Algebraic Expressions and Identities Test

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Algebraic Expressions and Identities Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

$$\displaystyle \left ( z^{2}+13 \right )\left ( z^{2}-5 \right )$$ is equal to

A
$$\displaystyle 2z^{4}+18z^{2}-8$$

B
$$\displaystyle z^{4}+8z^{2}-65$$

C
$$\displaystyle z^{4}-8z^{2}-65$$

D
$$\displaystyle z^{4}+8z^{2}+65$$

Solution

$$ ( z^{2}+13) ( z^{2}-5)$$

= $$( z^{2})^{2}+ ( 13+ ( -5 ) )z^{2}+ ( 13\times -5 )$$

= $$\displaystyle z^{4}+8z^{2}-65 $$
Question 2
1 / -0

Simplify:$$\displaystyle \left ( p-q \right )^{2}+4pq$$.

A
$$\displaystyle p^{2}-q^{2}$$

B
$$\displaystyle \left ( p+q \right )^{2}$$

C
$$\displaystyle \left ( 2p-q \right )^{2}$$

D
$$\displaystyle \left ( 2p-2q \right )^{2}$$

Solution

Given, $$(p-q)^2+4pq$$.

We know, $$(a-b)^2=a^2-2ab+b^2$$
and $$(a+b)^2=a^2+2ab+b^2$$.

Therefore, $$(p-q)^2+4pq$$
$$=p^2-2pq+q^2$$+ 4$$pq$$ [Since, $$(a-b)^2=a^2-2ab+b^2$$]
$$=p^2 + 2pq +q^2$$
$$=(p+q)^2$$. [Since, $$(a+b)^2=a^2+2ab+b^2$$]

Therefore option $$B$$ is correct.
Question 3
1 / -0

The product of $$\displaystyle 4a^{2},-6b^{2}$$ and $$\displaystyle 3a^{2}b^{2}$$ is

A
$$\displaystyle a^{2}b^{2}$$

B
$$\displaystyle 13a^{4}b^{4}$$

C
$$\displaystyle -72a^{4}b^{4}$$

D
$$\displaystyle a^{4}b^{4}$$

Solution

$$\displaystyle \left ( 4a^{2} \right )\left ( -6b^{2} \right )\left ( 3a^{2}b^{2} \right )$$
= $$\displaystyle 4\times \left ( -6 \right )\times 3\times a^{2+2}.b^{2+2}$$
= $$\displaystyle -72a^{4}b^{4}$$
Question 4
1 / -0

When $$\displaystyle a^{2}b\left ( a^{3}-a+1 \right )-ab\left ( a^{4}-2a^{2}+2a \right )-b\left ( a^{3}-a^{2}-1 \right )$$ is simplified, the answer is

A
$$\displaystyle -a^{2}b$$

B
$$ab$$

C
$$b$$

D
$$0$$

Solution

Given expression is,
$$a^2b(a^3-a+1)-ab(a^4-2a^2+2a)-b(a^3-a^2-1)$$
$$=(a^5b-a^3b+a^2b)-(a^5b-2a^3b+2a^2b)-(a^3b-a^2b-b)$$
$$=a^5b-a^3b+a^2b-a^5b+2a^3b-2a^2b-a^3b+a^2b+b$$
$$=(a^5b-a^5b)+(2a^3b-a^3b-a^3b)+(a^2b-2a^2b+a^2b)+b$$
$$=0+0+0+b$$
$$=b$$
Option C is correct.
Question 5
1 / -0

The value of $$\displaystyle \left ( x+4 \right )(x+3)-\left ( x-4 \right )\left ( x-3 \right )$$ is equal to

A
$$\displaystyle 2x^{2}-14x+24$$

B
$$\displaystyle 2x^{2}+14x-24$$

C
$$14x$$

D
$$24$$

Solution

$$(x + 4)(x + 3) - (x - 4) (x - 3)$$
= $$\displaystyle ( x^{2}+3x+4x+12)-( x^{2}-3x-4x+12 )$$
= $$\displaystyle ( x^{2}+7x+12 )- ( x^{2}-7x+12 )$$
= $$\displaystyle x^{2}+7x+12-x^{2}+7x-12$$
= $$7x + 7x = 14x$$
Question 6
1 / -0

The product of $$\displaystyle \left ( \frac{4p}{5}-3 \right )$$ and $$\displaystyle \left ( \frac{5p}{8}-6 \right )$$ is

A
$$\displaystyle \frac{p^{2}}{2}+\frac{267}{40}p-18$$

B
$$\displaystyle \frac{p^{2}}{2}-\frac{267}{40}p-18$$

C
$$\displaystyle \frac{p^{2}}{2}+\frac{267}{40}p+18$$

D
$$\displaystyle \frac{p^{2}}{2}-\frac{267}{40}p+18$$

Solution

$$\displaystyle \left ( \frac{4p}{5}-3 \right )\left ( \frac{5p}{8} -6\right )$$

$$\displaystyle =\frac{4p}{5}\times \left ( \frac{5p}{8}-6 \right )+(-3)\times \left ( \frac{5p}{8}-6 \right )$$

$$\displaystyle =\frac{p^{2}}{2}-\frac{24p}{5}-\frac{15p}{8}+18$$

$$\displaystyle =\frac{p^{2}}{2}-\frac{267}{40}p+18$$
Question 7
1 / -0

Simplify the following expression : $$x (y - z) - y (z - x) - z (x - y)$$

A
$$2x (y - z)$$

B
$$2y (z -x)$$

C
$$2x ( z- y)$$

D
None

Solution

$$x(y-z) - y(z-x) - z(x-y) = xy - xz -yz + xy -zx + zy$$
= $$ 2xy -2xz$$
= $$ 2x(y-z)$$
Question 8
1 / -0

$$\displaystyle \left ( mx-ny \right )\left ( mx-ny \right )$$=_____

A
$$\displaystyle m^{2}x^{2}+2mxny-n^{2}y^{2}$$

B
$$\displaystyle m^{2}x^{2}-2mxny-n^{2}y^{2}$$

C
$$\displaystyle m^{2}x^{2}-2mxny+n^{2}y^{2}$$

D
$$\displaystyle m^{2}x^{2}+2mxny+n^{2}y^{2}$$

Solution

$$\displaystyle \left ( mx-ny \right )\left ( mx-ny \right )=\left ( mx-ny \right )^{2}$$
= $$\displaystyle \left ( mx \right )^{2}-2mx\times ny+\left ( ny \right )^{2}$$
= $$\displaystyle m^{2}x^{2}-2mxny+n^{2}y^{2}$$
Question 9
1 / -0

The value of the product $$\displaystyle \left ( 4a^{2}+3b \right )\left ( 9b^{2}+4a \right )$$ at $$a = 1$$ and $$b = - 2$$ is

A
$$60$$

B
$$-80$$

C
$$70$$

D
$$-50$$

Solution

equired product $$=(4a^2+3b)(9b^2+4a)$$
$$=[4(1)^2+3(-2)][9(-2)^2+4(1)]$$
$$=[4(1)-6][9(4)+4]$$
$$=[4-6][36+4]$$
$$=[-2][40]$$
$$=-80$$
Option B is correct.
Question 10
1 / -0

The product of $$\displaystyle \left ( \frac{1}{5}x^{2}-\frac{1}{6}y^{2} \right )$$ and $$\displaystyle \left ( 5x^{2}+6y^{2} \right )$$ is

A
$$1$$

B
$$\displaystyle x^{4}+\frac{11}{60}x^{2}y^{2}+y^{4}$$

C
$$\displaystyle x^{4}+\frac{11}{30}x^{2}y^{2}-y^{4}$$

D
$$\displaystyle x^{4}-\frac{11}{30}x^{2}y^{2}-y^{4}$$

Solution

$$\displaystyle \left ( \frac{1}{5}x^{2}-\frac{1}{6}y^{2} \right )\left ( 5x^{2}+6y^{2} \right )$$

= $$\displaystyle \frac{1}{5}x^{2}\left ( 5x^{2}+6y^{2} \right )-\frac{1}{6}y^{2}\left ( 5x^{2}+6y^{2} \right )$$

= $$\displaystyle x^{4}+\frac{6}{5}x^{2}y^{2}-\frac{5}{6}x^{2}y^{2}-y^{4}$$

= $$\displaystyle x^{4}+\frac{36x^{2}y^{2}-25x^{2}y^{2}}{30}-y^{4}$$

= $$\displaystyle x^{4}+\frac{11x^{2}y^{2}}{30}-y^{4}$$

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Re-Attempt Weekly Quiz Competition

Algebraic Expressions and Identities Test - 27

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Algebraic Expressions and Identities Test - 27

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SHARING IS CARING

Question 1

$$\displaystyle 2z^{4}+18z^{2}-8$$

$$\displaystyle z^{4}+8z^{2}-65$$

$$\displaystyle z^{4}-8z^{2}-65$$

$$\displaystyle z^{4}+8z^{2}+65$$

Solution

Question 2

$$\displaystyle p^{2}-q^{2}$$

$$\displaystyle \left ( p+q \right )^{2}$$

$$\displaystyle \left ( 2p-q \right )^{2}$$

$$\displaystyle \left ( 2p-2q \right )^{2}$$

Solution

Question 3

$$\displaystyle a^{2}b^{2}$$

$$\displaystyle 13a^{4}b^{4}$$

$$\displaystyle -72a^{4}b^{4}$$

$$\displaystyle a^{4}b^{4}$$

Solution

Question 4

$$\displaystyle -a^{2}b$$

$$ab$$

$$b$$

$$0$$

Solution

Question 5

$$\displaystyle 2x^{2}-14x+24$$

$$\displaystyle 2x^{2}+14x-24$$

$$14x$$

$$24$$

Solution

Question 6

$$\displaystyle \frac{p^{2}}{2}+\frac{267}{40}p-18$$

$$\displaystyle \frac{p^{2}}{2}-\frac{267}{40}p-18$$

$$\displaystyle \frac{p^{2}}{2}+\frac{267}{40}p+18$$

$$\displaystyle \frac{p^{2}}{2}-\frac{267}{40}p+18$$

Solution

Question 7

$$2x (y - z)$$

$$2y (z -x)$$

$$2x ( z- y)$$

None

Solution

Question 8

$$\displaystyle m^{2}x^{2}+2mxny-n^{2}y^{2}$$

$$\displaystyle m^{2}x^{2}-2mxny-n^{2}y^{2}$$

$$\displaystyle m^{2}x^{2}-2mxny+n^{2}y^{2}$$

$$\displaystyle m^{2}x^{2}+2mxny+n^{2}y^{2}$$

Solution

Question 9

$$60$$

$$-80$$

$$70$$

$$-50$$

Solution

Question 10

$$1$$

$$\displaystyle x^{4}+\frac{11}{60}x^{2}y^{2}+y^{4}$$

$$\displaystyle x^{4}+\frac{11}{30}x^{2}y^{2}-y^{4}$$

$$\displaystyle x^{4}-\frac{11}{30}x^{2}y^{2}-y^{4}$$

Solution

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