MCQ Questions for Class 8 Maths Factorisation Quiz 3

Factorise :

$\displaystyle 13a+26$

0%
$\displaystyle a+13$
0%
$\displaystyle a+2$
0%
$\displaystyle 13(a+26)$
0%
$\displaystyle 13(a+2)$

Explanation

$\displaystyle 13a+26=\left( 13\times a \right) +\left( 13\times 2 \right)$

$=13\left( a+2 \right)$

Factorise :

$\displaystyle 8x-64$

0%
$\displaystyle 8(x-9)$
0%
$\displaystyle 8x-8$
0%
$\displaystyle 8(x-6)$
0%
$\displaystyle 8(x-8)$

Explanation

$\displaystyle 8x-64=\left( 8\times x \right) -\left( 8\times 8 \right)$

$=8\left( x-8 \right)$

Which of the following statements is true?

0%
$\displaystyle 5x+3=5(x+3)$
0%
$\displaystyle 5x+3{ x }^{ 2 }=x\left( 5+3x \right)$
0%
$\displaystyle 5x+3{ x }^{ 2 }=8x\left( x+{ x }^{ 2 } \right)$
0%
$\displaystyle 5x+3=5\left( 1+3x \right)$

Explanation

$\displaystyle 5x+3{ x }^{ 2 }=\left( 5\times x \right) +\left( 3\times x\times x \right)$

$=x\left( 5+3x \right)$

$\displaystyle \therefore$ The statement

$5x+3{ x }^{ 2 }=x\left( 5+3x \right)$ is true.

________ is a method of writing numbers as the product of their factors or divisors.

0%
Polynomial
0%
Factorisation
0%
Division algorithm
0%
Quadratic equation

Explanation

Factorisation is a method of writing numbers as the product of their factors or divisors.
Example:

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

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

Factorize

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

0%
$x(x^2-x-1)$
0%
$x(x-x^2-x^3)$
0%
$x(x^3-x^2)$
0%
$x^2(x^2-x-1)$

Explanation

Given,

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

Here all the terms are unlike.

Only

$x^2$ is common from all, so

$\Rightarrow x^2(x^2-x-1)$

The expression

$\displaystyle xy-xz$ is equivalent to:

0%
$\displaystyle x\left( y-z \right)$
0%
$y(z-x)$
0%
$x(y+z)$
0%
$z(-x+y)$

Explanation

$xy - xz$

$= x(y - z)$

A number besides a bracket means the number is multiplied with each term in the bracket.

$30x^{3} + 45x^{2} - 10x$ is divided by

$5x$ , find the resulting coefficient of

$x$

0%
$6$
0%
$9$
0%
$25$
0%
$40$

Explanation

The value of

$\dfrac{30x^{3}+45x^{2}-10x}{5x}$

$=$

$\dfrac{5x\left ( 6x^{2}+9x-10 \right )}{5x}$

$=$

$6x^{2}+9x-10$

In resultant equation, coefficient of

$x$ is

$9$ .

The expression

$\displaystyle abc+xyc$ is equivalent to:

0%
$\displaystyle c\left( ab+xy \right)$
0%
$c(ab-xy)$
0%
$x(ab+cy)$
0%
$x(ab-cy)$

Explanation

$abc + xyc$

$= c(ab + xy)$ .....

$c$ is common to both the terms

A number besides a bracket means the number is multiplied with each term in the bracket.

$\left( 49{ x }^{ 2 }yz+35p \right) \div 7=$

0%
$7{ x }^{ 2 }yz-35p$
0%
$7{ x }^{ 2 }yz+35p$
0%
$7{ x }^{ 2 }yz-5p$
0%
$7{ x }^{ 2 }yz+5p$

Explanation

$(49x^{2}yz+35p)\div 7$

$=(49x^{2}yz+35p)\times \frac{1}{7}$

$=\dfrac{49}{7}x^{2}yz+\dfrac{35}{7}p$

$=7x^{2}yz+5p$

Factorise completely by removing a monomial factor.

$3{y^2} - 7y$

0%
7y(3y-1)
0%
y(3y-7)
0%
3y(y-7)
0%
7(3y-1)

Explanation

$3y^2-7y$

$=y(3y-7)$ .

One of the factor of

$(25x^2-1)+(1+5x^2)$ is

0%
$5+x$
0%
$5-x$
0%
$5x-1$
0%
$10x$

Explanation

Given,

$(25x^2-1)+(1+5x^2)=30x^2=(3x)(10x)$
Clearly,

$10x$ is factor of

$(25x^2-1)+(1+5x^2)$
Option D is correct.

$\displaystyle \left( { 3x }^{ 2 }-x \right) \div \left( -x \right)$ is equal to

0%
$\displaystyle 3x+1$
0%
$\displaystyle -3x-1$
0%
$\displaystyle -3x+1$
0%
$\displaystyle 3x-1$

Explanation

$\displaystyle \left( { 3x }^{ 2 }-x \right) \div \left( -x \right)$

By separating denominators, we get

$=\dfrac { 3{ x }^{ 2 } }{ -x } +\dfrac { \left( -x \right) }{ \left( -x \right) } =-3x+1$
Hence, final result after given operation is

$-3x+1$ .

$\displaystyle (6ab-3abc+9abcd)\div \left( -\frac { 1 }{ 3 } ab \right)$ is equal to

0%
$\displaystyle 2+c+3cd$
0%
$\displaystyle -2+c-3cd$
0%
$\displaystyle -18+9c-27cd$
0%
$\displaystyle 18-9c-27cd$

Explanation

$\displaystyle \frac { 6ab-3abc+9abcd }{ -\frac { 1 }{ 3 } ab }$

By separating denominators, we get

$\displaystyle =\dfrac { 6ab }{ -\frac { 1 }{ 3 } ab } +\frac { \left( -3abc \right) }{ -\frac { 1 }{ 3 } ab } +\frac { 9abcd }{ -\frac { 1 }{ 3 } ab }$

$\displaystyle =\frac { 6ab }{ ab } \times (-3)+\frac { \left( -3abc \right) }{ ab } \times (-3)+\frac { \left( 9abcd \right) }{ ab } \times (-3)$

$\displaystyle =-18+9c-27cd$
Hence final value after given operation is

$-18+9c-27cd$

Factorize the below equation:

$b(a+d) -c(a+d)$

0%
$(a - c)(b + d)$
0%
$(b + c)(a - d)$
0%
$(b - c)(a + d)$
0%
None of these

Explanation

The common factor between

$b(a+d)$ and

$c(a+d)$ is

$(a+d)$ that is the HCF of

$b(a+d)$ and

$c(a+d)$ is

$(a+d)$ .

Therefore, we take

$(a+d)$ as a common factor in the expression

$b(a+d)-c(a+d)$ as shown below:

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

Hence, the factorization of

$b(a+d)-c(a+d)$ is

$(b-c)(a+d)$ .

Factorise :

$6xy^2 + 4x^2y$

0%
$2xy(3x+y)$
0%
$xy(3x+2y)$
0%
$2xy(2x+3y)$
0%
none of these

Explanation

The common factor between

$6xy^2$ and

$4x^2y$ is

$2xy$ that is the HCF of

$6xy^2$ and

$4x^2y$ is

$2xy$ .

Therefore, we take

$2xy$ as a common factor in the expression

$6xy^2+4x^2y$ as shown below:

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

Hence, the factors of

$6xy^2+4x^2y$ are

$2xy$ and

$(2x+3y)$ .

Factorize :

$4ax + 4ay$

0%
$4a(x + y)$
0%
$4(x + y)$
0%
$4a(x - y)$
0%
$4(x - y)$

Explanation

The common factor between

$4ax$ and

$4ay$ is

$4a$ that is the HCF of

$4ax$ and

$4ay$ is

$4a$ .

Therefore, we take

$4a$ as a common factor in the expression

$4ax+4ay$ as shown below:

$4ax+4ay=4a(x+y)$

Hence, the factorization of

$4ax+4ay$ is

$4a(x+y)$ .

Simplify :

$10 a^2-15 b^2 + 20 c^2$

0%
$5(2a^2 - 3b^2 + 4c^2)$
0%
$5(2a -3b^2 + 4c^2)$
0%
$5(10a^2 - 3b^2 + 4c^2)$
0%
$5(2a^2 + 3b^2 + 4c^2)$

Explanation

$10a^{ 2 }-15b^{ 2 }+20c^{ 2 }$
The common factor

$=5$
Thus,

$10a^{ 2 }-15b^{ 2 }+20c^{ 2 }$

$=5(2a^{ 2 }-3b^{ 2 }+4c^{ 2 })$

What is the possible expression for the dimension of the cuboid whose volume is given below?

$Volume: 3x^2-12x$

0%
$4\times x \times (x-4)$
0%
$3\times x \times (x-4)$
0%
$4\times x \times (x+4)$
0%
$3\times x \times (x+4)$

Explanation

Given,

$Volume= 3x^2-12x$

$=3x(x-4)=3\times x\times(x-4)$

$\therefore$ Possible dimensions are 3 units, x units and (x-4) units

Factorise

$ax^2y + bxy^2 + cxyz$

0%
$x(ax+by+cz)$
0%
$y(ax-by-cz)$
0%
$xy(ax+by+cz)$
0%
$xy(a+by+cz)$

Explanation

We have,

$ax^2y =a \times x \times x \times y$

$bxy^2 = b \times x \times y \times y$
and,

$cxyz= c \times x \times y \times z$
The three terms have x and y as common factors

$\therefore ax^2y + bxy^2 + cxyz = (a \times x \times x \times y) + (b \times x \times y \times y) +(c \times x \times y \times z)$

$= x \times y \times (a \times x +b \times y + c \times z)$

$=xy(ax+by+ cz)$

Divide:

$8x^2y^2-6xy^2 +10x^2y^3$ by

$2xy$

0%
$4y -3y + 5xy^3$
0%
$4xy -3y - 5xy^3$
0%
$4x -3y - 5xy^2$
0%
$4xy -3y + 5xy^2$

Explanation

$8x^2y^2-6xy^2 +10x^2y^3 \div 2xy$

$\displaystyle \frac { 8x^{ 2 }y^{ 2 }-6xy^{ 2 }+10x^{ 2 }y^{ 3 } }{ 2xy }$

$=\displaystyle \frac { 2xy(4xy-6y+10xy^{ 2 }) }{ 2xy }$

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

Which of the following is factorization of

$(-56mnp^2 + 7mnp).$

0%
$7mnp(-8p + 1)$
0%
$7mnp(8p + 1)$
0%
$7mnp(8p - 1)$
0%
None of these

Explanation

$-56mnp^2+7mnp$

$=-7\times 8mnp^2+7mnp$

$=7mnp(-8p+1)$

Factorise :

$\displaystyle 121ac-16a^{2}b^{2}$

0%
$a(121c-16ab^{2})$
0%
$a(121c+16ab^{2})$
0%
$a(121c-16ac^{2})$
0%
none of these

Explanation

$121ac-16a^2b^2$

$=121\times a\times c-16\times a\times a\times b\times b$

$=a(121c-16ab^2)$

Which of the following is an example of factorisation?

0%
$x^2+2x=x(x+2)$
0%
$x^2+2x=x(x+1)$
0%
$x^2+2x=x(x+3)$
0%
None of the above

Explanation

The common factor between

$x^2$ and

$2x$ is

$x$ that is the HCF of

$x^2$ and

$2x$ is

$x$ .

Therefore, we take

$x$ as a common factor in the expression

$x^2+2x$ as shown below:

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

Hence, the factorization of

$x^2+2x$ is

$x(x+2)$ .

Factorise

$\displaystyle 4m^{3}n^{2}+12m^{2}n^{2}+18m^{4}n^{3}$

0%
$\displaystyle m^{3}n^{3}(m+3mn+4m^{2}n)$
0%
$\displaystyle m^{2}n^{2}(5m+2n+6m^{2}n)$
0%
$\displaystyle 2m^{2}n^{2}(2m+6+9m^{2}n)$
0%
None of these

Explanation

The factorization of

$\displaystyle 4m^{3}n^{2}+12m^{2}n^{2}+18m^{4}n^{3}$ is

$2m^{2}n^{2}(2m+6+9m^{2}n)$ ....Taking

$2m^2n^2$ common

Factorise :

$5mn+15mnp$

0%
$5mn(1 + 3p)$
0%
$3mn(1 + 5p)$
0%
$5mn(1 - 3p)$
0%
none of these

Explanation

The common factor between

$5mn$ and

$15mnp$ is

$5mn$ that is the HCF of

$5mn$ and

$15mnp$ is

$5mn$ .

Therefore, we take

$5mn$ as a common factor in the expression

$5mn+15mnp$ as shown below:

$5mn+15mnp=5mn(1+3p)$

Hence, the factorization of

$5mn+15mnp$ is

$5mn(1+3p)$ .

Factorise

$10a^2 -15b^2 + 20c^2$

0%
$5(a^2 -3b^2 + 4c^2)$
0%
$10(2a^2 +3b^2 - 4c^2)$
0%
$5(2a^2 -3b^2 + 4c^2)$
0%
$10(a^2 -b^2 + 4c^2)$

Explanation

We have,

$10a^2= 2 \times 5 \times a \times a,$

$15b^2 = 3 \times 5 \times b \times b$
and,

$20c^2= 2 \times 2 \times 5 \times c \times c$
The three terms have 5 as a common factor

$10a^2 - 15b^2+ 20c^2 = (2 \times 5 \times a \times a)-(3 \times 5 \times b \times b) +(2 \times 2 \times 5\times c \times c)$
=

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

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

Evaluate

$8 (x^3 y^2 z^2 + x^2y^3z^2 + x^2y^2z^3) \div 4x^2y^2z^2$

0%
$2(x+y+z)$
0%
$2(x-y+z)$
0%
$2(x+y-z)$
0%
$2(x+2y+2z)$

Explanation

$8(x^3 y^2 z^2 + x^2 y^3 z^2 + x^2 y^2 z^3) \div 4x^2 y^2 z^2$

$= \displaystyle \frac{8 (x^3 y^2 z^2 + z^2 y^3 z^2 + x^2 y^2 z^3)}{4 x^2 y^2 z^2}$

$\displaystyle = \frac{8 \times x^2 y^2 z^2 (x + y + z)}{4 x^2 y^2 z^2} = 2 (x + y + z)$

Evaluate

$(p^3 q^6 -p^6 q^3) \div p^3 q^3$

0%
$q^3+p^3$
0%
$q^3-p^3$
0%
$q^2-p^2$
0%
$q^2+p^2$

Explanation

$\begin{aligned}{}\left( {{p^3}{q^6} - {p^6}{q^3}} \right) \div {p^3}{q^3} &= \frac{{{p^3}{q^6} - {p^6}{q^3}}}{{{p^3}{q^3}}}\\ &= \frac{{{p^3}{q^3}\left( {{q^3} - {p^3}} \right)}}{{{p^3}{q^3}}}\\& = {q^3} - {p^3}\end{aligned}$

Hence, option

$B$ is correct.

Evaluate:

$\displaystyle \left( 21x-35 \right) \div 7$

0%
$\displaystyle 2x-5$
0%
$\displaystyle 4x-5$
0%
$\displaystyle 3x-5$
0%
$\displaystyle 5x-3$

Explanation

$\displaystyle \left( 21x-35 \right) \div 7=\frac { 21x-35 }{ 7 }$

$=\dfrac { 7\left( 3x-5 \right) }{ 7 } =3x-5$

Simplify:

$\displaystyle \left( { 8m }^{ 2 }-9m \right) \div 3m$

0%
$\displaystyle \frac { 1 }{ 3 } \left( 8m-9 \right)$
0%
$\displaystyle 8m-9$
0%
$\displaystyle \frac { 1 }{ 3 } \left( 8m-3 \right)$
0%
$\displaystyle 5m-3$

Explanation

$\displaystyle \left( { 8m }^{ 2 }-9m \right) \div 3m=\frac { { 8m }^{ 2 }-9m }{ 3m }$

$\displaystyle =\frac { m\left( 8m-9 \right) }{ 3m }$

$\displaystyle =\frac { 8m-9 }{ 3 }$

$\displaystyle = \frac { 1 }{ 3 } \left( 8m-9 \right)$

CBSE Questions for Class 8 Maths Factorisation Quiz 3 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Factorisation Quiz 3 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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