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CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 6 - MCQExams.com
CBSE
Class 8 Maths
Algebraic Expressions And Identities
Quiz 6
Find the missing term in the following
problem.
$$\displaystyle {\left (\frac{3x}{4}\, -\, \frac{4y}{3} \right )^2\, =\, \frac{9x^2}{16}\, +\, .........\, +\, \frac{16y^2}{9}}$$
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$$2xy$$
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$$-2xy$$
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$$12xy$$
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$$-12xy$$
Explanation
The L.H.S. of the algebraic expression is of the form $${ \left( a-b \right) }^{ 2 }$$
where, $$a=\dfrac { 3x }{ 4 } \quad \& \quad b=\dfrac { 4y }{ 3 } $$.
$${ \left( a-b \right) }^{ 2 }={ a }^{ 2 }-2ab+{ b }^{ 2 }\\ \implies \left(\dfrac{3x}{4}-\dfrac{4y}{3}\right)={ \left( \dfrac { 3x }{ 4 } \right) }^{ 2 }-2\times \dfrac { 3x }{ 4 } \times \dfrac { 4y }{ 3 } +{ \left( \dfrac { 4y }{ 3 } \right) }^{ 2 }\\ =\dfrac { 9{ x }^{ 2 } }{ 16 } -2xy+\dfrac { 16{ y }^{ 2 } }{ 9 } $$.
So the middle term $$-2xy$$ is missing.
Find the areas of rectangles with the following pairs of monomials as their lengths and breadths
respectively
$$(p, q); (10m, 5n); (20x^2, 5y^2); (4x, 3x^2); (3mn, 4np)$$
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$$100x^2y^2 ;12x^3 ; 12mn^2p$$
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$$100x^2y^2 ;12x^2 ; 12mn^2p$$
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$$100x^2y ;12x^3 ; 12mn^2p$$
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$$100x^2y^2 ;12x^3 ; 12mnp$$
Explanation
We know that the area of a rectangle = I $$\times $$ b, where I = length and b = breadth.
Therefore, the areas of rectangles with pair of monomials $$(p, q), (10m, 5n); (20x^2, 5y^2); (4x, 3x^2)$$and$$ (3mn, 4np) $$as their lengths and breadths are given by $$p\times q = pq$$
1.$$ 10m \times n =(10 \times5) \times (m \times n) = 50mn$$
2. $$20x^2 \times 5y^2= (20 \times 5) \times (x^2 \times y^2) = 100x^2y^2$$
3. $$4x \times 3x^2 = (4 \times 3) \times (x \times x^2)$$
= $$12x^3$$
and,
4. $$3mn \times 4np = (3\times4) \times (m \times n \times n \times p) $$
= $$12mn^2p$$
Multiply the binomials
$$(2x + 5)$$ and $$(4x - 3)$$
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$$ 8x^2+ 14x -15$$
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$$ 7x^2-14x -15$$
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$$ x^2+ 14x -18$$
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None of these
Explanation
$$(2x + 5)\times (4x - 3)= 2x(4x - 3) + 5(4x - 3)$$
$$ = 8x^2- 6x + 20x-15$$
$$ = 8x^2+ 14x -15$$
Obtain the product of:
$$rn, -mn, mnp$$
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$$-m^3n^3p$$
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$$-m^2n^2p$$
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$$m^2n^2p$$
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$$-m^3n^2p$$
Explanation
product of
rn,mn,mnp
$$rn \times -mn \times mnp = (1\times -1 \times 1) \times (r \times m \times m \times n \times n \times n \times p)$$
=$$ -1 \times r \times m^2 \times n^3 \times p$$
=$$-rm^2n^3p$$
Using identities, evaluate
$$8.9^2$$.
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$$74.53$$
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$$73.42$$
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$$79.21$$
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$$73.21$$
Explanation
Given, $$8.9^2$$
$$= (9 -0.1)^2$$.
We know,
$$(a-b)^2 =a^2-2ab+b^2$$.
Then,
$$8.9^2$$
$$= (9 -0.1)^2$$
$$ = (9)^2 -2(9) (0.1) + (0.1)^2$$
$$ = 81 -1.8 + 0.01 = 79.20+0.01=79.21$$.
Therefore, option $$C$$ is correct.
Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
$$a, 2b, 3c$$
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$$6abc$$
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$$3ab$$
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$$2abc$$
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$$9ac$$
Explanation
volume of rectangular box= length $$\times$$ breadth $$\times$$ height
volume = $$a \times 2b \times 3c $$
= $$(1 \times 2 \times 3) \times (a \times b \times c) $$
$$= 6abc$$
Obtain the product of
$$xy, yz, zx$$.
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$$x^2y^2z^3$$
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$$x^2y^2z^2$$
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$$2x^2y^2z^2$$
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$$2x^2y^2z^3$$
Explanation
product of
xy, yz, zx is
$$ xy \times yz \times zx = x \times x \times y \times y \times z \times z $$
= $$x^{1 +1} \times y^{1+1} \times z^{1+1}$$
= $$x^2y^2z^2$$
Carry out the multiplication of the expressions in the following pair:
$$a^2 -9, 4a$$
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$$ a^3 -12a$$
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$$ 4a^3 -22a$$
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$$ a^3 -16$$
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$$ 4a^3 -36a$$
Explanation
multiplication of $$a^2 -9, 4a$$
=$$ (a^2-9) \times 4a = a^2 \times 4a -9 \times 4a$$
$$ 4a^3 -36a$$
Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
$$ 5a, 3a^2, 7a^4$$
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$$102a^7$$
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$$105a^7$$
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$$192a^7$$
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$$606a^7$$
Simplify: $$(7m -8n)^2 + (7m + 8n)^2$$.
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$$49m^2+164n^2$$
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$$98m^2+128n^2$$
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$$49m^2+64n^2$$
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$$128m^2+64n^2$$
Explanation
Given, $$(7m -8n)^2 + (7m + 8n)^2$$.
We know, $$(a-b)^2=a^2-2ab+b^2$$
and $$(a+b)^2=a^2+2ab+b^2$$.
Then,
$$(7m -8n)^2 + (7m + 8n)^2$$
$$ = ((7m)^2-2(7m)(8n) + (8n)^2)+((7m)^2 + 2(7m)(8n) + (8n)^2)$$
$$ = (49m^2-112mn + 64n^2)+(49m^2 + 112mn + 64n^2)$$
$$ = 49m^2-112mn + 64n^2+49m^2 + 112m + 64n^2$$
$$ = 98m^2 + 128n^2$$.
Therefore, option $$B$$ is correct.
Using identities, evaluate
$$5.2^2$$.
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$$26.04$$
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$$2704$$
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$$27.04$$
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$$2604$$
Explanation
Given, $$(5.2)^2$$
We know, $$(a+b)^2=a^2+b^2+2ab$$.
Then,
$$(5.2)^2$$
$$=(5+0.2)^2$$
$$=5^2+(0.2)^2+2(5)(0.2)$$
$$=25+0.04+2(1)$$
$$=25+0.04+2$$
$$=27.04$$.
Therefore, option $$C$$ is correct.
The product of (x -y) and (y -x) is equal to
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$$x^2+y^2 -2xy$$
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$$2xy + x^2+y^2$$
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$$2xy-x^2-y^2$$
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$$x^2-2xy+y^2$$
Explanation
$$(x -y) (y -x) = x(y -x) -y(y -x)$$
$$=xy-x^2-y^2+yx$$
$$=2xy-x^2-y^2$$
The product of (2x + 3) and (3x + 2) is equal to
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$$6x^2+6$$
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$$6x^2+5x+6$$
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$$6(x^2+x+1)$$
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$$6x^2+13x+6$$
Explanation
$$(2x + 3) (3x + 2) =2x(3x + 2) + 3(3x + 2)$$
$$= 6x^2 + 4x + 9x +6$$
$$=6x^2 + 13x +6$$
Using $$(x + a) (x + b) = x^2+(a+b)x+ab$$, find
$$5.1 \times 5.2$$.
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$$17.61$$
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$$22.41$$
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$$13.32$$
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$$26.52$$
Explanation
Given,
$$5.1 \times 5.2 = (5 + 0.1)(5 + 0.2)$$.
We know, $$ (x+a)(x+b)=x^2+(a+b)x+ab$$.
Then,
$$5.1 \times 5.2 $$
$$= (5 + 0.1)(5 + 0.2)$$
$$ = (5)^2 + (0.1 + 0.2) (5)+ (0.1)(0.2)$$
$$ = 25 + (0.3) (5)+ 0.02$$
$$=25 + 1.5 + 0.02$$
$$= 26.52$$.
Therefore, option $$D$$ is correct.
Using identities, evaluate
$$12.1^2 -7.9^2$$.
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$$84$$
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$$72$$
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$$69$$
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$$87$$
Explanation
Given, $$(12.1)^2-(7.9)^2$$.
We know, $$(a^2-b^2)=(a+b)(a-b)$$.
Then,
$$(12.1)^2-(7.9)^2$$
$$=(12.1+7.9)(12.1-7.9)$$
$$=20\times 4.2$$
$$=84$$.
Therefore, option $$A$$ is correct.
The value of $$(x + 3y)^2 + (x - 3y)^2$$ is:
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$$2x^2 + 18 y^2$$
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$$2x^2 - 18 y^2$$
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$$2x^2 + 18y^2 - 2xy$$
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None of these
Explanation
We know, $$(a+b)^2=a^2+2ab+b^2$$ and $$(a-b)^2=a^2-2ab+b^2$$.
Then,
$$(x + 3y)^2 + (x - 3y)^2=[(x)^2+(3y)^2+2\times x\times3y]+[(x)^2+(3y)^2-2\times x\times3y]$$
$$=x^2+9y^2+6xy+x^2+9y^2-6xy$$
$$=2x^2+18y^2$$.
Hence, Option $$A$$ is correct.
Using $$(x + a) (x + b) = x^2+(a+b)x+ab$$, find
$$103 \times 104$$.
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$$14982$$
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$$11992$$
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$$10712$$
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$$11482$$
Explanation
Given, $$103 \times 104 $$
$$=(100+3)(100+4)$$.
Using, $$ (x+a)(x+b)=x^2+(a+b)x+ab$$.
Then,
$$103 \times 104 $$
$$=(100+3)(100+4)$$
$$ =(100)^2+(3+4)(100) +(3)(4)$$
$$ =(100)^2+(7)(100) +12$$
$$ = 10000 +700 + 12$$
$$ = 10712$$.
Therefore, option $$C$$ is correct.
Simplify: $$(x + 5)(x + 4)$$
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$$x^2+27x+20$$
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$$x^2+9x+20$$
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$$x^2+18x+20$$
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$$x^2+x+20$$
Explanation
$$(x + 5)(x + 4)$$, using the formula $$(x+a)(x+b)=x^2+x(a+b)+ab$$
$$=x^2+x(5+4)+5\times 4$$
$$=x^2+9x+20$$
$$(x + 2)(x + 3) -(x + 3)(x -2)=$$
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12
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0
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6x
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4x+12
Explanation
$$\textbf{Step 1: Multiply the factors by distribution law}$$
$$(x+2)(x+3)-(x+3)(x-2)$$
$$\Rightarrow x(x+3)+2(x+3)-(x(x-2)+3(x-2))$$
$$\Rightarrow x^2+3x+2x+6-(x^2-2x+3x-6)$$
$$\Rightarrow x^2+3x+2x+6-x^2+2x-3x+6$$
$$\text{[multiplying with $$(-)$$ will change the sign]}$$
$$\Rightarrow x^2+5x+6-x^2-x+6$$
$$\Rightarrow 4x+12$$
$$\textbf{Hence, Option D is correct}$$
Find the product of $$\left( 3x-5 \right)$$ and $$\left( x+5 \right) $$.
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$${ x }^{ 2 }-25x+10$$
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$${ x }^{ 2 }+10x-22$$
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$$3{ x }^{ 2 }+10x-25$$
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$$3{ x }^{ 2 }-22x-25$$
Explanation
Product of $$(3x-5)(x+5)=3x(x+5)-5(x+5)$$
$$=3x^2+15x-5x-25$$
$$=3x^2+10x-25$$
Option C is correct.
$$(a -b)^2 -(a + b)^2$$ is equal to:
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$$2ab$$
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$$4ab$$
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$$6ab$$
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$$-4ab$$
Explanation
Given, $$(a -b)^2-(a + b)^2 $$.
We know, $$(x+y)^2=x^2+2xy+y^2$$
and
$$(x-y)^2=x^2-2xy+y^2$$.
Then,
$$(a -b)^2-(a + b)^2 $$
$$= (a^2 + b^2 -2ab) -(a^2 + b^2 + 2ab) $$
$$= a^2 + b^2 -2ab -a^2 - b^2 - 2ab) $$
$$= -4ab$$.
Therefore, option $$D$$ is correct.
$$(3x+2y)(x+y)$$
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$$3x^2+y^2+5xy$$
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$$3x^2+2y^2+5xy$$
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$$3x^2+2y^2+6xy$$
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$$3x^2+2y+5xy$$
Explanation
$$=(3x+2y)(x+y)$$
$$=3x\times x+2y\times y+3x\times y+2y\times x$$
$$= 3x^2+2y^2+5xy$$
The product of a monomial and a trinomial is a
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monomial
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binomial
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trinomial
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none of these
Explanation
The product of a monomial and trinomial results trinomial.
For example: Suppose a monomial $$2x$$ and a trinomial $$2x+3y+3$$.
Their product will be $$2x(2x+3y+3)=4x^2+6xy+6x$$
The result is a trinomial.
Hence, option C is correct.
Find the value of $$1.03^2$$ using identity.
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$$1.0409$$
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$$1.0609$$
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$$1.0009$$
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$$1.0309$$
Explanation
Given, $$1.03^2 = (1+0.03)^2$$.
We know, $$(a+b)^2=a^2+2ab+b^2$$.
Then,
$$1.03^2 = (1+0.03)^2$$
$$=1^2+(0.03)^2+2\times 1\times 0.03$$
$$=1+0.0009+0.06$$
$$=1.0609$$.
Therefore, option $$B$$ is correct.
$$3x \times (4x + 2)$$
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$$12x^{2} + 6x$$
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$$6 (2x^{2} + x)$$
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$$x (12x + 6)$$
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All of the above
Explanation
$$3x \times (4x + 2)$$
$$= 12x^{2} + 6x$$
Hence option A is correct.
$$3x \times (4x + 2)$$
$$=3x \times 2 (2x+1)$$
$$=6(2x^2+x)$$
Hence option B is correct.
$$3x \times (4x + 2)$$
$$= 12x^{2} + 6x$$
$$=x(12x+6)$$
Hence option C is correct.
$$(4x + 1)(4x - 1) =$$
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$$16x^{2} + 8x + 2$$
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$$16x^{2} + 8x + 1$$
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$$16x^{2} - 1$$
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$$16x^{2} + 1$$
Explanation
Given that: $$ (4x+1)\times(4x-1)$$
We can see that:
$$ (4x+1)\times(4x-1)=$$
$$ 4x\times(4x-1)+1\times (4x-1)$$
$$\Rightarrow (4x\times 4x)-(4x\times 1)+(1\times 4x)-(1\times 1)$$
$$\Rightarrow 16x^2-4x+4x-1$$
$$\Rightarrow 16x^2-1$$
$$-2x\times 10y\times -3z=$$
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$$-60xyz$$
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$$60xz$$
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$$-60xz$$
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$$60xyz$$
Explanation
$$-2x\times 10y\times -3z= -2\times 10\times -3\times x\times y\times z$$
$$=60xyz$$
$$-8yz\times -2xy=$$
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$$16xyz$$
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$$-16xy^2z$$
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$$16xy^2z$$
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$$-16x^2y^2z^2$$
Explanation
$$-8yz\times -2xy = (-8\times -2)\times yz\times xy$$
$$= 16\times y^2zx=16xy^2z$$
$$2xy\times 7yz\times 9xz$$
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$$126xyz$$
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$$126x^2y^2z^2$$
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$$126xyz^2$$
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$$126x^2yz^2$$
Explanation
$$2xy\times 7yz\times 9xz= (2\times 7\times 9)\times xy\times yx\times xz$$
$$=126x^2y^2z^2$$
Find the product of $$x^2yz\times xy^2z^3$$.
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$$x^3y^3z^4$$
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$$x^3y^3z^3$$
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$$x^3y^4z^3$$
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$$x^3y^3z$$
Explanation
$$x^2yz\times xy^2z^3 = x^2\times y\times z\times x\times y^2\times z^3$$
$$x^2\times x\times y\times y^2\times z\times z^3=x^3y^3z^4$$
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