Polynomials Test

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Polynomials Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

If $$p(x)=x^2-2\sqrt 2x+1$$, then $$p(2\sqrt 2)$$ is equal to :

A
$$0$$

B
$$1$$

C
$$4\sqrt 2$$

D
$$8\sqrt 2+1$$

Solution

Given, $$p(x)=x^2-2\sqrt 2x+1$$
$$\therefore p(2\sqrt2)=(2\sqrt2)^2-2\sqrt 2(2\sqrt2)+1$$
$$=8-8+1=1$$
Option B is correct.
Question 2
1 / -0

Which out of the following options is a trinomial, having degree 7?

A
$$x^7 + 6x - 8$$

B
$$5x^2 + 12x^{-7} + x$$

C
$$y^3 - 4x^2 + 12xy$$

D
$$y^7 - y^6 + 5y^4 + 6 - 8y^5 + \sqrt y$$

Solution

The polynomial $$x^7+6x-8$$ has three terms. The first one is $$x^7$$, the second is $$6x$$, and the third is $$-8$$.

The exponent of the first term is $$7$$.

The exponent of the second term is $$1$$ because $$6x = 6x^1$$.

The exponent of the third term is $$0$$ because $$-8 = -8x^0$$.

Since the highest exponent is $$7$$, therefore, the degree of $$x^7+6x-8$$ is $$7$$.
Question 3
1 / -0

The degree of trinomial $$\displaystyle ax^{5}-bx^{4}+c$$ is

A
$$9$$

B
$$5$$

C
$$4$$

D
$$1$$

Solution

The polynomial $$ax^5-bx^4+c$$ has three terms. The first one is $$ax^5$$, the second is $$-bx$$, and the third is $$c$$.

The exponent of the first term is $$5$$.

The exponent of the second term is $$1$$ because $$-bx = -bx^1$$.

The exponent of the third term is $$0$$ because $$c = cx^0$$.

Since the highest exponent is $$5$$, therefore, the degree of $$ax^5-bx^4+c$$ is $$5$$.
Question 4
1 / -0

The value of $$25x^2\, +\, 16y^2\, +\, 40xy$$ at $$x = 1$$ and $$y = -1$$ is :

A
$$81$$

B
$$- 49$$

C
$$1$$

D
None of these

Solution

Substitute $$x=1$$ and $$y=-1$$ in the given polynomial.
$$\therefore 25x^2 + 16y^2 + 40xy = 25(1)^2 + 16(-1)^2 + 40(1)(-1)$$
=$$25(1) + 16(1) + 40(-1)$$
=$$25 + 16 - 40$$
=$$1$$
Question 5
1 / -0

The value of $$(501)^2\, -\, (500)^2$$ is :

A
$$1$$

B
$$101$$

C
$$1,001$$

D
None of these

Solution

Given, $$(501)^{2}- (500)^{2}$$.

We know, $$a^{2}- b^{2}= (a+b)(a-b)$$.

Then,
$$(501)^{2}- (500)^{2}$$
$$=(501+500)(501-500)$$
$$=(1001)(1) $$
$$= 1001$$.

Therefore, option $$C$$ is correct.
Question 6
1 / -0

$$\displaystyle \left ( a+b \right )^{2}-\left ( b-a \right )^{2}$$=______

A
$$\displaystyle \left ( 2a+2b \right )$$

B
$$\displaystyle \left ( 2a-2b \right )$$

C
$$\displaystyle 4ab$$

D
$$\displaystyle -4ab$$

Solution

Given, $$\displaystyle \left ( a+b \right )^{2}-\left ( b-a \right )^{2}$$.

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

Then,
$$\displaystyle \left ( a+b \right )^{2}-\left ( b-a \right )^{2}$$
= $$\displaystyle \left \{ (a+b)+\left ( b-a \right ) \right \}\left \{ (a+b)-\left ( b-a \right ) \right \}$$
$$=(a+b+b-a)(a+b-b+a)$$
$$= 2b \times 2a = 4ab$$.

Therefore, option $$C$$ is correct.
Question 7
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 8
1 / -0

$$\displaystyle \left ( x+4 \right )\left ( x-4 \right )\left ( x^{2}+16 \right )$$ is equal to:

A
$$\displaystyle x^{2}-64$$

B
$$\displaystyle x^{4}-64$$

C
$$\displaystyle x^{4}-256$$

D
$$\displaystyle x^{2}-256$$

Solution

Given, $$\displaystyle \left ( x+4 \right )\left ( x-4 \right )\left ( x^{2}+16 \right )$$.

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

Then,
$$\displaystyle \left ( x+4 \right )\left ( x-4 \right )\left ( x^{2}+16 \right )$$
= $$\displaystyle \left ( x^{2}-4^{2} \right )\left ( x^{2}+16 \right )$$
= $$\displaystyle \left ( x^{2}-16 \right )\left ( x^{2}+16 \right )=\left ( x^{2} \right )^{2}-\left ( 16 \right )^{2}$$
= $$\displaystyle x^{4}-256 $$.

Therefore, option $$C$$ is correct.
Question 9
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 10
1 / -0

$$a^{2} + 4a + 4$$ =

A
$$(a + 2)^{2}$$

B
$$(a + 1)^{2}$$

C
$$(a - 2)^{2}$$

D
$$(a - 1)^{2}$$

Solution

$$a^{2} + 4a + 4 = a^{2} + 2(a) (2)+ (2)^{2}$$
$$=(a + 2)^{2}$$

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Re-Attempt Weekly Quiz Competition

Polynomials Test - 31

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Polynomials Test - 31

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SHARING IS CARING

Question 1

$$0$$

$$1$$

$$4\sqrt 2$$

$$8\sqrt 2+1$$

Solution

Question 2

$$x^7 + 6x - 8$$

$$5x^2 + 12x^{-7} + x$$

$$y^3 - 4x^2 + 12xy$$

$$y^7 - y^6 + 5y^4 + 6 - 8y^5 + \sqrt y$$

Solution

Question 3

$$9$$

$$5$$

$$4$$

$$1$$

Solution

Question 4

$$81$$

$$- 49$$

$$1$$

None of these

Solution

Question 5

$$1$$

$$101$$

$$1,001$$

None of these

Solution

Question 6

$$\displaystyle \left ( 2a+2b \right )$$

$$\displaystyle \left ( 2a-2b \right )$$

$$\displaystyle 4ab$$

$$\displaystyle -4ab$$

Solution

Question 7

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

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

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

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

Solution

Question 8

$$\displaystyle x^{2}-64$$

$$\displaystyle x^{4}-64$$

$$\displaystyle x^{4}-256$$

$$\displaystyle x^{2}-256$$

Solution

Question 9

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

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

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

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

Solution

Question 10

$$(a + 2)^{2}$$

$$(a + 1)^{2}$$

$$(a - 2)^{2}$$

$$(a - 1)^{2}$$

Solution

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