^2-6(a+b)$ ,

$2(a+b)$ is the common in both the terms,

therefore, the given expression can be factorised as follows:

$4(a+b)^2-6(a+b)$

$=2(a+b)[2(a+b)-3]$

$=2(a+b)(2a+2b-3)$

$3x^4 + 6x^3y + 9x^z$

Let us first find the HCF of all the terms of the given polynomial

$3x^4+6x^3y+9x^2$ by factoring the terms as follows:

$3x^{ 4 }=3\times x\times x\times x\times x\\ 6x^{ 3 }y=2\times 3\times x\times x\times x\times y\\ 9x^{ 2 }=3\times 3\times x\times x$

Therefore,

$HCF=3\times x\times x=3x^{ 2 }$

Now, we factor out the HCF from each term of the polynomial

$3x^4+6x^3y+9x^2$ as shown below:

$3x^{ 4 }+6x^{ 3 }y+9x^{ 2 }=3x^{ 2 }(x^{ 2 }+2xy+3)$

Hence,

$3x^{ 4 }+6x^{ 3 }y+9x^{ 2 }=3x^{ 2 }(x^{ 2 }+2xy+3)$ .

Solve: $8x^{2}-72xy+12x$

We first factorize each term of the given polynomial

$8x^2-72xy+12x$ as shown below:

$8x^{ 2 }=2\times 2\times 2\times x\times x\\ 72xy=2\times 2\times 2\times 3\times 3\times x\times y\\ 12x=2\times 2\times 3\times x$

The common factors of

$8x^2,72xy$ and

$12x$ are

$2,2$ and

$x$ , therefore, the HCF is:

$2\times 2\times x=4x$

Thus, taking out the HCF, we have:

$8x^{ 2 }-72xy+12x=4x(2x-18y+3)$

Hence,

$8x^{ 2 }-72xy+12x=4x(2x-18y+3)$ .

Factories the given algebraic expression by taking out the common terms. Also, write the common term
$5pq+20p^{3}q^{3}-15p^{2}q$

We first factorize each term of the given expression

$5pq+20p^3q^3−15p^2q$ as shown below:

$5pq=5\times p\times q\\ 20p^{ 3 }q^{ 3 }=2\times 2\times 5\times p\times p\times p\times q\times q\times q\\ 15p^{ 2 }q=3\times 5\times p\times p\times q$

The common factors are

$5,p$ and

$q$ , therefore, the HCF is

$5\times p\times q=5pq$ .

Now, by taking out the common term(HCF), we have:

$5pq+20p^{ 3 }q^{ 3 }−15p^{ 2 }q=5pq(1+4p^{ 2 }q^{ 2 }-3p)$

Hence, the common term is

$5pq$ .

$3a^2bc + 6ab^2c + 9abc^2$

Let us first find the HCF of all the terms of the given polynomial

$3a^2bc+6ab^2c+9abc^2$ by factoring the terms as follows:

$3a^{ 2 }bc=3\times a\times a\times b\times c\\ 6ab^{ 2 }c=3\times 2\times a\times b\times b\times c\\ 9abc^{ 2 }=3\times 3\times a\times b\times c\times c$

Therefore,

$HCF=3\times a\times b\times c=3abc$

Now, we factor out the HCF from each term of the polynomial

$3a^2bc+6ab^2c+9abc^2$ as shown below:

$3a^{ 2 }bc+6ab^{ 2 }c+9abc^{ 2 }=3abc(a+2b+3c)$

Hence,

$3a^{ 2 }bc+6ab^{ 2 }c+9abc^{ 2 }=3abc(a+2b+3c)$ .

$4p^2 + 5pq - 6pq^2$

Let us first find the HCF of all the terms of the given polynomial

$4p^2+5pq-6pq^2$ by factoring the terms as follows:

$4p^{ 2 }=2\times 2\times p\times p\\ 5pq=5\times p\times q\\ 6pq^{ 2 }=2\times 3\times p\times q\times q$

Therefore,

$HCF=p$

Now, we factor out the HCF from each term of the polynomial

$4p^2+5pq-6pq^2$ as shown below:

$4p^{ 2 }+5pq-6pq^{ 2 }=p(4p+5q-6q^{ 2 })$

Hence,

$4p^{ 2 }+5pq-6pq^{ 2 }=p(4p+5q-6q^{ 2 })$ .

Factorise $3x^2 + 6x^2 y + 9xy^2$

Let us first find the HCF of all the terms of the given polynomial

$3x^2+6x^2y+9xy^2$ by factoring the terms as follows:

$3x^{ 2 }=3\times x\times x\\ 6x^{ 2 }y=3\times 2\times x\times x\times y\\ 9xy^{ 2 }=3\times 3\times x\times y\times y$

Therefore,

$HCF=3\times x=3x$

Now, we factor out the HCF from each term of the polynomial

$3x^2+6x^2y+9xy^2$ as shown below:

$3x^{ 2 }+6x^{ 2 }y+9xy^{ 2 }=3x(x+2xy+3y^{ 2 })$

Hence,

$3x^{ 2 }+6x^{ 2 }y+9xy^{ 2 }=3x(x+2xy+3y^{ 2 })$ .

Factorize the following expressions:
$6x^{3}y - 12x^{2}y + 15x^{4}$

Let us first find the HCF of all the terms of the given polynomial

$6x^3y-12x^2y+15x^4$ by factoring the terms as follows:

$6x^{ 3 }y=2\times 3\times x\times x\times x\times y\\ 12x^{ 2 }y=2\times 2\times 3\times x\times x\times y\\ 15x^{ 4 }=3\times 5\times x\times x\times x\times x$

Therefore,

$HCF=3\times x\times x=3x^{ 2 }$

Now, we factor out the HCF from each term of the polynomial

$6x^3y-12x^2y+15x^4$ as shown below:

$6x^{ 3 }y-12x^{ 2 }y+15x^{ 4 }=3x^{ 2 }(2xy-4y+5x^{ 2 })$

Hence,

$6x^{ 3 }y-12x^{ 2 }y+15x^{ 4 }=3x^{ 2 }(2xy-4y+5x^{ 2 })$ .

Factorise:
$x^2-17x+60$

In the given polynomial

$x^2-17x+60$ ,

The first term is

$x^2$ and its coefficient is

$1$ .

The middle term is

$-17x$ and its coefficient is

$-17$ .

The last term is a constant term

$60$ .

Multiply the coefficient of the first term by the constant

$1\times 60=60$ .

We now find the factor of

$60$ whose sum equals the coefficient of the middle term, which is

$-17$ and then factorize the polynomial

$x^2-17x+60$ as shown below:

$x^{ 2 }-17x+60\\ =x^{ 2 }-12x-5x+60\\ =x(x-12)-5(x-12)\\ =(x-5)(x-12)$

Hence,

$x^{ 2 }-17x+60=(x-5)(x-12)$ .

Find integers 'a' and 'b' such that $(x^{2} - x- 1)$ divides $ax^{17} + bx^{16} + 1$ .

Let

$f(x)=a\cdot x^{17}+b\cdot x^{16}+1$

Given that

$x^2-x-1$ divides

$f(x)$ , hence we can represent

$f(x)$ as

$f(x)=(x^2-x-1)\cdot p(x)$ Eq.

$(1)$

where

$p(x)=c_1.x^{15}+c_2.x^{14}+ ...........c_{15}.x+c_{16}$ in which

$c_1,c_2,...c_{16}$ are constant coefficients.

On comparing both sides of Eq.

$(1)$ , we see that

$f(x)$ has its constant term as

$+1$ , which can only be achieved via multiplication of

$x^2-x-1$ and

$p(x)$ , if and only if constant term

$c_{16}=-1$

Now,

$f(x)=(x^2-x-1)(c_1.x^{15}+c_2.x^{14}+ ...........c_{15}.x-1)$

As,

$f(x)$ has no terms of

$x$ , we can say

$c_{15}=1$

Similarly, on comparing the coefficients of

$f(x)$ and the resultant from the multiplcation of

$p(x)$ with

$x^2-x-1$ , we can easily evaluate coefficients

$c_{14}, c_{13}.....b , a$

On solving , we see that

$c_{14}=-2, c_{13}=3,c_{12}=-5$ and so on

which gives a pattern with alternate + and - signs and with each next terms magnitude is equal to the sum of magnitude of last two terms. whiuch is the property for Fibbonaci series.

Therefore, for magnitude

$a=c_1=F_{17}$ and

$b=c_1+c_2 =-F_{16}$

where

$F_{16}$ and

$F_{17}$ are the sixteenth and seventeenth Fibbonaci numbers.

And sign we can adjust accordingly as the alternate sign is changing.

$\Rightarrow$

$a=F_{17}= 1597$ and

$b=-F_{16}=-987$

Factorise $4x^2+4\sqrt{3}x+3=0$ .

$\begin{matrix} 4{ x^{ 2 } }+4\sqrt { 3 } x+3=0 \\ Factor\, \, of\, 12\, \, is: \\ 4\times 3=12=2\sqrt { 3 } \times 2\sqrt { 3 } \\ Then,\, \, \, \\ 4{ x^{ 2 } }+2\sqrt { 3 } x+2\sqrt { 3 } x+3=0 \\ 2x\left( { 2x+\sqrt { 3 } } \right) +\sqrt { 3 } \left( { 2x+\sqrt { 3 } } \right) =0 \\ \left( { 2x+\sqrt { 3 } } \right) \, \, \, \left( { 2x+\sqrt { 3 } } \right) \\ { \left( { 2x+\sqrt { 3 } } \right) ^{ 2 } }=0 \\ \therefore \left( { 2x+\sqrt { 3 } } \right) =0 \\ x=\sqrt { \frac { 3 }{ 2 } } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, Ans. \\ \end{matrix}$

Factorisation - Class 8 Maths - Extra Questions

Factorise ab(x2+y2)−xy(a2+b2)ab({x^2} + {y^2}) - xy({a^2} + {b^2}).

Divide the polynomial 3x^{4} -4x^{3} -3x - 13x^{4} -4x^{3} -3x - 1 by x -1x -1.

Find the root of given equation {x}^{2}+6x+5{x}^{2}+6x+5

Factorize x\left( {x + z} \right) - y\left( {y + z} \right)x\left( {x + z} \right) - y\left( {y + z} \right)

Factorise the expression and divide them as directed.(5p^2-25p+20)\div (p-1)(5p^2-25p+20)\div (p-1).

Factousation, 16{p^2}{q^2}{r^3} - 24{p^3}{q^2}{r^2} + 8{p^3}{q^2}{r^2}16{p^2}{q^2}{r^3} - 24{p^3}{q^2}{r^2} + 8{p^3}{q^2}{r^2}

factorise 3 x ^ { 2 } + 6 x ^ { 3 }3 x ^ { 2 } + 6 x ^ { 3 }

Factorise:ab^{2}+b(a-1)-1ab^{2}+b(a-1)-1

Factorize:4 a ^ { 2 } - 8 a b4 a ^ { 2 } - 8 a b

Factorize x ^ { 2 } + 7 x + 12 = 0x ^ { 2 } + 7 x + 12 = 0

Factorize:2 x ^ { 3 } b ^ { 2 } - 4 x ^ { 5 } b ^ { 4 }2 x ^ { 3 } b ^ { 2 } - 4 x ^ { 5 } b ^ { 4 }

Factorize: 15 x ^ { 4 } y ^ { 3 } - 20 x ^ { 3 } y15 x ^ { 4 } y ^ { 3 } - 20 x ^ { 3 } y

Factorize:.15 x + 515 x + 5

Factorise completely by removing a monomial factor.5m+5n

Factorise completely by removing a monomial factor.5a+8b

Factorise completely by removing a monomial factor.2x-4

Factorize: a ^ { 3 } b - a ^ { 2 } b ^ { 2 } - b ^ { 3 }a ^ { 3 } b - a ^ { 2 } b ^ { 2 } - b ^ { 3 }

Factorise completely by removing a monomial factor.5x+10

Solve : (x^3+2x^2+3x) \div 2x(x^3+2x^2+3x) \div 2x

Factorise completely by removing a monomial factor.-3m-15n

Factorise completely by removing a monomial factor.7x-14y

Factorise completely by removing a monomial factor.3y-93y-9

Factorize:6 x ^ { 2 } y + 9 x y ^ { 2 } + 4 y ^ { 3 }6 x ^ { 2 } y + 9 x y ^ { 2 } + 4 y ^ { 3 }

Factorise completely by removing a monomial factor.-7p-14q

Factorise completely by removing a monomial factor.ax+bx

Factorise completely by removing a monomial factor.4 + 12{x^2}4 + 12{x^2}

Factorise completely by removing a monomial factor.{x^2}{y^{}} + x{y^2}{x^2}{y^{}} + x{y^2}

Factorize:6{x^2} - 11x6{x^2} - 11x

Factorise :a{x^2} + a{b^3}a{x^2} + a{b^3}

Factorise completely by removing a monomial factor{x^2}{y^2} + {x^2}{x^2}{y^2} + {x^2}

divide({y^2} + 10y + 24) \div (y + 4)({y^2} + 10y + 24) \div (y + 4)

Factorise completely by removing a monomial factorax+ay+az

Factories:-6n+12n^26n+12n^2

Factories:-6n+12n^2+21n6n+12n^2+21n

Factorize x^3+13x^2+32x+20x^3+13x^2+32x+20, if it is given that x+2x+2 is its factor.

Factorise12{x^2} - 9x12{x^2} - 9x

12xy(9x^{2}-16y^{2})\div 4xy(3x+4y)12xy(9x^{2}-16y^{2})\div 4xy(3x+4y)

Factorize: {x^3} + 6{x^2}y + 9x{y^2}{x^3} + 6{x^2}y + 9x{y^2}.

Factorize:(a^2-b^2)x^2+2ax+1(a^2-b^2)x^2+2ax+1.

Factorise : 1. 2x^{2}-x-32x^{2}-x-32. -2x^{2}+3x+9-2x^{2}+3x+9

Solve:z^2-16z+55=0z^2-16z+55=0.

Factorize: {x}^{2}+10x+24{x}^{2}+10x+24

(4x^4-5x^3-7x+1)\div (4x-1)(4x^4-5x^3-7x+1)\div (4x-1).

Divide \left( { 2y }^{ 3 }+{ 4y }^{ 2 }+3 \right) \div { 2y }^{ 2 }\left( { 2y }^{ 3 }+{ 4y }^{ 2 }+3 \right) \div { 2y }^{ 2 }

Factorise : 150 -6x^2 150 -6x^2

Factorise : 8ab^2 - 18a^3 8ab^2 - 18a^3

Factorise : x^2 +\dfrac {1}{x^2} - 2 -3x +\dfrac {3}{x} x^2 +\dfrac {1}{x^2} - 2 -3x +\dfrac {3}{x}

Factorise : a (a-2b-c) +2bc a (a-2b-c) +2bc

Factorise : 4a^2 -9b-2bc -c^2 4a^2 -9b-2bc -c^2

Factorise the following expressions:(i) 32a^2b - 72b^3 32a^2b - 72b^3 (ii) 9 (a +b)^3 - 25 ( a+b) 9 (a +b)^3 - 25 ( a+b)

18 m +16 n 18 m +16 n

20x^2 -45y^2 20x^2 -45y^2

x^3 -25x x^3 -25x

14 (a-3b)^3 - 21p (a -3b) 14 (a-3b)^3 - 21p (a -3b)

Factorise 27a^3b^3 - 18 a^2b^3 + 75a^3b^2 27a^3b^3 - 18 a^2b^3 + 75a^3b^2

10 (2p +q)^3 -15b( 2p +q)^2 + 35(2p +q) 10 (2p +q)^3 -15b( 2p +q)^2 + 35(2p +q)

150 -6a^2 150 -6a^2

32x^2 -18y^2 32x^2 -18y^2

3a(x^2 +y^2) +6b ( x^2 +y^2 ) 3a(x^2 +y^2) +6b ( x^2 +y^2 )

x(x^2+y^2-x^2) +y(-x^2-y^2 +z^2) -z (x^2 +y -z^2) x(x^2+y^2-x^2) +y(-x^2-y^2 +z^2) -z (x^2 +y -z^2)

\pi a^5 - \pi^3 ab^2 \pi a^5 - \pi^3 ab^2

Factorise m^2+m+1=0m^2+m+1=0.

Factorize:{a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3}+8{a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3}+8

Factorise:x^{2}+20x-69x^{2}+20x-69

Divide \sqrt{2}a^3+3\sqrt{2}a^2+6a\sqrt{2}a^3+3\sqrt{2}a^2+6a by 2a2a.

Factorize \dfrac {1}{6}a^{2}-a+\dfrac {4}{3}\dfrac {1}{6}a^{2}-a+\dfrac {4}{3}.

Factorise:4u^2+8u4u^2+8u

Factorise - 3{x^2} + 3xy - 4xza - 3{x^2} + 3xy - 4xza

Factorise:\left( {a - b} \right) + {\left( {a - b} \right)^2}\left( {a - b} \right) + {\left( {a - b} \right)^2}

Factorise: - 5{a^2} + 4a + 5\left( { - {a^2} + 5} \right) - 5{a^2} + 4a + 5\left( { - {a^2} + 5} \right)

Factorize x^{3}+13x^{2}+32x+20x^{3}+13x^{2}+32x+20

Resolve 2{b}^{2}-5b-32{b}^{2}-5b-3 in to factors.

Factorise: { 7x }^{ 3 }z-{ 21x }z^{ 2 }{ 7x }^{ 3 }z-{ 21x }z^{ 2 }

{ ab }^{ 2 }+(a-1)b-1{ ab }^{ 2 }+(a-1)b-1

Factorise: { x }^{ 2 }+18x+45{ x }^{ 2 }+18x+45

Factorize p^2+1.5p+0.5p^2+1.5p+0.5.

Find the quotient and remainder on dividing the polynomials a{x^2} + \left( {b + ac} \right)x + bc\;by\;x + ca{x^2} + \left( {b + ac} \right)x + bc\;by\;x + c

Resolve into factors 3x^2-6x3x^2-6x

Divide : a^2 + 7a +12a^2 + 7a +12 by (a+4)(a+4).

Factorise $ab({x^2} + {y^2}) - xy({a^2} + {b^2})$ .

Divide the polynomial $3x^{4} -4x^{3} -3x - 1$ by $x -1$ .

Find the root of given equation ${x}^{2}+6x+5$

Factorize $x\left( {x + z} \right) - y\left( {y + z} \right)$

Factorise the expression and divide them as directed.
$(5p^2-25p+20)\div (p-1)$ .

Factousation, $16{p^2}{q^2}{r^3} - 24{p^3}{q^2}{r^2} + 8{p^3}{q^2}{r^2}$

factorise
$3 x ^ { 2 } + 6 x ^ { 3 }$

Factorise:
$ab^{2}+b(a-1)-1$

Factorize:
$4 a ^ { 2 } - 8 a b$

Factorize
$x ^ { 2 } + 7 x + 12 = 0$

Factorize:
$2 x ^ { 3 } b ^ { 2 } - 4 x ^ { 5 } b ^ { 4 }$

Factorize: $15 x ^ { 4 } y ^ { 3 } - 20 x ^ { 3 } y$

Factorize:. $15 x + 5$

Factorise completely by removing a monomial factor.
5m+5n

Factorise completely by removing a monomial factor.
5a+8b

Factorise completely by removing a monomial factor.
2x-4

Factorize:
$a ^ { 3 } b - a ^ { 2 } b ^ { 2 } - b ^ { 3 }$

Factorise completely by removing a monomial factor.
5x+10

Solve : $(x^3+2x^2+3x) \div 2x$

Factorise completely by removing a monomial factor.
-3m-15n

Factorise completely by removing a monomial factor.
7x-14y

Factorise completely by removing a monomial factor.
$3y-9$

Factorize:
$6 x ^ { 2 } y + 9 x y ^ { 2 } + 4 y ^ { 3 }$

Factorise completely by removing a monomial factor.
-7p-14q

Factorise completely by removing a monomial factor.
ax+bx

Factorise completely by removing a monomial factor.
$4 + 12{x^2}$

Factorise completely by removing a monomial factor.
${x^2}{y^{}} + x{y^2}$

Factorize:
$6{x^2} - 11x$

Factorise :
$a{x^2} + a{b^3}$

Factorise completely by removing a monomial factor
${x^2}{y^2} + {x^2}$

divide
$({y^2} + 10y + 24) \div (y + 4)$

Factorise completely by removing a monomial factor
ax+ay+az

Factories:-
$6n+12n^2$

Factories:-
$6n+12n^2+21n$

Factorize $x^3+13x^2+32x+20$ , if it is given that $x+2$ is its factor.

Factorise
$12{x^2} - 9x$

$12xy(9x^{2}-16y^{2})\div 4xy(3x+4y)$

Factorize: ${x^3} + 6{x^2}y + 9x{y^2}$ .

Factorize:
$(a^2-b^2)x^2+2ax+1$ .

Factorise :
1. $2x^{2}-x-3$
2. $-2x^{2}+3x+9$

Solve:
$z^2-16z+55=0$ .

Factorize: ${x}^{2}+10x+24$

$(4x^4-5x^3-7x+1)\div (4x-1)$ .

Divide $\left( { 2y }^{ 3 }+{ 4y }^{ 2 }+3 \right) \div { 2y }^{ 2 }$

Factorise : $150 -6x^2$

Factorise : $8ab^2 - 18a^3$

Factorise : $x^2 +\dfrac {1}{x^2} - 2 -3x +\dfrac {3}{x}$

Factorise : $a (a-2b-c) +2bc$

Factorise : $4a^2 -9b-2bc -c^2$

Factorise the following expressions:
(i) $32a^2b - 72b^3$
(ii) $9 (a +b)^3 - 25 ( a+b)$

$18 m +16 n$

$20x^2 -45y^2$

$x^3 -25x$

$14 (a-3b)^3 - 21p (a -3b)$

Factorise $27a^3b^3 - 18 a^2b^3 + 75a^3b^2$

$10 (2p +q)^3 -15b( 2p +q)^2 + 35(2p +q)$

$150 -6a^2$

$32x^2 -18y^2$

$3a(x^2 +y^2) +6b ( x^2 +y^2 )$

$x(x^2+y^2-x^2) +y(-x^2-y^2 +z^2) -z (x^2 +y -z^2)$

$\pi a^5 - \pi^3 ab^2$

Factorise $m^2+m+1=0$ .

Factorize:
${a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3}+8$

Factorise:
$x^{2}+20x-69$

Divide $\sqrt{2}a^3+3\sqrt{2}a^2+6a$ by $2a$ .

Factorize $\dfrac {1}{6}a^{2}-a+\dfrac {4}{3}$ .

Factorise:
$4u^2+8u$

Factorise
$- 3{x^2} + 3xy - 4xza$

Factorise:
$\left( {a - b} \right) + {\left( {a - b} \right)^2}$

Factorise:
$- 5{a^2} + 4a + 5\left( { - {a^2} + 5} \right)$

Factorize $x^{3}+13x^{2}+32x+20$

Resolve $2{b}^{2}-5b-3$ in to factors.

Factorise: ${ 7x }^{ 3 }z-{ 21x }z^{ 2 }$

${ ab }^{ 2 }+(a-1)b-1$

Factorise: ${ x }^{ 2 }+18x+45$

Factorize $p^2+1.5p+0.5$ .

Find the quotient and remainder on dividing the polynomials $a{x^2} + \left( {b + ac} \right)x + bc\;by\;x + c$

Resolve into factors $3x^2-6x$

Divide : $a^2 + 7a +12$ by $(a+4)$ .

Find the quotient and remainder.
$(2p^{2}+7p-9)\div (p-6)$