Polynomials Test

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Polynomials Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

For what value of $$k$$, $$-2$$ is a zero of the polynomial $$3x^2 + 4x + 2k$$?

A
$$5$$

B
$$3$$

C
$$-8$$

D
$$-2$$

Solution

If $$-2$$ is a zero of $$f(x)=3x^2+4x+2k$$
Thus,
$$f(-2)=0$$
$$\therefore 3(-2)^2+4(-2)+2k=0$$
$$=>12-8+2k=0$$
$$=>2k=-4$$
$$=>k=-2$$
Question 2
1 / -0

The value of the polynomial $$x^{3} -27$$ when $$x = 3$$ is

A
$$26$$

B
$$25$$

C
$$0$$

D
$$1$$

Solution

Let $$f(x)=x^3-27.$$

Therefore, at $$x=3$$,
$$f(3) = 3^3-27$$
$$ = 27-27$$
$$= 0$$

Hence, option $$C$$ is correct.
Question 3
1 / -0

$$(a -b)^2 -(a + b)^2$$ is equal to:

A
$$2ab$$

B
$$4ab$$

C
$$6ab$$

D
$$-4ab$$

Solution

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.
Question 4
1 / -0

Find the value of $$99\times 101$$ using standard identity.

A
$$9999$$

B
$$9989$$

C
$$9979$$

D
$$1009$$

Solution

Given, $$99\times 101=(100-1)(100+1)$$.

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

Then,
$$99\times 101=(100-1)(100+1)$$
$$=(100)^2-1^2$$
$$=10000-1$$
$$=9999$$.

Therefore, option $$A$$ is correct.
Question 5
1 / -0

Using standard identity, find the value of $$(a+2b)^2$$.

A
$$a^2+2b^2+4ab$$

B
$$a^2+4b^2+2ab$$

C
$$a^2+4b^2+4ab$$

D
$$2a^2+4b^2+4ab$$

Solution

Given, $$(a+2b)^2$$.

We know, $$(x+y)^2=x^2+2xy+y^2$$.

Then,
$$(a+2b)^2$$ $$=a^2+(2b)^2 +2\times a\times 2b$$
$$=a^2+4b^2+4ab$$.

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

Using standard identity, find the value of $$102^2$$.

A
$$10402$$

B
$$10408$$

C
$$10204$$

D
$$10404$$

Solution

Given, $$102^2$$
$$=(100+2)^2$$.

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

Then,
$$102^2$$
$$=(100+2)^2$$
$$=100^2+2^2+2(2)(100) $$
$$=10000+4+400 $$
$$= 10404$$.

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

Using standard identity, find $$(2x-y)^2$$.

A
$$4x^2+y^2-4xy$$

B
$$4x^2+y^2+4xy$$

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

D
$$4x^2+y^2-2xy$$

Solution

Given, $$(2x-y)^2$$

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

Then,
$$(2x-y)^2$$
$$=(2x)^2+y^2-2\times 2x\times y$$
$$= 4x^2+y^2-4xy$$.

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

The value of the polynomial $$2x^{5} - 5x^{3} - 10x + 9$$ when $$x = -1$$

A
$$21$$

B
$$22$$

C
$$23$$

D
None of the above

Solution

$$f(x) = 2x^5-5x^3-10x+9$$
$$f(-1) = 2(-1)^5-5(-1)^3-10(-1)+9$$
$$f(-1) = 2(-1)-5(-1)+10+9$$
$$f(-1) = -2+5+19$$
$$f(-1) = 22$$
Question 9
1 / -0

Using standard identity, find the value of $$99^2$$.

A
9901

B
9801

C
1001

D
9701

Solution

Given,
$$=99^2$$
$$=(100-1)^2$$

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

Then,
$$99^2$$
$$=(100-1)^2$$
$$=100^2+1^2 - 2\times 100\times 1$$
$$= 10000+1- 200$$
$$=10001-200$$
$$=9801$$

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

Find the value of $$995^2-5^2$$.

A
$$99000$$

B
$$9900$$

C
$$99005$$

D
$$990000$$

Solution

Given, $$995^2-5^2$$.

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

Then,
$$995^2-5^2=(995+5)(995-5)$$
$$=1000\times 990$$
$$=990000$$.

Therefore, option $$D$$ is correct.

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Re-Attempt Weekly Quiz Competition

Polynomials Test - 33

Result

Polynomials Test - 33

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Question 1

$$5$$

$$3$$

$$-8$$

$$-2$$

Solution

Question 2

$$26$$

$$25$$

$$0$$

$$1$$

Solution

Question 3

$$2ab$$

$$4ab$$

$$6ab$$

$$-4ab$$

Solution

Question 4

$$9999$$

$$9989$$

$$9979$$

$$1009$$

Solution

Question 5

$$a^2+2b^2+4ab$$

$$a^2+4b^2+2ab$$

$$a^2+4b^2+4ab$$

$$2a^2+4b^2+4ab$$

Solution

Question 6

$$10402$$

$$10408$$

$$10204$$

$$10404$$

Solution

Question 7

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

$$4x^2+y^2+4xy$$

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

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

Solution

Question 8

$$21$$

$$22$$

$$23$$

None of the above

Solution

Question 9

9901

9801

1001

9701

Solution

Question 10

$$99000$$

$$9900$$

$$99005$$

$$990000$$

Solution

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