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CBSE Questions for Class 8 Maths Factorisation Quiz 2 - MCQExams.com
CBSE
Class 8 Maths
Factorisation
Quiz 2
Factorize by taking out the common factors:
$$3a^{2}\, -\, 9ab$$
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$$3a (a - 3b)$$
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$$3a (a + 3b)$$
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$$2a (a - 3b)$$
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$$2a (a + 3b)$$
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 $$7ab^2 c (5a^2 + 6c)$$.
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True
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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 $$3x-5y-2x^2y^2$$.
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True
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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 :
$$y^2 + 100y$$
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$$y(y+100)$$
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$$y^2(y+100)$$
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$$y(y^2+100)$$
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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
$$-16z + 20z^3$$
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$$-z (4 - 5z^3) $$
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$$-z (4 - 5z^2) $$
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$$4z (4 - 5z^2) $$
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$$-4z (4 - 5z^2) $$
Explanation
$$-16z+20z^3$$
$$=-4\times 4\times z+4\times 5\times z\times z\times z$$
$$=4z(5z^2-4)$$
Factorise
$$6p -12q$$
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$$(p-2q)$$
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$$6(p-2q)$$
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$$6(p-q)$$
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$$(p-q)$$
Explanation
$$6p-12q$$
$$=2\times3\times p-2\times 2\times3\times q$$
$$=2\times3(p-2q)$$
$$=6(p-2q)$$
Factorise :
$$4x+12$$
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$$4(x+2)$$
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$$4(x+3)$$
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$$3(x+4)$$
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none of these
Explanation
$$4x+12=4(x+3)$$
Factorise:
$$20l^2m + 30alm$$
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$$10 l m ( 2l +3a)$$
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$$10 l m ( l +3a)$$
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$$20 l m ( 2l +3a)$$
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$$30 l m ( 2l +a)$$
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:
$$7x - 42$$
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$$7x$$
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$$x-7$$
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$$7$$
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$$7(x-6)$$
Explanation
$$7x-42$$
$$=7(x-6)$$
Factorise:
$$7a^2 +14a$$
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$$7a (a + 2)$$
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$$14a (a + 1)$$
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$$7a (a - 2)$$
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$$14 (a + 2)$$
Explanation
Take $$7a$$ common from the given expression.
$$7a^{2}+14a$$ $$=7a(a+2)$$
Factorise
$$x^2yz + xy^2z + xyz^2$$
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$$yz(x+y-z)$$
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$$xyz(x-y+z)$$
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$$xyz(x+y+z)$$
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$$xyz(2x+y+2z)$$
Explanation
Take $$xyz$$ common from every term,
$$x^2yz+xy^2z+xyz^2=xyz(x+y+z)$$
Factorise
$$5x^2y -15xy^2$$
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$$2xy ( 2x - 15y )$$
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$$5y ( x - 15y )$$
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$$5xy ( x - 3y )$$
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$$xy (5 x - 3y )$$
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
$$-4a^2 + 4ab -4ca$$
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$$ -a(a-b+c) $$
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$$ -4a(a-b+c) $$
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$$ -4(2a-b+c) $$
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$$ -4a(2a-b+2c) $$
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:
$$15xy-6x+5y -2$$
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$$(x-1)(y+2)$$
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$$(x+1)(y-2)$$
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$$(3x+1)(5y-2)$$
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$$(3x-1)(5y+2)$$
Explanation
$$15xy-6x+5y -2$$
$$=15xy+5y-6x-2$$
$$=5y(3x+1)-2(3x+1)$$
$$=(3x+1)(5y-2)$$
Factorise the expression:
$$7p^2 + 21q^2$$
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$$7(p^2+q^2)$$
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$$(p^2-3q^2)$$
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$$7(p^2+3q^2)$$
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$$(pq^2+3q^2)$$
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:
$$ax^2+ bx$$
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$$x(ax+b)$$
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$$x(a+b)$$
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$$x(x+ba)$$
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$$(ax+b)$$
Explanation
$$ax^2+bx $$
$$= x(ax+b)$$
Factorise:
$$ax+bx -ay-by$$
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$$(x+y)(a+b)$$
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$$(x-a)(a+y)$$
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$$(x-y)(a+b)$$
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$$(x+a)(a-y)$$
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:
$$2x^3 + 2xy^2 + 2xz^2$$
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$$(x^2+y^2-z^2)$$
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$$2x(x^2-y^2+z^2)$$
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$$(x^2+y^2+z^2)$$
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$$2x(x^2+y^2+z^2)$$
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:
$$15pq+15+9q+25p$$
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$$(5+3q)(3+5p)$$
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$$q(3+5p)$$
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$$(5-3q)(5p)$$
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$$(5-3q)(3+5p)$$
Explanation
$$15pq+15+9q+25p$$
$$=15pq+9q+25p+15$$
$$=3q(5p+3)+5(5p+3)$$
$$=(5p+3)(3q+5)$$
Factorise:
$$x^2+xy+8x +8y$$
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$$(x+8)(x-y)$$
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$$(x+8)(x-8)$$
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$$(x-y)(x+y)$$
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$$(x+8)(x+y)$$
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: $$\displaystyle 2x+6$$
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$$\displaystyle 2(x+6)$$
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$$\displaystyle 2(x+3)$$
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$$\displaystyle x+3$$
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$$\displaystyle 2x+3$$
Explanation
$$2x+6 = (2\times x) + (2 \times 3)$$
$$=2 (x + 3)$$
Divide the given polynomial by the given monomial.
$$(3y^8 -4y^6+ 5y^4) \div y^4$$
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$$3y^4-4y^2+5$$
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$$3y^3-4y^2-5$$
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$$3y^3-4y^2+5$$
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$$3y^4-y^2+5$$
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:
$$ab^2-bc^2 -ab +c^2$$
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$$(b-1)(ab-c^2)$$
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$$(b+1)(2ab-c^2)$$
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$$(b-2)(ab-c^2)$$
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$$(b-2)(ab-ac^2)$$
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......
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factors
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multiples
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HCF
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LCM
Explanation
Factorization means writing an expression as a product of its factors.
Which of the following expression is correct?
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$$\displaystyle 5\left( x-6 \right) =5x-6$$
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$$\displaystyle 4x+7x=11{ x }^{ 2 }$$
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$$\displaystyle x+4x+5x=8x$$
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$$\displaystyle x+4{ x }^{ 2 }=x\left( 1+4x \right) $$
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: $$\displaystyle 5x-115$$
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$$\displaystyle x-115$$
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$$\displaystyle 5x-23$$
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$$\displaystyle 5(x-23)$$
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$$\displaystyle 5(x-3)$$
Explanation
$$\displaystyle 5x-115=\left( 5\times x \right) -\left( 5\times 23 \right) $$
$$ =5(x-23)$$
Factorise: $$\displaystyle { x }^{ 3 }+4{ x }^{ 2 }+5x$$
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$$\displaystyle x\left( { x }^{ 2 }+4x+5 \right) $$
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$$\displaystyle x\left( { x }^{ 2 }+4x+5x \right) $$
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$$\displaystyle { x }^{ 3 }+4x+5$$
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$$\displaystyle { x }^{ 2 }+4x+5$$
Explanation
$$\displaystyle { x }^{ 3 }+4{ x }^{ 2 }+5x=x\left( { x }^{ 2 }+4x+5 \right) $$
Which of the following statements is true ?
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$$\displaystyle 6x+3x=9{ x }^{ 2 }$$
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$$\displaystyle 4(x+5)=4x+5$$
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$$\displaystyle 4(x+5)=4x+20$$
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$$\displaystyle x\left( 4x+5 \right) =4{ x }^{ 2 }+5$$
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 $$ab\neq 0$$)
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$$6b^2 - a$$
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$$2b^2 - 3a$$
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$$4b^2-2a$$
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$$3b^2 - 2a$$
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 : $$\displaystyle 5xy+25xyz$$
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$$\displaystyle 5xy(1+5xyz)$$
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$$\displaystyle 5xy(1+z)$$
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$$\displaystyle 5xy(1+5z)$$
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$$\displaystyle 5xyz$$
Explanation
$$\displaystyle 5xy+25xyz=\left( 5\times x\times y \right) +\left( 5\times 5\times x\times y\times z \right) $$
$$=5xy(1+5z)$$
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