MCQ Questions for Class 8 Maths Factorisation Quiz 1

Evaluate

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

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

Explanation

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

$=(x\times x\times x)+(2\times x\times x)+(3\times x)$

$=\frac{x(x^2+2x+3)}{2x}$

$=\frac{1}{2}(x^2+2x+3)$

Factorise:

$16a^2-24ab$

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

Explanation

$16a^2-24ab$
Taking 8a as a common,

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

$8a(2a-3b)$

The process of finding the factors of given number (or expression) is called ...........

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multiplication
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factorisation
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division
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addition

Evaluate

$(10x - 25) \div 5$

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$2x-5$
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$2x+6$
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$3x-5$
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$3x+6$

Explanation

$(10 x - 25) \div 5$

$= \displaystyle \frac{10 x - 25}{5}$

$= \dfrac{5 \times (2x - 5)}{5}$

$= 2x - 5$

Factorise the following expression:

$7a^2 + 14a$

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$7a(a + 2)$
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$7(a - 2)$
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$14(7a + 1)$
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$7a(a - 2)$

Explanation

Given expression is

$7a^2 + 14a$

Taking common factor

$7a$ from both terms we get,

$7a(a+2)$

Hence, option A is correct.

Factorise the following expressions:

$a x^2 y + b x y^2 + c x y z$

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xy(ax + by + cz)
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axy(x + by + cz)
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bxy(ax + y + cz)
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xy(ax + by - cz)

Explanation

$a x^2 y + b x y^2 + c x y z$
In the above expression, the common factor is

$=xy$

$\therefore$

$a x^2 y + b x y^2 + c x y z$ =

$xy(ax+by+cz)$

Factorise the expression:

$ax^2 + bx$

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$x(ax + b)$
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$x(ax - b)$
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$x(a + b)$
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$x(a - b)$

Explanation

To factorise:

$ax^2+bx$

Now,

$ax^2+bx$

$=ax\times x+b\times x$

$=x(ax+b)$

Hence, the factorised form is

$x(ax+b)$ .

Factorise the following expression:

$16 z + 20 z^3$

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$4z( 2 - 5z^2)$
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$4z( 4 + 5z)$
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$4z( 4 + 5z^2)$
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$4z( 2 + 5z^2)$

Explanation

$16 z + 20 z^3$

Taking common factor 4z from both the terms we get,

$=4z(4+5z^2)$

Factorise the following expression:

$4 a^2 + 4 ab -4 ca$

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$4a( a + b - c)$
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$4b( a + b - c)$
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$4c( a + b - c)$
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$4a(a+b+c)$

Explanation

$4a^{ 2 }+4ab-4ca$
The common factor

$=2\times 2\times a$
Thus,

$4a^{ 2 }+4ab-4ca=4a(a+b-c)$

Factorise the following expressions.

$20l^2 m + 30 a l m = 10lm(2l+3a)$

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True
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False

Explanation

$20 { l}^{2 } m + 30 a l m$

$=10lm(2l+3a)$

Factorize:

$5 x^2 y -15 xy^2$

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

Explanation

$5{x }^{ 2}y-15x{y}^{2}$

Taking common factor

$5xy$ for both the terms we get,

$=5xy(x-3y)$

Factorise the expression.

$7p^2 + 21q^2$

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$7(p^2 + 7q^2)$
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$7(p^2 + 2q^2)$
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$7(p^2 + 3q^2)$
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$7(p^2 + 4q^2)$

Explanation

$\begin{aligned}{}7{p^2} + 21{q^2}& = 7{p^2} + 7 \times 3{q^2}\\ &= 7\left( {{p^2} + 3{q^2}} \right)\end{aligned}$

Divide the given polynomial by the given monomial.

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

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

Explanation

$(3y^{ 8 }-4y^{ 6 }+5y^{ 4 })\div y^{ 4 }$

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

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

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

Divide the given polynomial by the given monomial.

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

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$\dfrac{1}{2}(x^2+2x+3)$
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$\dfrac{1}{4}(x^2-2x+3)$
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$\dfrac{1}{2}(x^2-2x+3)$
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$\dfrac{1}{2}(x^2+2x-3)$

Explanation

$(x^{ 3 }+2x^{ 2 }+3x)\div 2x$

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

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

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

Divide the given polynomial by the given monomial.

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

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

Explanation

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

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

$=\dfrac{8x^2y^2z^2(x+y+z)}{4x^{ 2 }y^{ 2 }z^{ 2 }}$

$=2(x+y+z)$

Factorize

$3a^{2}-9ab$ by taking common factors.

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$3a\left ( a-3b \right )$
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$3a\left ( a-b \right )$
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$3a\left ( 3a-b \right )$
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$3a\left ( a+b \right )$

Explanation

$3a^{2}-9ab$

$3$ and

$a$ are common in the above two terms in the addition, so taking these as common.

$3a^2-9ab=3a(a-3b)$

Option A is correct.

Factorise:

$5t+25t^2$

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$5t(1+5t)$
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$t(1-5t)$
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$5t(1-5t)$
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$t(1+5t)$

Explanation

$5t+25t^2$

$=5t(1+5t)$

Factorise:

$ab-4ac$

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$a(b+ 4c)$
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$(b- 4c)$
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$a(b- 4c)$
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$a(b- c)$

Explanation

The common variable in the two terms is

$a$ . So, we can take

$a$ common from both the terms as follows:

$ab-4ac =a(b-4c)$

Hence, option C is correct.

Factorise:

$4a + 12b$

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$4(a + 3b)$
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$a(4+3b)$
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$4a(1+3b)$
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$4(a+4b)$

Explanation

$4a+12b$

$= (2\times 2 \times a) + (2\times 2\times 3 \times b)$

$= (2\times 2) (a + 3 \times b)$

$= 4 (a+3b)$

Factorise

$a^3 - a^2 + a$

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

Factorise:

$4x-8y$

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

Explanation

$4x-8y$

$=4(x-2y)$

Factorise

$15 x + 5$

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

Explanation

$15x+5$

$= (5\times 3 \times x) + (5\times1)$

$= 5(3\times x +1)$

$= 5(3x+1)$

Factorise

$9b-12x$

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

Explanation

$Factorizing\quad 9b-12x,\\ 3*\frac { 9b-12x }{ 3 } \quad =3(3b-4x)$

State whether True or False.

Divide:

$-4a^2b^3-8ab^2+6ab$ by

$-2ab$ , then answer is

$2ab^2+4b-3$ .

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True
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False

Explanation

$\Rightarrow \frac{-4a^2b^3-8ab^2+6ab}{-2ab}$
In

$-4a^2b^3-8ab^2+6ab$ we take

$-2ab$ common then

$\Rightarrow \frac{-2ab\left(2ab^2+4b-3 \right)}{-2ab}$

$\Rightarrow 2ab^2+4b-3$

State whether True or False.

Divide:

$15a^3b^4- 10a^4b^3-25a^3b^6$ by

$-5a^3b^2$ , then answer is

$-3b^2+2ab+5b^4$ .

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True
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False

Explanation

$\Rightarrow 15a^3b^4- 10a^4b^3-25a^3b^6 \div -5a^3b^2$

$\Rightarrow \frac{15a^3b^4- 10a^4b^3-25a^3b^6}{-5a^3b^2}$
we take

$5a^3b^2$ common

$\Rightarrow \frac{5a^3b^2\left (3b^2-2ab-5b^4 \right )}{-5a^3b^2}$

$\Rightarrow -3b^2+2ab+5b^4$

Factorise

$4a^2 - 8 ab$

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$4a(a + 2b)$
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$4a(a - b)$
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$4a(a - 2b)$
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$a(a - 2b)$

Explanation

Given expression is

$4a^2-8ab$ .

Taking

$4a$ common from both terms of

$4a^2-8ab$ , we get:

$4a(a-2b)$

Hence,

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

Factorise:

$3x^2 + 6x^3$

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

Explanation

$Factorizing:\quad { 3x }^{ 2 }+{ 6x }^{ 3 },\\ we\quad take\quad 3{ x }^{ 2 }\quad common,\\ =3{ x }^{ 2 }(1+2x)$

State whether True or False.

Factorization of

$6x^3 - 8x^2$ is

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

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True
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False

Explanation

$\\ 6{ x }^{ 2 }+8{ x }^{ 3 }\\ =2{ x }^{ 2 }(3-4x)\\ \\$

State whether True or False

Factorization of

$36 x^2 y^2 - 30 x^3 y^3 + 48 x^3 y^2$ is

$6x^2 y^2 (6 - 5xy + 8x)$ .

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True
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False

Explanation

$Factorizing\quad 36x^{ 2 }y^{ 2 }-30x^{ 3 }y^{ 3 }+48x^{ 3 }y^{ 2 },\\ =6x^{ 2 }y^{ 2 }(6-5xy+8x)$

State whether True or False.

Divide:

$-14x^6y^3-21x^4y^5+7x^5y^4$ by

$7x^2y^2$ , then answer is

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

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True
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False

Explanation

$\Rightarrow -14x^6y^3-21x^4y^5+7x^5y^4 \div 7x^2y^2$

$\Rightarrow \frac{-14x^6y^3-21x^4y^5+7x^5y^4}{7x^2y^2}$
We take

$7x^2y^2$ common

$\Rightarrow \frac{7x^2y^2\left (-2x^4y-3x^2y^3+x^3y^2 \right )}{7x^2y^2}$

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

CBSE Questions for Class 8 Maths Factorisation Quiz 1 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Factorisation Quiz 1 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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