MCQ Questions for Class 9 Maths Polynomials Quiz 8

$g (x) = \displaystyle 3x^{2}-2x-5$ , what is the value of

$g(-1)$ ?

0%
$-4$
0%
$-10$
0%
$-6$
0%
$0$

Explanation

Given,

$g(x)=x^{2}-2x-5$

Then

$g(-1)=3(-1)^{2}-2(-1)-5$

$=3+2-5$

$=0$

For what value of

$k$ ,

$-2$ is a zero of the polynomial

$3x^2 + 4x + 2k$ ?

0%
$5$
0%
$3$
0%
$-8$
0%
$-2$

Explanation

$-2$ is a zero of

$f(x)=3x^2+4x+2k$

Thus,

$f(-2)=0$

$\therefore 3(-2)^2+4(-2)+2k=0$

$=>12-8+2k=0$

$=>2k=-4$

$=>k=-2$

The value of the polynomial

$x^{3} -27$ when

$x = 3$ is

0%
$26$
0%
$25$
0%
$0$
0%
$1$

Explanation

Let

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

Therefore, at

$x=3$ ,

$f(3) = 3^3-27$

$= 27-27$

$= 0$

Hence, option

$C$ is correct.

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

0%
$2ab$
0%
$4ab$
0%
$6ab$
0%
$-4ab$

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.

Say true or false:

The zeros of the polynomial

$x^{2} - 14x + 49$ is equal to

$7$ .

0%
True
0%
False

Explanation

The given polynomial

$x^2-14x+49=0$ can be factorized as follows:

$x^{ 2 }-14x+49=0\\ \Rightarrow x^{ 2 }-7x-7x+49=0\\ \Rightarrow x(x-7)-7(x-7)=0\\ \Rightarrow (x-7)(x-7)=0\\ \Rightarrow (x-7)=0,\quad (x-7)=0\\ \Rightarrow x=7,\quad x=7$

Hence, the zeroes of

$x^2-14x+49$ is

$7$ .

Say true or false:

The zero of

$x-5$ is

$5$ .

0%
True
0%
False

Explanation

To find the zeroes of any polynomial, we need to equate the polynomial expression to zero.

Here, the polynomial is

$x-5$ that is

$x-5=0$

$\implies x=5$ .

Hence, the zero of

$x-5$ is

$5$ .

The value of the polynomial

$7x^{4} + 6x^{2} + x$ when

$x = -2$ is

0%
$131$
0%
$132$
0%
$134$
0%
$136$

Explanation

given,

$f(x) = 7x^4+6x^2+x$

$f(-2) = 7(-2)^4+6(-2)^2+(-2)$

$f(-2) = 7(16)+6(4)-2$

$f(-2) = 112+22 = 134$

The value of the polynomial

$4x^{2} + 3x - 7$ at

$x=1$ is:

0%
$1$
0%
$2$
0%
$0$
0%
None of the above

Explanation

$f(x) = 4x^2+3x-7$

So at

$x=1$ ,

$f(1) = 4(1)^2+3(1)-7$

$f(1) = 4+3-7 = 0$

State the following statement is True or False

The zero of the polynomial

$x + 2 + 4$ is

$-2$ .

0%
True
0%
False

Explanation

Rewrite the polynomial

$x+2+4$ as

$x+6$ and equate it to

$0$ as follows:

$x+6=0\\ \Rightarrow x=-6$

Hence, the only zero of the polynomial

$x+2+4$ is

$-6$ .

Using standard identity, find the value of

$102^2$ .

0%
$10402$
0%
$10408$
0%
$10204$
0%
$10404$

Explanation

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.

Using standard identity, find

$(2x-y)^2$ .

0%
$4x^2+y^2-4xy$
0%
$4x^2+y^2+4xy$
0%
$2x^2+y^2-4xy$
0%
$4x^2+y^2-2xy$

Explanation

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.

Using standard identity, find the value of

$(a+2b)^2$ .

0%
$a^2+2b^2+4ab$
0%
$a^2+4b^2+2ab$
0%
$a^2+4b^2+4ab$
0%
$2a^2+4b^2+4ab$

Explanation

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.

The value of the polynomial

$2x^{5} - 5x^{3} - 10x + 9$ when

$x = -1$

0%
$21$
0%
$22$
0%
$23$
0%
None of the above

Explanation

$f(x) = 2x^5-5x^3-10x+9$

$f(-1) = 2(-1)^5-5(-1)^3-10(-1)+9$

$f(-1) = 2(-1)-5(-1)+10+9$

$f(-1) = -2+5+19$

$f(-1) = 22$

Using standard identity, find the value of

$99^2$ .

0%
9901
0%
9801
0%
1001
0%
9701

Explanation

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.

Which of the following is correct?

0%
$(x-y)^2=x^2+2xy-y^2$
0%
$(x-y)^2=x^2-2xy+y^2$
0%
$(x-y)^2=x^2-y^2$
0%
$(x+y)^2=x^2+2xy-y^2$

Explanation

We know,

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

Then,

$(x-y)^2$

$=x^2-2(x)(y)+y^2$

$= x^2-2xy+y^2$ .

To verify this, let us consider:

$(x-y)^2$

$=(x-y)$

$(x-y)$

$=x(x-y)-y(x-y)$

$=x^2-xy-yx+y^2$

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

$=x^2-2xy+y^2$ .

Hence,

$(x-y)^2$

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

Therefore, option

$B$ is correct.

Find the value of

$49^2$ using standard identity.

0%
$2391$
0%
$2401$
0%
$2411$
0%
$2421$

Explanation

As we know that,

$(a-b)^2=a^2-2ab+b^2$

Substitute

$50$ for

$a$ and

$1$ for

$b$ in above formula,

$\begin{aligned}{}{(50 - 1)^2}& = {50^2} - 2 \times 50 \times 1 + {1^2}\\{49^2}& = 2500 - 100 + 1\\ &= 2401\end{aligned}$

Therefore,

$B$ is correct.

Find the value of

$99\times 101$ using standard identity.

0%
$9999$
0%
$9989$
0%
$9979$
0%
$1009$

Explanation

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.

Square of

$3a-4b$ is:

0%
$9a^2+16b^2-24ab$
0%
$6a^2-8b^2$
0%
$9a^2-16b^2$
0%
$9a^2+16b^2+24ab$

Explanation

Given, square of

$3a-4b$

i.e.

$(3a-4b)^2$ .

We know,

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

Then,

$(3a-4b)^2$

$= (3a)^2-2(3a)(4b)+(4 b)^2$

$=9a^2-24ab+16b^2$ .

Therefore, option

$A$ is correct.

Find the value of

$47\times 53$ .

0%
$2451$
0%
$2471$
0%
$2491$
0%
$2501$

Explanation

Given,

$47\times 53$

$= (50-3)(50+3)$ .

We know,

$(a+b)(a-b)=a^2-b^2$ .

Then,

$47\times 53 = (50-3)(50+3)$

$=(50)^2-3^2=2500-9=2491$ .

Therefore, option

$C$ is correct.

Find the value of

$87^2-13^2$ .

0%
$7300$
0%
$7350$
0%
$7400$
0%
$7450$

Explanation

Given,

$87^2-13^2$ .

We know,

$a^2-b^2=(a+b)(a-b)$ .

Then,

$=87^2-13^2$

$=(87+13)(87-13)$

$=100\times 74=7400$ .

Therefore, option

$C$ is correct.

Find the value of

$52^2$ using standard identity.

0%
$2604$
0%
$2704$
0%
$2804$
0%
$2904$

Explanation

Given,

$52^2$

$=(50+2)^2$ .

We know,

$(a+b)^2=a^2+b^2+2ab$ .

Then,

$52^2=(50+2)^2$

$=50^2+2^2+2\times 50\times 2$

$=2500+4+200$

$=2704$ .

Therefore, option

$B$ is correct.

Find the value of

$199\times 201$ .

0%
$39989$
0%
$40001$
0%
$39999$
0%
$400011$

Explanation

Given,

$199\times 201$

$=(200-1)(200+1)$ .

We know,

$a^2-b^2=(a-b)(a+b)$ .

Then,

$199\times 201$

$=(200-1)(200+1)$

$= 200^2-1^2$

$= 40000-1$

$=39999$ .

Therefore, option

$C$ is correct.

Find the value of

$995^2-5^2$ .

0%
$99000$
0%
$9900$
0%
$99005$
0%
$990000$

Explanation

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.

Find the value of

$(a+b)^2-(a-b)^2$ .

0%
$ab$
0%
$2ab$
0%
$3ab$
0%
$4ab$

Explanation

Given,

$(a+b)^2-(a-b)^2$ .

We know,

$(a+b)^2$

$=a^2+b^2+2ab$

and

$(a-b)^2$

$a^2+b^2-2ab$ .

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.

The zero of the polynomial

$2x+1$ is:

0%
$x=-\dfrac{1}{2}$
0%
$x=-1$
0%
$x=-2$
0%
None of the above

Explanation

Let

$p(x)=2x+1$

Consider

$p(x)=0$

$\Rightarrow 2x+1=0$

$\Rightarrow x=\dfrac{-1}{2}$

Therefore,

$x=\dfrac{-1}{2}$ is the zero of the polynomial

$2x+1=0$

A zero of a polynomial:

0%
needs to be $0$ .
0%
can be any number including $0$ .
0%
needs to be an integer.
0%
None of the above

Explanation

Consider the polynomial,

$p(x) = ax^2 + bx +c$ .

To find the zero of a polynomial, we write

$p(x) = 0$ .

Hence, a real number

$c$ is said to be a zero of the polynomial

$p(x)$ , if

$p(c)=0$ .

E.g. : Suppose,

$p(x)=x^2$ .

When

$x=0$ , we get

$p(x)=(0)^2=0$ .

Clearly, a zero of the polynomial can be any number including

$0$ .

Therefore, option

$B$ is correct.

The zero of the polynomial

$3x+5$ is :

0%
$\dfrac{-5}{3}$
0%
$\dfrac{5}{3}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{-1}{2}$

Explanation

We find the zeros of the given polynomial using:

$3x+5=0$

$3x=-5$

$x=\dfrac{-5}{3}$ .

$\left( 3-\sqrt { 7 } \right) \left( 3+\sqrt { 7 } \right) =$ ?

0%
$4$
0%
$2$
0%
$6$
0%
$8$

Explanation

Given,

$({3}-{\sqrt { 7 } })$

$({3}+{\sqrt { 7 } })$ .

We know,

$(x-y)(x+y)$

$={x}^{2}-{y}^{2}$ .

Thus,

$({3}-{\sqrt { 7 } })$

$({3}+{\sqrt { 7 } })$

$={3}^{2}-{(\sqrt { 7 } )}^{2}$

$=9-7=2$ .

Therefore, option

$B$ is correct.

The value of

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

0%
$4ab-b^2$
0%
$2ab-b^2$
0%
$3ab-b^2$
0%
$6ab-2b^2$

Explanation

Given,

$(a+b)^2-2(a-b)^2+(a-b)(a+b)$ .

We know,

$(a+b)^{2}=a^2+2ab+b^2$ ,

$(a-b)^{2}=a^2-2ab+b^2$

and

$(a+b)(a-b)=a^2-b^2$ .

Then,

$(a+b)^2-2(a-b)^2+(a-b)(a+b)$

$=(a^2+b^2+2ab)-2(a^2+b^2-2ab)+(a^2-b^2)$

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

$=2a^2+b^2+6ab-2a^2-3b^2$

$=6ab-2b^2$ .

Therefore, option

$D$ is correct.

Find the value of

$1.05\times 0.95$ using standard identity.

0%
$0.9985$
0%
$0.9975$
0%
$0.9875$
0%
$0.9995$

Explanation

Given,

$1.05\times 0.95$

$=(1+0.05)(1-0.05)$ .

We know,

$(a+b)(a-b)=a^2-b^2$ .

Then,

$1.05\times 0.95$

$=(1+0.05)(1-0.05)$

$=1^2-(0.05)^2$

$=1-0.0025$

$=0.9975$ .

Therefore, option

$B$ is correct.

CBSE Questions for Class 9 Maths Polynomials Quiz 8 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Polynomials Quiz 8 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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