MCQ Questions for Class 9 Maths Polynomials Quiz 2

Use the identity

$(a+b)(a-b) = a^2-b^2$ to evaluate:

$21\times 19$ .

0%
$389$
0%
$399$
0%
$289$
0%
$429$

Explanation

We know,

$21 \times 19 = (20+ 1) \times (20 - 1)$ .

Applying the formula

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

$a = 20 , b = 1$ ,

we get,

$21 \times 19 = (20+ 1) \times (20 - 1) = { 20 }^{ 2 }-{ 1 }^{ 2 } = 400 - 1 = 399$ .
Therefore, option

$B$ is correct.

What is the degree of the following polynomial expression:

$\dfrac{4}{3}x^{7} - 3x^{5} + 2x^{3} + 1$

0%
7
0%
4
0%
5
0%
3

Explanation

Clearly, the degree of the polynomial expression

$\dfrac{4}{3}x^{7} - 3x^{5} + 2x^{3} + 1$ is

$7$ .

The degree of a polynomial is the highest degree of its monomials (individual terms) with non-zero coefficients. The degree of a term is the sum of the exponents of the variables that appear in it, and thus is a non-negative integer.

Find the degree of the following polynomial

$x^9-x^4+x^{12}+x-2$

0%
12
0%
9
0%
4
0%
2

Explanation

The degree is the value of the greatest exponent of any expression (except the constant) in the polynomial.
Given,

$x^9-x^4+x^{12}+x-2$
Since there are 4 terms

$x^9-x^4+x^{12}+x-2$ , this is a polynomial and has the highest degree

$x^{12}$ of all the terms.
Therefore,
Degree is

$12$

Find the degree of the following polynomial

$3y+4y^2$

0%
y
0%
0
0%
1
0%
2

Explanation

We know that, the degree is the value of the greatest exponent of any expression (except the constant) in the polynomial.
Given polynomial is

$3y+4y^2$
It is a polynomial in

$y$ having highest exponent value as

$2$ .

Hence, the degree of the given polynomial is

$2$ .

What is the degree of the following polynomial expression:

$\dfrac{5}{3} x^{3} + 7x + 16$

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

Explanation

Clearly, the degree of the polynomial expression

$\frac{5}{3} x^{3} + 7x + 16$ is

$3$ .7.

Find the degree of the following polynomial:

$-\dfrac{3}{2}$

0%
$1$
0%
$2$
0%
$0$
0%
Cannot be determined

Explanation

Given polynomial can be rewritten as

$-\dfrac32 x^0$ so, the highest degree of the variable present in the polynomial is

$0$ hence the degree of the polynomial is equal to

$0.$

Find the degree of following polynomial

$4x-\sqrt{5}$

0%
$\dfrac{1}{2}$
0%
1
0%
2
0%
0

Explanation

The degree is the value of the greatest exponent of any expression (except the constant) in the polynomial.
Given,

$4x-\sqrt{5}$
There are 2 terms

$4p-\sqrt{5}$ ,so it is a binomial and the polynomial has variable

$x$ as the highest degree of all the terms.
Therefore,
Degree is

$1$

Find the degree of following polynomial

$5$

0%
$5$
0%
$1$
0%
$0$
0%
Constant

Explanation

The degree is the value of the greatest exponent of any expression (except the constant) in the polynomial.
Given,

$5$
Since there is

$1$ term

$5$ , it is a monomial and the polynomial has no variable.
Therefore,
Degree is

$0$

State true or false:

A binomial may have degree 5.

0%
True
0%
False

Explanation

A binomial is an algebraic expression with two terms in addition or subtraction. There is no other condition. So, it is true that a binomial may have degree 5.

$e.g.\space p(x)=x^{5}-7$ , is a binomial.
Therefore, option A is correct.

Use identities to solve:

$(97)^{2}$

0%
$9,659$
0%
$9,409$
0%
$9,009$
0%
$9,209$

Explanation

$(97)^2$

$=(100-3)^2$
using,

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

$=(100)^2-2(100)3+(3)^2$

$=10000-600+9$

$=9409$

Find the zero of the polynomial given below:

$p(x) = 9x - 3$ .

0%
$\dfrac{6}{7}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{7}{3}$

Explanation

Given,

$p(x) = 9x - 3$ .
to find the zero of a polynomial we need to equal that polynomial to zero.
thus,

$p(x)=0$

$\Rightarrow9x-3=0$

$\Rightarrow x=\dfrac{3}{9}$

$\Rightarrow x=\dfrac{1}{3}$ .

Therefore, option

$C$ is correct.

If p(x)=

$x^3-4x^2+5x-2$ , then p(2) is:

0%
$0$
0%
$3$
0%
$5$
0%
$8$

Explanation

Let

$f(x)=x^3-4x^2+5x-2$
Now,

$f(2)=2^3-4\times2^2+5\times2-2$

$=8-16+10-2=0$
Since,

$f(2)=0$ .
Then we can say that,

$2$ is a zero of

$f(x)=x^3-4x^2+5x-2$ .

Therefore, option

$A$ is correct.

Solve for

$x$ :

$(502)^{2}$

0%
$2,52,004$
0%
$2,62,104$
0%
$2,22,004$
0%
$2,52,864$

Explanation

$(502)^2$

$=(500+2)^2$
using,

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

$=(500)^2+2(500)(2)+(2)^2$

$=250000+2000+4$

$=252004$

Degree of the polynomial

$4x^4+0x^3+0x^5+5x+7$ is

0%
$4$
0%
$5$
0%
$3$
0%
$7$

Explanation

Clearly, the highest power of

$x$ is

$4$ .

Hence, the degree of given polynomial is

$4$ .
Option A is correct.

Use identities to evaluate:

$(101)^{2}$

0%
$11,601$
0%
$12,761$
0%
$10,111$
0%
$10,201$

Explanation

$(101)^2$

$=(100+1)^2$
using,

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

$=(100)^2+2(100)(1)+1^2$

$=10000+200+1$

$=10201$

Evaluate using expansion of

$(a+b)^2$ or

$(a-b)^2$ :

$(9.4)^2$

0%
88.36
0%
88.46
0%
89.16
0%
89.56

Explanation

${ (a+b) }^{ 2 }\quad =\quad a^{ 2 }+b^{ 2 }+2ab\\ a=9,\quad b=0.4\\ =\quad 9^{ 2 }+0.4^{ 2 }+2*9*0.4\\ =\quad 88.36$

Evaluate using the expansion of

$(a+b)^2$ or

$(a-b)^2$ :

$(92)^2$

0%
8444
0%
8464
0%
8474
0%
8414

Explanation

Given:

${92}^{2} = {(90 + 2)}^{2}$

It is in the form of

${(a+b)}^{2}$ , where

$a = 90, b = 2$ .

Applying the formula

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

${92}^{2} = {(90 + 2)}^{2}$

$= {90}^{2} + { 2 }^{ 2 } + 2\times 90 \times 2$

$= 8100 + 4 + 360 = 8464$

Evaluate using expansion of

$(a+b)^2$ or

$(a-b)^2$ :

$(188)^2$

0%
35444
0%
35484
0%
35344
0%
35384

Explanation

${188}^{2} = {(200 - 12)}^{2}$

It is the form of

${(a - b)}^{2}$ , where

$a = 200, b = 12$

Applying the formula

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

${188}^{2} = {(200 - 12)}^{2} = {200}^{2} + { 12 }^{ 2 } - 2\times 200 \times 12 = 40000 + 144 - 4800 = 35344$

State true or false:

Zero of a polynomial is always

$0$ .

0%
True
0%
False

Explanation

Zero of a polynomial is not always

$0$ .
e.g. Let

$p(x)=x+3$
Zero of

$x+3$ is

$-3$ .

That is, the given statement is false.
Hence, option

$B$ $ is correct.

Find the zeroes of

$p(x)=x-4$ .

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Given,

$p(x)=x-4$ .
The zero of

$p(x)$ is given by

$p(x)=0$
Thus,

$x-4=0$

$=>x=4$ .

Hence, option

$D$ is correct.

Find the zeroes of the following polynomial:

$g(x)=2x+1$ .

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

Explanation

To find zero of

$g(x)=2x+1$ :

Consider,

$g(x)=0$ .

$\implies 2x+1=0$

$\implies 2x=-1$

$\implies x=\dfrac{-1}2$ .

Therefore, zero of

$g(x)=2x+1$ is

$\dfrac{-1}2$ .

Hence, option

$A$ is correct.

The degree of

$5t-\sqrt 7$ is

0%
$5$
0%
$1$
0%
$7$
0%
$0$

Explanation

Clearly, the given the given expression

$5t-\sqrt7$ is a linear polynomial, means the highest power of

$t$ is 1.

$\therefore$ degree of the given polynomial is 1.

State whether True or False, if the following are zeros of the polynomial, indicated against them:

$p(x)=lx+m, \ x=-\dfrac {m}{l}$ .

0%
True
0%
False

Explanation

Given,

$p(x)=lx+m$ .

On substituting

$x=\dfrac{-m}l$ in

$p(x)$ , we get,

$p\left(\dfrac{-m}l\right)=l\times\left(\dfrac{-m}l\right)+m=-m+m=0$ ,

Since

$p(x)=0$ at

$x=\dfrac{-m}l$ ,

Therefore,

$x=\dfrac{-m}l$ is zero of polynomial

$p(x)=lx+m$ .

Hence, it is true and option

$A$ is correct.

State whether True or False, if the following are zeros of the polynomial, indicated against them:

$p(x)=(x+1)(x-2), \ x=-1, 2$ .

0%
True
0%
False

Explanation

Given,

$p(x)=(x+1)(x-2)$ .

On substituting

$x=-1$ in

$p(x)$ , we get,

$p(-1)=(-1+1)(-1-2)=0$ .

Again on substituting

$x=2$ in

$p(x)$ , we get,

$p(2)=(2+1)(2-2)=0$ .

Since

$p(x)=0$ at

$x=-1,2$ .

Therefore,

$x=-1,2$ are zeros of polynomial

$p(x)=(x+1)(x-2)$ .

That is, it is true and option

$A$ is correct.

The degree of

$5x^3+4x^2+7x$ is

0%
$1$
0%
$2$
0%
$4$
0%
$3$

Explanation

$5x^3+4x^2+7x$
The degree is the value of the greatest exponent of any expression (except the constant ) in the polynomial.
Here, term with greatest exponent

$= 5x^3$
exponent of this term

$=3$
Thus, degree

$=3$

Simplify

$a (a^2 + a + 1) + 5$ and find its value for a = 0

0%
5
0%
4
0%
3
0%
2

Explanation

$a (a^2 + a + 1) + 5$

$=a^3+a^2+a+5$
At

$a=0$

$=0^3+0^2+0+5$

$=5$

State True or False, if the following are zeros of the polynomial, indicated against them:

$p(x)=3x+1, \ x=-\dfrac {1}{3}$ .

0%
True
0%
False

Explanation

Given,

$p(x)=3x+1$ .

On substituting

$x=\dfrac{-1}3$ in

$p(x)$ , we get,

$p\left(\dfrac{-1}3\right)=3\left(\dfrac{-1}3\right)+1=-1+1=0$ .

Since

$p(x)=0$ at

$x=\dfrac{-1}3$ ,

therefore,

$x=\dfrac{-1}3$ is zero of polynomial

$p(x)=3x+1$ .

Hence, it is true and option

$A$ is correct.

Find the zero of the polynomial

$p(x)=x+5$ .

0%
$-5$
0%
$5$
0%
$0$
0%
$3$

Explanation

Here,

$p(x)=x+5$ .
In order to find zero of

$p(x)$ , we equate

$p(x)$ to

$0$ .

$p(x)=0$

$x+5=0$

$\Rightarrow x=-5$ .

$\therefore -5$ is the zero of

$p(x)$ .

Therefore, option

$A$ is correct.

Simplify

$a (a^2 + a + 1) + 5$ and find its value for a = -1

0%
0
0%
4
0%
2
0%
6

Explanation

$a (a^2 + a + 1) + 5$

$=a^3+a^2+a+5$
At

$a=-1$

$=(-1^3)+(-1^2)+(-1)+5$

$=-1+1-1+5$

$=4$

Substituting

$x= -3$ in

$x^2+5x +4$ gives

0%
$-2$
0%
$2$
0%
$15$
0%
$-15$

Explanation

For

$x=-3$

$x^2+5x+4$

$=(-3)^2+5(-3)+4$

$=9-15+4$

$=-2 \neq R.H.S$
The correct statement is

$x^2+5x+4=-2$

CBSE Questions for Class 9 Maths Polynomials Quiz 2 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Polynomials Quiz 2 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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