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CBSE Questions for Class 8 Maths Factorisation Quiz 3 - MCQExams.com
CBSE
Class 8 Maths
Factorisation
Quiz 3
Factorise : $$\displaystyle 13a+26$$
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$$\displaystyle a+13$$
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$$\displaystyle a+2$$
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$$\displaystyle 13(a+26)$$
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$$\displaystyle 13(a+2)$$
Explanation
$$\displaystyle 13a+26=\left( 13\times a \right) +\left( 13\times 2 \right) $$
$$ =13\left( a+2 \right) $$
Factorise : $$\displaystyle 8x-64$$
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$$\displaystyle 8(x-9)$$
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$$\displaystyle 8x-8$$
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$$\displaystyle 8(x-6)$$
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$$\displaystyle 8(x-8)$$
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?
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$$\displaystyle 5x+3=5(x+3)$$
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$$\displaystyle 5x+3{ x }^{ 2 }=x\left( 5+3x \right) $$
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$$\displaystyle 5x+3{ x }^{ 2 }=8x\left( x+{ x }^{ 2 } \right) $$
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$$\displaystyle 5x+3=5\left( 1+3x \right) $$
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.
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Polynomial
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Factorisation
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Division algorithm
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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 $$x^4-x^3-x^2$$
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$$x(x^2-x-1)$$
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$$x(x-x^2-x^3)$$
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$$x(x^3-x^2)$$
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$$x^2(x^2-x-1)$$
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:
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$$\displaystyle x\left( y-z \right) $$
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$$y(z-x)$$
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$$x(y+z)$$
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$$z(-x+y)$$
Explanation
$$xy - xz $$
$$= x(y - z)$$
A number besides a bracket means the number is multiplied with each term in the bracket.
If $$30x^{3} + 45x^{2} - 10x$$ is divided by $$5x$$, find the resulting coefficient of $$x$$
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$$6$$
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$$9$$
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$$25$$
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$$40$$
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:
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$$\displaystyle c\left( ab+xy \right) $$
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$$c(ab-xy)$$
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$$x(ab+cy)$$
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$$x(ab-cy)$$
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.
$$\left( 49{ x }^{ 2 }yz+35p \right) \div 7=$$
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$$7{ x }^{ 2 }yz-35p$$
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$$7{ x }^{ 2 }yz+35p$$
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$$7{ x }^{ 2 }yz-5p$$
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$$7{ x }^{ 2 }yz+5p$$
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.
$$3{y^2} - 7y$$
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7y(3y-1)
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y(3y-7)
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3y(y-7)
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7(3y-1)
Explanation
$$3y^2-7y$$
$$=y(3y-7)$$.
One of the factor of $$(25x^2-1)+(1+5x^2)$$ is
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$$5+x$$
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$$5-x$$
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$$5x-1$$
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$$10x$$
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
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$$\displaystyle 3x+1$$
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$$\displaystyle -3x-1$$
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$$\displaystyle -3x+1$$
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$$\displaystyle 3x-1$$
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
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$$\displaystyle 2+c+3cd$$
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$$\displaystyle -2+c-3cd$$
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$$\displaystyle -18+9c-27cd$$
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$$\displaystyle 18-9c-27cd$$
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:
$$b(a+d) -c(a+d)$$
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$$(a - c)(b + d)$$
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$$(b + c)(a - d)$$
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$$(b - c)(a + d)$$
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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 :
$$6xy^2 + 4x^2y$$
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$$2xy(3x+y)$$
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$$xy(3x+2y)$$
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$$2xy(2x+3y)$$
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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 : $$4ax + 4ay$$
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$$4a(x + y)$$
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$$4(x + y)$$
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$$4a(x - y)$$
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$$4(x - y)$$
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 :
$$10 a^2-15 b^2 + 20 c^2$$
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$$5(2a^2 - 3b^2 + 4c^2)$$
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$$5(2a -3b^2 + 4c^2)$$
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$$5(10a^2 - 3b^2 + 4c^2)$$
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$$5(2a^2 + 3b^2 + 4c^2)$$
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?
$$Volume: 3x^2-12x$$
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$$4\times x \times (x-4)$$
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$$3\times x \times (x-4)$$
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$$4\times x \times (x+4)$$
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$$3\times x \times (x+4)$$
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
$$ax^2y + bxy^2 + cxyz$$
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$$x(ax+by+cz)$$
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$$y(ax-by-cz)$$
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$$xy(ax+by+cz)$$
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$$xy(a+by+cz)$$
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 $$2xy$$
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$$4y -3y + 5xy^3$$
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$$4xy -3y - 5xy^3$$
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$$4x -3y - 5xy^2$$
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$$4xy -3y + 5xy^2$$
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 $$(-56mnp^2 + 7mnp).$$
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$$7mnp(-8p + 1)$$
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$$7mnp(8p + 1)$$
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$$7mnp(8p - 1)$$
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None of these
Explanation
$$-56mnp^2+7mnp$$
$$=-7\times 8mnp^2+7mnp$$
$$=7mnp(-8p+1)$$
Factorise :
$$\displaystyle 121ac-16a^{2}b^{2}$$
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$$a(121c-16ab^{2})$$
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$$a(121c+16ab^{2})$$
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$$a(121c-16ac^{2})$$
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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?
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$$x^2+2x=x(x+2)$$
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$$x^2+2x=x(x+1)$$
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$$x^2+2x=x(x+3)$$
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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 $$\displaystyle 4m^{3}n^{2}+12m^{2}n^{2}+18m^{4}n^{3}$$
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$$\displaystyle m^{3}n^{3}(m+3mn+4m^{2}n)$$
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$$\displaystyle m^{2}n^{2}(5m+2n+6m^{2}n)$$
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$$\displaystyle 2m^{2}n^{2}(2m+6+9m^{2}n)$$
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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 : $$5mn+15mnp$$
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$$5mn(1 + 3p)$$
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$$3mn(1 + 5p)$$
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$$5mn(1 - 3p)$$
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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
$$10a^2 -15b^2 + 20c^2$$
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$$5(a^2 -3b^2 + 4c^2)$$
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$$10(2a^2 +3b^2 - 4c^2)$$
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$$5(2a^2 -3b^2 + 4c^2)$$
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$$10(a^2 -b^2 + 4c^2)$$
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
$$8 (x^3 y^2 z^2 + x^2y^3z^2 + x^2y^2z^3) \div 4x^2y^2z^2$$
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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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$$2(x+2y+2z)$$
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
$$(p^3 q^6 -p^6 q^3) \div p^3 q^3$$
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$$q^3+p^3$$
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$$q^3-p^3$$
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$$q^2-p^2$$
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$$q^2+p^2$$
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: $$\displaystyle \left( 21x-35 \right) \div 7$$
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$$\displaystyle 2x-5$$
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$$\displaystyle 4x-5$$
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$$\displaystyle 3x-5$$
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$$\displaystyle 5x-3$$
Explanation
$$\displaystyle \left( 21x-35 \right) \div 7=\frac { 21x-35 }{ 7 } $$
$$=\dfrac { 7\left( 3x-5 \right) }{ 7 } =3x-5$$
Simplify: $$\displaystyle \left( { 8m }^{ 2 }-9m \right) \div 3m$$
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$$\displaystyle \frac { 1 }{ 3 } \left( 8m-9 \right) $$
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$$\displaystyle 8m-9$$
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$$\displaystyle \frac { 1 }{ 3 } \left( 8m-3 \right) $$
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$$\displaystyle 5m-3$$
Explanation
$$\displaystyle \left( { 8m }^{ 2 }-9m \right) \div 3m=\frac { { 8m }^{ 2 }-9m }{ 3m } $$
$$\displaystyle =\frac { m\left( 8m-9 \right) }{ 3m } $$
$$\displaystyle =\frac { 8m-9 }{ 3 } $$
$$\displaystyle = \frac { 1 }{ 3 } \left( 8m-9 \right) $$
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