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CBSE Questions for Class 8 Maths Factorisation Quiz 5 - MCQExams.com
CBSE
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: $$\displaystyle 80{ p }^{ 2 }-72p$$
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$$\displaystyle 8p(5p-9)$$
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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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$$l(2l^2 - 3l + 3)$$
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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,
$$\dfrac { \dfrac { 2 }{ 3 } a^{ 2 }b^{ 2 }c^{ 2 }+\dfrac { 4 }{ 3 } ab^{ 2 }c^{ 2 } }{ \dfrac { 1 }{ 2 } abc } =\dfrac { 4 }{ 3 } (abc+2bc)$$
.
$$(25x^5 - 15x^4) \div 5x^3$$
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$$x(5x + 3)$$
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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
$$16x + 64x^2 y$$
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$$16x (1 + 4xy)$$
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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
$$pq - pr - 3ps$$
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$$p(q + r - 3s)$$
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$$p(q + r + 3s)$$
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$$q(q + r - 3s)$$
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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:
$$2a^3 - 3a^2 b + 2a^2 c$$
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$$a^2 (2a - 3b + 2c)$$
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$$a^2 (2a + 3b + 2c)$$
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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:
$$10x^3 - 25 x^4 y$$
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$$5x^3 (2 - 5xy)$$
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$$5x^3 (2 + 5xy)$$
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$$x^3 (2 - 5xy)$$
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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, $$(3x^3+2x^2+7x-5)\div (x+3)$$
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$$3x^2-7x+28, -89$$
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$$3x^2-7x+40, 125$$
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$$3x^2-11x+40, -15$$
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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 $$-89$$.
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
$$-\dfrac { 41 }{ 2 }$$
.
Find the quotient and remainder of the given expression,
$$(8x^4-2x^2+6x-5)\div (4x+1)$$
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$$2x^3-\dfrac{x^2}{2}-\dfrac{3}{8}x+\dfrac{51}{32}, -\dfrac{211}{32}$$
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$$2x^3-\dfrac{x^2}{2}-\dfrac{3}{8}x+\dfrac{51}{32}, \dfrac{211}{32}$$
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$$2x^3+\dfrac{x^2}{2}-\dfrac{3}{8}x+\dfrac{51}{32}, -\dfrac{211}{32}$$
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None of these
The factorized form of $$3x — 24$$ is
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$$ 3x \times 24 $$
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$$ 3(x - 8) $$
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$$ 24 (x - 3)$$
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$$ 3(x - 12)$$
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
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$$ (23x + 54) (y - 2)$$
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$$ (23x + 54y) (y - 2)$$
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$$ (23xy + 54y) (- 46x - 108)$$
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$$ (23x + 54) (y + 2)$$
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 $$3a(-a-b-c)$$.
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True
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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 $$4$$.
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True
0%
False
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