MCQ Questions for Class 8 Maths Factorisation Quiz 2

Factorize by taking out the common factors:

$3a^{2}\, -\, 9ab$

0%
$3a (a - 3b)$
0%
$3a (a + 3b)$
0%
$2a (a - 3b)$
0%
$2a (a + 3b)$

Explanation

$3a^2-9ab$

$=3a(a-3b)$

State whether True or False.

Factorization of

$35 a^3 b^2 c + 42 ab^2 c^2$ is

$7ab^2 c (5a^2 + 6c)$ .

0%
True
0%
False

Explanation

$Factorizing\quad 35a^{ 3 }b^{ 2 }c+42ab^{ 2 }c^{ 2 }\\ =\quad 7ab^{ 2 }c(5a^{ 2 }+6c)$

State whether True or False.

Divide:

$9x^3y-15x^2y^2-6x^4y^3$ by

$3x^2y$ , then answer is

$3x-5y-2x^2y^2$ .

0%
True
0%
False

Explanation

$\Rightarrow 9x^3y-15x^2y^2-6x^4y^3 \div 3x^2y$

$\Rightarrow \frac{9x^3y-15x^2y^2-6x^4y^3}{3x^2y}$
we take

$3x^2y$ common

$\Rightarrow \frac{3x^2y\left (3x-5y-2x^2y^2 \right )}{3x^2y}$

$\Rightarrow 3x-5y-2x^2y^2$

Factorize :

$y^2 + 100y$

0%
$y(y+100)$
0%
$y^2(y+100)$
0%
$y(y^2+100)$
0%
none of these

Explanation

We need to find value of

$y^2+100y$

Taking

$y$ common, we get

$y(y+100)$ .

Hence, option A is correct.

Factorise

$-16z + 20z^3$

0%
$-z (4 - 5z^3)$
0%
$-z (4 - 5z^2)$
0%
$4z (4 - 5z^2)$
0%
$-4z (4 - 5z^2)$

Explanation

$-16z+20z^3$

$=-4\times 4\times z+4\times 5\times z\times z\times z$

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

Factorise

$6p -12q$

0%
$(p-2q)$
0%
$6(p-2q)$
0%
$6(p-q)$
0%
$(p-q)$

Explanation

$6p-12q$

$=2\times3\times p-2\times 2\times3\times q$

$=2\times3(p-2q)$

$=6(p-2q)$

Factorise :

$4x+12$

0%
$4(x+2)$
0%
$4(x+3)$
0%
$3(x+4)$
0%
none of these

Explanation

$4x+12=4(x+3)$

Factorise:

$20l^2m + 30alm$

0%
$10 l m ( 2l +3a)$
0%
$10 l m ( l +3a)$
0%
$20 l m ( 2l +3a)$
0%
$30 l m ( 2l +a)$

Explanation

We have,

$20l^2 m =2 \times 2 \times 5 \times l \times l \times m$
and,

$30alm = 3 \times 2 \times 5 \times a \times l \times m$
The two terms have

$2, 5, l$ and

$m$ as common factors

$\therefore 20l^2m + 30alm = (2 \times 2 \times 5 \times l \times l \times m)+ (3 \times 2 \times 5 \times a \times l \times m) =2 \times 5 \times l \times m \times (2 \times l \times +3 \times a)$

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

Factorise:

$7x - 42$

0%
$7x$
0%
$x-7$
0%
$7$
0%
$7(x-6)$

Explanation

$7x-42$

$=7(x-6)$

Factorise:

$7a^2 +14a$

0%
$7a (a + 2)$
0%
$14a (a + 1)$
0%
$7a (a - 2)$
0%
$14 (a + 2)$

Explanation

Take

$7a$ common from the given expression.

$7a^{2}+14a$

$=7a(a+2)$

Factorise

$x^2yz + xy^2z + xyz^2$

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

Explanation

Take

$xyz$ common from every term,

$x^2yz+xy^2z+xyz^2=xyz(x+y+z)$

Factorise

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

0%
$2xy ( 2x - 15y )$
0%
$5y ( x - 15y )$
0%
$5xy ( x - 3y )$
0%
$xy (5 x - 3y )$

Explanation

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

$=5\times x\times x\times y-5\times 3\times x\times y\times y$

$=5xy(x-3y)$

Factorise

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

0%
$-a(a-b+c)$
0%
$-4a(a-b+c)$
0%
$-4(2a-b+c)$
0%
$-4a(2a-b+2c)$

Explanation

We have,

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

$4ab= 2 \times 2 \times a \times b$
and,

$4ca=2 \times 2 \times c \times a$
The three terms have 2, 2 and a as common facotrs

$-4a^2 + 4ab -4ca = -(2 \times 2 \times a \times a ) + (2 \times 2 \times a \times b ) - ( 2 \times 2 \times c \times a )$

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

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

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

Factorise:

$15xy-6x+5y -2$

0%
$(x-1)(y+2)$
0%
$(x+1)(y-2)$
0%
$(3x+1)(5y-2)$
0%
$(3x-1)(5y+2)$

Explanation

$15xy-6x+5y -2$

$=15xy+5y-6x-2$

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

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

Factorise the expression:

$7p^2 + 21q^2$

0%
$7(p^2+q^2)$
0%
$(p^2-3q^2)$
0%
$7(p^2+3q^2)$
0%
$(pq^2+3q^2)$

Explanation

$7p^2+21q^2$

$=7\times p\times p + 7\times 3\times q\times q$

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

Factorise the expression:

$ax^2+ bx$

0%
$x(ax+b)$
0%
$x(a+b)$
0%
$x(x+ba)$
0%
$(ax+b)$

Explanation

$ax^2+bx$

$= x(ax+b)$

Factorise:

$ax+bx -ay-by$

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

Explanation

$ax+bx -ay-by$

$= (ax+ bx) -(ay + by)$
[Grouping the terms]

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

$= (a+b)(x-y)$
[Taking (a + b) common]

Factorise the expression:

$2x^3 + 2xy^2 + 2xz^2$

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

Explanation

$2x^3+2xy^2+2xz^2$

$=2\times x\times x\times x + 2\times x\times y\times y+2\times x\times z\times z$

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

Factorise:

$15pq+15+9q+25p$

0%
$(5+3q)(3+5p)$
0%
$q(3+5p)$
0%
$(5-3q)(5p)$
0%
$(5-3q)(3+5p)$

Explanation

$15pq+15+9q+25p$

$=15pq+9q+25p+15$

$=3q(5p+3)+5(5p+3)$

$=(5p+3)(3q+5)$

Factorise:

$x^2+xy+8x +8y$

0%
$(x+8)(x-y)$
0%
$(x+8)(x-8)$
0%
$(x-y)(x+y)$
0%
$(x+8)(x+y)$

Explanation

$x^2+xy+8x +8y = (x^2+xy)+(8x+8y)$

$= x(x+y)+8(x+y)$

$=(x + y) (x + 8)$ [Taking

$(x + y)$ common]

Factorize:

$\displaystyle 2x+6$

0%
$\displaystyle 2(x+6)$
0%
$\displaystyle 2(x+3)$
0%
$\displaystyle x+3$
0%
$\displaystyle 2x+3$

Explanation

$2x+6 = (2\times x) + (2 \times 3)$

$=2 (x + 3)$

Divide the given polynomial by the given monomial.

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

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

Explanation

$(3y^84y^6+5y^4)\div y^4$

$3y^8=3\times y\times y\times y\times y\times y\times y\times y\times y$

$4y^6=4\times y\times y\times y\times y\times y\times y$

$5y^4=5\times y\times y\times y\times y$

$=\frac{(3\times y\times y\times y\times y\times y\times y\times y\times y)-(4\times y\times y\times y\times y\times y\times y)+(5\times y\times y\times y\times y)}{y^4}$

$=\frac { y^ 4[(3\times y\times y\times y\times y)-(4\times y\times y)+(5) ]}{ y^{ 4 } }$

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

Factorise:

$ab^2-bc^2 -ab +c^2$

0%
$(b-1)(ab-c^2)$
0%
$(b+1)(2ab-c^2)$
0%
$(b-2)(ab-c^2)$
0%
$(b-2)(ab-ac^2)$

Explanation

$ab^2-bc^2 -ab +c^2$

$=ab^2 -ab-bc^2+c^2$

$=ab(b-1)-c^2(b-1)$

$=(b-1)(ab-c^2)$

Factorization means , writing an expression as a product of its......

0%
factors
0%
multiples
0%
HCF
0%
LCM

Which of the following expression is correct?

0%
$\displaystyle 5\left( x-6 \right) =5x-6$
0%
$\displaystyle 4x+7x=11{ x }^{ 2 }$
0%
$\displaystyle x+4x+5x=8x$
0%
$\displaystyle x+4{ x }^{ 2 }=x\left( 1+4x \right)$

Explanation

$\displaystyle x+4{ x }^{ 2 }=\left( x \right) +\left( 4\times x\times x \right)$

$\displaystyle =x\left( 1+4\times x \right)$

$\displaystyle =x\left( 1+4x \right)$

$\therefore$ The statement

$x+4x^2=x(1+4x)$ is correct.

Factorise:

$\displaystyle 5x-115$

0%
$\displaystyle x-115$
0%
$\displaystyle 5x-23$
0%
$\displaystyle 5(x-23)$
0%
$\displaystyle 5(x-3)$

Explanation

$\displaystyle 5x-115=\left( 5\times x \right) -\left( 5\times 23 \right)$

$=5(x-23)$

Factorise:

$\displaystyle { x }^{ 3 }+4{ x }^{ 2 }+5x$

0%
$\displaystyle x\left( { x }^{ 2 }+4x+5 \right)$
0%
$\displaystyle x\left( { x }^{ 2 }+4x+5x \right)$
0%
$\displaystyle { x }^{ 3 }+4x+5$
0%
$\displaystyle { x }^{ 2 }+4x+5$

Explanation

$\displaystyle { x }^{ 3 }+4{ x }^{ 2 }+5x=x\left( { x }^{ 2 }+4x+5 \right)$

Which of the following statements is true ?

0%
$\displaystyle 6x+3x=9{ x }^{ 2 }$
0%
$\displaystyle 4(x+5)=4x+5$
0%
$\displaystyle 4(x+5)=4x+20$
0%
$\displaystyle x\left( 4x+5 \right) =4{ x }^{ 2 }+5$

Explanation

$\displaystyle 4\left( x+5 \right) =4 \times x+ 4\times 5$

$= 4x+20$

$\displaystyle \therefore 4\left( x+5 \right) =4x+20$ is a correct statement.

Simplify:

$\displaystyle \frac{12ab^3-6a^2b}{3ab}$ (given that

$ab\neq 0$ )

0%
$6b^2 - a$
0%
$2b^2 - 3a$
0%
$4b^2-2a$
0%
$3b^2 - 2a$

Explanation

$(2b^2 - a)$ or

$4b^2 -2a$ : First, factor out common terms in the numerator. Then, cancel terms in both the numerator and denominator:

$\displaystyle \frac{6ab(2b^2 -a)}{3ab} = 2(2b^2 - a)$ or

$4b^2-2a$

Factorise :

$\displaystyle 5xy+25xyz$

0%
$\displaystyle 5xy(1+5xyz)$
0%
$\displaystyle 5xy(1+z)$
0%
$\displaystyle 5xy(1+5z)$
0%
$\displaystyle 5xyz$

Explanation

$\displaystyle 5xy+25xyz=\left( 5\times x\times y \right) +\left( 5\times 5\times x\times y\times z \right)$

$=5xy(1+5z)$

CBSE Questions for Class 8 Maths Factorisation Quiz 2 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Factorisation Quiz 2 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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