MCQ Questions for Class 8 Maths Factorisation Quiz 5

Factorisation of the expression

$\displaystyle { x }^{ 4 }yz+x{ y }^{ 4 }z+xy{ z }^{ 4 }$ results in:

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$\displaystyle xyz\left( { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right)$
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$\displaystyle xyz\left( { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } \right)$
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$\displaystyle { x }^{ 3 }{ y }^{ 3 }{ z }^{ 3 }\left( x+y+z \right)$
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$\displaystyle { x }^{ 2 }{ y }^{ 2 }{ z }^{ 2 }\left( { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right)$

Explanation

$\displaystyle { x }^{ 4 }yz+x{ y }^{ 4 }z+xy{ z }^{ 4 }$

$=\displaystyle xyz( { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } )$

Factorise :

$\displaystyle 12{ x }^{ 2 }-24{ y }^{ 2 }+36{ z }^{ 2 }$

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$\displaystyle 12{ x }^{ 2 }-2{ y }^{ 2 }+3{ z }^{ 2 }$
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$\displaystyle 12\left( { x }^{ 2 }-2{ y }^{ 2 }+3{ z }^{ 2 } \right)$
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$\displaystyle 12\left( { x }^{ 2 }-{ y }^{ 2 }+{ z }^{ 2 } \right)$
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$\displaystyle 12\left( { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right)$

Explanation

$\displaystyle 12{ x }^{ 2 }-24{ y }^{ 2 }+36{ z }^{ 2 }=12x^2 - 12\times 2y^2 + 12\times 3z^2$

$=12\left( { x }^{ 2 }-2{ y }^{ 2 }+3{ z }^{ 2 } \right)$

Factorisation of the expression :

$\displaystyle 7{ a }^{ 2 }-49{ b }^{ 2 }$

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$\displaystyle 7\left( { a }^{ 2 }-7{ b }^{ 2 } \right)$
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$\displaystyle 7{ a }^{ 2 }-7{ b }^{ 2 }$
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$\displaystyle 7\left( { a }^{ 2 }-{ b }^{ 2 } \right)$
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$\displaystyle 7\left( 7{ a }^{ 2 }-{ b }^{ 2 } \right)$

Explanation

$\displaystyle 7{ a }^{ 2 }-49{ b }^{ 2 }=( 7\times a\times a) -( 7\times 7\times b\times b )$

$\displaystyle = 7( { a }^{ 2 }-7{ b }^{ 2 } )$

Factorise:

$\displaystyle 80{ p }^{ 2 }-72p$

0%
$\displaystyle 8p(5p-9)$
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$\displaystyle 16p(5p-4)$
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$\displaystyle 8p(10p-8)$
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$\displaystyle 8p(10p-9)$

Explanation

$\displaystyle 80{ p }^{ 2 }-72p=\left( 8\times 10\times p\times p \right) -\left( 8\times 9\times p \right)$

$= 8p(10p-9)$

Factorisation of the expression

$\displaystyle 17{ p }^{ 2 }q-102p{ q }^{ 2 }$ results in:

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$\displaystyle 17pq(p-8q)$
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$\displaystyle 17pq(p-q)$
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$\displaystyle 17pq(p-4q)$
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$\displaystyle 17pq(p-6q)$

Explanation

$\displaystyle 17{ p }^{ 2 }q-102p{ q }^{ 2 }$

$\displaystyle =\left( 17\times p\times p\times q \right) -\left( 17\times 6\times p\times q\times q \right)$

$\displaystyle =17pq(p-6q)$

The factorisation of

$\left (21a^2+3a \right )$ is

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$3a(7a+1)$
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$7a(3a+1)$
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$3a(7a+3a)$
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$3a(a+7)$

Explanation

The factorisation of

$21a^2+3a$ is

$3a(7a+1)$
Taking common terms out, we get

$3a(7a+1)$ .

Multiplying factors is an example of

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polynomial
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quadratic equation
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division algorithm
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factorisation

Explanation

Multiplying factors is an example of factorisation.
Example:

$4x^2+2x$ is a factor

$2x(2x+1)$
By multiplying the factor we get

$2x(2x+1) = 4x^2+2x$

$\left( 14{ x }^{ 2 }yz-28{ x }^{ 2 }{ y }^{ 2 }{ z }^{ 3 }+32{ y }^{ 2 }{ z }^{ 2 } \right) \div \left( -4xy \right)$ is equal to

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$\dfrac { 7 }{ 2 } yz+7xy{ z }^{ 2 }+8xyz$
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$-\dfrac { 7 }{ 2 } xz+7xy{ z }^{ 3 }-\dfrac { 8y{ z }^{ 2 } }{ x }$
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$-\dfrac { 7 }{ 2 } xz-7xy{ z }^{ 3 }+\dfrac { 8y{ z }^{ 2 } }{ x }$
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$\dfrac { 7 }{ 2 } xz-7xy{ z }^{ 2 }-\dfrac { 8y{ z }^{ 2 } }{ x }$

Explanation

$(14x^{2}yz-28x^{2}y^{2}z^{3}+32y^{2}z^{2})\div (-4xy)$

$=\dfrac{14x^{2}yz}{-4xy}-\dfrac{28x^{2}y^{2}z^{3}}{-4xy}+\dfrac{32y^{2}z^{2}}{-4xy}$

$=-\dfrac{7}{2}xz+7xyz^{3}-\dfrac{8yz^{2}}{x}$

Hence, B is correct.

$4a + 12b$ is equal to

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$4a$
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$12b$
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$4(a+3b)$
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$3a$

Explanation

Given:

$4a+12b$

$=4\cdot a+4\cdot 3\cdot b$

$=4\cdot a+4\cdot 3b$

$=4(a+3b)$ , (common factor

$4$ is taken out)

$(3x^2 - 2x) \div x$

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$3x$
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$3x - 2$
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$3x + 2$
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None

Explanation

Let us first factorize the given polynomial

$3x^2-2x$ by finding the HCF of all the terms as follows:

$3x^{ 2 }=3\times x\times x\\ 2x=2\times x$

Therefore,

$HCF=x$

Now, we factor out the HCF from each term of the polynomial

$3x^2-2x$ as shown below:

$3x^{ 2 }-2x=x(3x-2)$

Let us now divide the polynomial

$3x^2-2x$ by the monomial

$x$ :

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

Hence,

$(3x^{ 2 }-2x)\div x=3x-2$ .

$(5a^3 b - 7ab^3) \div ab$

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$5a^2 - 7b^2$
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$5a - 7b^2$
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$5a^2 - b^2$
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None

Explanation

We divide the given polynomial

$5a^3b-7ab^3$ by the monomial

$ab$ as shown below:

$\dfrac { 5a^{ 3 }b-7ab^{ 3 } }{ ab } \\ =\dfrac { 5a^{ 3 }b }{ ab } -\dfrac { 7ab^{ 3 } }{ ab } \\ =\dfrac { 5a^{ 3 }b }{ ab } -\dfrac { 7ab^{ 3 } }{ ab } \\ =\dfrac { 5\times a\times a\times a\times b }{ a\times b } -\dfrac { 7\times a\times b\times b\times b }{ a\times b } \\ =5a^{ 2 }-7b^{ 2 }$

Hence,

$\dfrac { 5a^{ 3 }b-7ab^{ 3 } }{ ab } =5a^{ 2 }-7b^{ 2 }$ .

$(4l^5 - 6l^4 + 8l^3) \div 2l^2$

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$(2l^2 - 3l + 4)$
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$l(2l^2 - 3l + 4)$
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$l(2l^2 - 3l + 3)$
0%
None

Explanation

We divide the given polynomial

$4l^5-6l^4+8l^3$ by the monomial

$2l^2$ as shown below:

$\dfrac { 4l^{ 5 }-6l^{ 4 }+8l^{ 3 } }{ 2l^{ 2 } } \\ =\dfrac { 4l^{ 5 } }{ 2l^{ 2 } } -\dfrac { 6l^{ 4 } }{ 2l^{ 2 } } +\dfrac { 8l^{ 3 } }{ 2l^{ 2 } } \\ =\left( \dfrac { 2\times 2\times l^{ 5 } }{ 2l^{ 2 } } \right) -\left( \dfrac { 2\times 3\times l^{ 4 } }{ 2l^{ 2 } } \right) +\left( \dfrac { 2\times 2\times 2\times l^{ 3 } }{ 2l^{ 2 } } \right) \\ =\left( \dfrac { 2\times l^{ 5 }\times l^{ -2 } }{ 1 } \right) -\left( \dfrac { 3\times l^{ 4 }\times l^{ -2 } }{ 1 } \right) +\left( \dfrac { 2\times 2\times l^{ 3 }\times l^{ -2 } }{ 1 } \right)$

$=(2\times l^{ (5-2) })-(3\times l^{ (4-2) })+(4\times l^{ (3-2) })\quad \quad \quad \quad \quad \quad \left( \because \quad a^{ x }+a^{ y }=a^{ x+y } \right) \\ =2l^{ 3 }-3l^{ 2 }+4l\\ =l(2l^{ 2 }-3l+4)$

Hence,

$\dfrac { 4l^{ 5 }-6l^{ 4 }+8l^{ 3 } }{ 2l^{ 2 } } =l(2l^{ 2 }-3l+4)$ .

Factorize

$5xyz^2-3zy$

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$yz(5xz-3)$
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$yz(5yz-3)$
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$xz(5xz-3y)$
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None of these

Explanation

We know that

$a(b+c)=ab+ac$ and the reverse process

$ab+ac=a(b+c)$ is called taking out the common factor.

Consider the factorization of the expression

$5xyz^2-3zy$ :

$5xyz^{ 2 }=5\times x\times y\times z\times z\\ 3zy=3\times z\times y$

$\Rightarrow HCF(5xyz^{ 2 },3zy)=z\times y=yz$

Therefore, we have:

$5xyz^{ 2 }-3zy\\ =(5\times x\times y\times z\times z)-(3\times z\times y)\\ =yz(5xz-3)$

Hence,

$5xyz^{ 2 }-3zy=yz(5xz-3)$ .

$15(a^3 b^2 c^2 - a^2 b^3 c^2 + a^2b^2c^3) \div 3abc$

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$abc(a - b + c)$
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$5abc(a + b + c)$
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$5abc(a - b + c)$
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None

Explanation

We divide the given polynomial

$15(a^3b^2c^2-a^2b^3c^2+a^2b^2c^3)$ by the monomial

$3abc$ as shown below:

$\dfrac { 15(a^{ 3 }b^{ 2 }c^{ 2 }-a^{ 2 }b^{ 3 }c^{ 2 }+a^{ 2 }b^{ 2 }c^{ 3 }) }{ 3abc } \\ =\dfrac { 15a^{ 3 }b^{ 2 }c^{ 2 }-15a^{ 2 }b^{ 3 }c^{ 2 }+15a^{ 2 }b^{ 2 }c^{ 3 } }{ 3abc } \\ =\dfrac { 15a^{ 3 }b^{ 2 }c^{ 2 } }{ 3abc } -\dfrac { 15a^{ 2 }b^{ 3 }c^{ 2 } }{ 3abc } +\dfrac { 15a^{ 2 }b^{ 2 }c^{ 3 } }{ 3abc } \\ =\left( \dfrac { 3\times 5\times a\times a\times a\times b\times b\times c\times c }{ 3abc } \right) -\left( \dfrac { 3\times 5\times a\times a\times b\times b\times b\times c\times c }{ 3abc } \right) +\left( \dfrac { 3\times 5\times a\times a\times b\times b\times c\times c\times c }{ 3abc } \right)$

$=5a^{ 2 }bc-5ab^{ 2 }c+5abc^{ 2 }\\ =5abc(a-b+c)$

Hence,

$\dfrac { 15(a^{ 3 }b^{ 2 }c^{ 2 }-a^{ 2 }b^{ 3 }c^{ 2 }+a^{ 2 }b^{ 2 }c^{ 3 }) }{ 3abc } =5abc(a-b+c)$ .

$(3p^3 - 9p^2q - 6pq^2) \div (-3p)$

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$(2q^2 + 3pq + p^2)$
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$(2q^2 + pq - p^2)$
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$(2q^2 + 3pq - p^2)$
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None

Explanation

We divide the given polynomial

$3p^3-9p^2q-6pq^2$ by the monomial

$-3p$ as shown below:

$\Rightarrow \dfrac { 3p^{ 3 }-9p^{ 2 }q-6pq^{ 2 } }{ -3p } \\ =\dfrac { 3p^{ 3 } }{ -3p } -\dfrac { 9p^{ 2 }q }{ -3p } -\dfrac { 6pq^{ 2 } }{ -3p } \\ =-\dfrac { 3p^{ 3 } }{ 3p } +\dfrac { 9p^{ 2 }q }{ 3p } +\dfrac { 6pq^{ 2 } }{ 3p } \\ =-\left( \dfrac { 3\times p\times p\times p }{ 3p } \right) -\left( \dfrac { 3\times 3\times p\times p\times q }{ 3p } \right) +\left( \dfrac { 2\times 3\times p\times q\times q }{ 3p } \right) \\ =-p^{ 2 }+3pq+2q^{ 2 }$

Hence,

$\dfrac { 3p^{ 3 }-9p^{ 2 }q-6pq^{ 2 } }{ -3p } =2q^{ 2 }+3pq-p^{ 2 }$ .

Divide the given polynomial by the given monomial:

$\left(\dfrac{2}{3} a^2 b^2 c^2 + \dfrac{4}{3} ab^2 c^2 \right ) \div \dfrac{1}{2} abc$

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$\dfrac{4}{3} (abc + bc)$
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$\dfrac{2}{3} (abc + 2bc)$
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$\dfrac{4}{3} (abc + 2bc)$
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None

Explanation

We divide the given polynomial

$\dfrac { 2 }{ 3 } a^{ 2 }b^{ 2 }c^{ 2 }+\dfrac { 4 }{ 3 } ab^{ 2 }c^{ 2 }$ by the monomial

$\dfrac { 1 }{ 2 } abc$ as shown below:

$\dfrac { \dfrac { 2 }{ 3 } a^{ 2 }b^{ 2 }c^{ 2 }+\dfrac { 4 }{ 3 } ab^{ 2 }c^{ 2 } }{ \dfrac { 1 }{ 2 } abc } \\ =\dfrac { \dfrac { 2 }{ 3 } a^{ 2 }b^{ 2 }c^{ 2 } }{ \dfrac { 1 }{ 2 } abc } +\dfrac { \dfrac { 4 }{ 3 } ab^{ 2 }c^{ 2 } }{ \dfrac { 1 }{ 2 } abc }$

$=\dfrac { \dfrac { 2 }{ 3 } \times a\times a\times b\times b\times c\times c }{ \dfrac { 1 }{ 2 } \times a\times b\times c } +\dfrac { \dfrac { 4 }{ 3 } \times a\times b\times b\times c\times c }{ \dfrac { 1 }{ 2 } \times a\times b\times c } \\ =\left( \dfrac { 2 }{ 3 } \times 2\times a\times b\times c \right) +\left( \dfrac { 4 }{ 3 } \times 2\times b\times c \right) \\ =\dfrac { 4 }{ 3 } abc +\dfrac { 8 }{ 3 } bc \\ =\dfrac { 4 }{ 3 } (abc+2bc)$

Hence,

$\dfrac { \dfrac { 2 }{ 3 } a^{ 2 }b^{ 2 }c^{ 2 }+\dfrac { 4 }{ 3 } ab^{ 2 }c^{ 2 } }{ \dfrac { 1 }{ 2 } abc } =\dfrac { 4 }{ 3 } (abc+2bc)$ .

$(25x^5 - 15x^4) \div 5x^3$

0%
$x(5x + 3)$
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$(5x - 3)$
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$x(5x - 3)$
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None

Explanation

We divide the given polynomial

$25x^5-15x^4$ by the monomial

$5x^3$ as shown below:

$\dfrac { 25x^{ 5 }-15x^{ 4 } }{ 5x^{ 3 } } \\ =\dfrac { 25x^{ 5 } }{ 5x^{ 3 } } -\dfrac { 15x^{ 4 } }{ 5x^{ 3 } } \\ =\left( \dfrac { 5\times 5\times x^{ 5 } }{ 5x^{ 3 } } \right) -\left( \dfrac { 3\times 5\times x^{ 4 } }{ 5x^{ 3 } } \right) \\ =\left( \dfrac { 5\times x^{ 5 }\times x^{ -3 } }{ 1 } \right) -\left( \dfrac { 3\times x^{ 4 }\times x^{ -3 } }{ 1 } \right)$

$=(5\times x^{ (5-3) })-(3\times x^{ (4-3) })\quad \quad \quad \quad \quad \quad \left( \because \quad a^{ x }+a^{ y }=a^{ x+y } \right) \\ =5x^{ 2 }-3x\\ =x(5x-3)$

Hence,

$\dfrac { 25x^{ 5 }-15x^{ 4 } }{ 5x^{ 3 } } =x(5x-3)$ .

Factorize the following expression

$16x + 64x^2 y$

0%
$16x (1 + 4xy)$
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$6x (1 + 4xy)$
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$16x (1 - 4xy)$
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None of these

Explanation

$16x+64x^{2}y$

$=16x+4\times (16x)(xy)$

Taking

$16x$ common

$=16x(1+4xy)$

Factorize the following

$pq - pr - 3ps$

0%
$p(q + r - 3s)$
0%
$p(q + r + 3s)$
0%
$q(q + r - 3s)$
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None of these

Explanation

$pq-pr-3ps$

(To factorise means to make factor)

Taking

$p$ common from all

$3$ equations, we have

$p(q-r-3s)$

Factorise the expression:

$2a^3 - 3a^2 b + 2a^2 c$

0%
$a^2 (2a - 3b + 2c)$
0%
$a^2 (2a + 3b + 2c)$
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$a^2 (2a - 3b - 2c)$
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None of these

Explanation

$2a^{3}-3a^{2}b+2a^{2}c$

Taking

$a^{2}$ common from all

$3$ terms

$=a^{2}(2a-3b+2c)$

Factorize the following expression:

$10x^3 - 25 x^4 y$

0%
$5x^3 (2 - 5xy)$
0%
$5x^3 (2 + 5xy)$
0%
$x^3 (2 - 5xy)$
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None of these

Explanation

Taking

$5x^{3}$ common from all the terms

$5x^{3}(2-5xy)$

Find the quotient and remainder of the given expression,

$(3x^3+4x^2-5)\div (3x+1)$

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$x^2+x-\dfrac{1}{3}, -\dfrac{14}{3}$
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$x^2-3x+\dfrac{5}{9}, \dfrac{14}{3}$
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$x^2+3x+\dfrac{5}{9}, -\dfrac{14}{3}$
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None of these

The quotient when

$x^3-5x^2+7x-4$ is divided by

$x-1$ is.

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$x^2+4x+3$
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$x^2-4x+3$
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$x^2-4x-3$
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$x^2+4x-3$

Find the quotient and remainder of the given expression,

$(3x^3+2x^2+7x-5)\div (x+3)$

0%
$3x^2-7x+28, -89$
0%
$3x^2-7x+40, 125$
0%
$3x^2-11x+40, -15$
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None of these

Explanation

We divide

$3x^3+2x^2+7x-5$ by

$(x+3)$ as shown in the above image:

From the division, we observe that the quotient is

$3x^2-7x+28$ and the remainder is

$-89$ .

Hence, the quotient is

$3x^2-7x+28$ and the remainder is

$-89$ .

1217362_621739_ans_9a734738c16f41e3bfd2bf64c6de8b3b.jpg

Find the quotient and remainder of the given expression,

$(2x^4-7x^3-13x^2+63x-48)\div (2x-1)$

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$x^3-3x^2-8x+\dfrac{55}{2}, -\dfrac{41}{2}$
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$x^3-3x^2-8x+\dfrac{55}{2}, \dfrac{41}{2}$
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$x^3+3x^2-8x+\dfrac{55}{2}, -\dfrac{41}{2}$
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None of these

Explanation

We divide

$2x^4-7x^3-13x^2+63x-48$ by

$(2x-1)$ as shown in the above image:

From the division, we observe that the quotient is

$x^3-3x^2-8x+\dfrac { 55 }{ 2 }$ and the remainder is

$-\dfrac { 41 }{ 2 }$ .

Hence, the quotient is

$x^3-3x^2-8x+\dfrac { 55 }{ 2 }$ and the remainder is

$-\dfrac { 41 }{ 2 }$ .

1217379_621743_ans_60fe8ebf768548979eff42c18e416b05.jpg

Find the quotient and remainder of the given expression,

$(8x^4-2x^2+6x-5)\div (4x+1)$

0%
$2x^3-\dfrac{x^2}{2}-\dfrac{3}{8}x+\dfrac{51}{32}, -\dfrac{211}{32}$
0%
$2x^3-\dfrac{x^2}{2}-\dfrac{3}{8}x+\dfrac{51}{32}, \dfrac{211}{32}$
0%
$2x^3+\dfrac{x^2}{2}-\dfrac{3}{8}x+\dfrac{51}{32}, -\dfrac{211}{32}$
0%
None of these

The factorized form of

$3x — 24$ is

0%
$3x \times 24$
0%
$3(x - 8)$
0%
$24 (x - 3)$
0%
$3(x - 12)$

Explanation

The given polynomial is

$3x-24$

Now,

$3x-24$

$=3\times x-3\times 8$

Taking out

$3$ as common,

$= 3 (x – 8)$

Hence,

$Op-B$ is correct.

Factorised form of

$23xy - 46x + 54y - 108$ is

0%
$(23x + 54) (y - 2)$
0%
$(23x + 54y) (y - 2)$
0%
$(23xy + 54y) (- 46x - 108)$
0%
$(23x + 54) (y + 2)$

Explanation

We have

$23xy - 46x + 54y - 108$

Factoring out

$23$ from the first two terms and

$54$ from the last two,

$\\=23x(y-2)+54(y-2)$

Now, factoring out

$(y-2),$

$= (23x+54)(y-2)$

$\therefore 23xy - 46x + 54y - 108 = (23x+54)(y-2)$

Hence, option (A) is correct.

State whether the statement is true (T) or false (F).
Factorization of

$-3a^{2} + 3ab + 3ac$ is

$3a(-a-b-c)$ .

0%
True
0%
False

Explanation

Given,

$-3a^{2} + 3ab + 3ac$

Factoring out

$3a$ , we get

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

$\therefore -3a^{2} + 3ab + 3ac=3a(-a+b+c)$

Hence, the given statement is false.

State whether the statement is true (T) or false (F).
Common factor of

$12a^{2}b^{2} + 4ab^{2} - 32$ is

$4$ .

0%
True
0%
False

CBSE Questions for Class 8 Maths Factorisation Quiz 5 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Factorisation Quiz 5 - MCQExams.com

0:0:1

Practice Class 8 Maths Quiz Questions and Answers

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