MCQ Questions for Class 9 Maths Polynomials Quiz 9

$2$ is a root of

$kx^4-11x^3+kx^2+13x+2$ , what is the value of

$k$ ?

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

Explanation

$f(2) = k\times (2)^4-11\times (2)^3+k\times (2)^2+13\times 2+2$

$f(2)=0$

$16k-88+4k+26+2=0$

$20k -88+28=0$

$20k-60=0$

$20k =60$

$k=3$

Find the value of

$1.03^2$ using identity.

0%
$1.0409$
0%
$1.0609$
0%
$1.0009$
0%
$1.0309$

Explanation

Given,

$1.03^2 = (1+0.03)^2$ .

We know,

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

Then,

$1.03^2 = (1+0.03)^2$

$=1^2+(0.03)^2+2\times 1\times 0.03$

$=1+0.0009+0.06$

$=1.0609$ .

Therefore, option

$B$ is correct.

The zero of a polynomial

$P(x)$ is:

0%
A number $c$ such that $P(c) = 0$
0%
A number $c$ such that $P(-c) = 1$
0%
A number $c$ such that $P(c)+P(-c) = 0$
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

$p(c)=0$ .

Therefore, option

$A$ is correct.

Find the value of

$(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)$ .

0%
$abc$
0%
$a^2+b^2+c^2$
0%
$0$
0%
$1$

Explanation

We know,

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

$\therefore (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)$

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

$=a^2-b^2+b^2-c^2+c^2-a^2$

$=0$ .

Therefore, option

$C$ is correct.

$p(x) = 5x - 4$ , then

$p(2) =$

0%
$1$
0%
$6$
0%
$0$
0%
$3$

Explanation

The polynomial is

$p(x)=5x-4$ we substitute

$x=2$ in the polynomial:

$p(2)=(5\times 2)-4=10-4=6$

Hence,

$p(2)=6$ .

The _________ power of the variable in a polynomial is called its degree.

0%
highest
0%
lowest
0%
negative
0%
none of these

Explanation

The polynomial

$x^2+x+a$ is in the standard form.

The power is simply the number in the exponent.

In the polynomial,

$x^2+x+a$ , the power of the first term is

$2$ . Since the polynomial has the largest exponent of the variable

$x$ as

$2$ , it is the degree of the polynomial.

Hence, the highest power of the variable in a polynomial is called its degree.

$p(x) = x^3 - 8x^2 + 4$ , then

$p(4) =$

0%
$68$
0%
$60$
0%
$-60$
0%
$-68$

Explanation

The polynomial is

$p(x)=x^3-8x^2+4$ and substitute

$x=4$ in the polynomial:

$p(4)=(4)^3-8(4)^2+4=64-(8\times 16)+4=64-128+4=68-128=-60$

Hence,

$p(4)=-60$ .

$p(-2) = 24$ , then

$p(x) =$

0%
$3x^2 + 5x + 3$
0%
$3x^2 - 5x + 3$
0%
$3x^2 + 5x + 2$
0%
$3x^2 - 5x + 2$

Explanation

Let

$p(x)=3x^2-5x+2$ and substitute

$x=-2$ as shown below:

$p(-2)=3(-2)^{ 2 }-\left( 5\times -2 \right) +2=\left( 3\times 4 \right) +10+2=12+10+2=24$

Hence,

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

$p(-3) = 27$ , then

$p(x) =$

0%
$4x^2 - 3x$
0%
$4x - 3x$
0%
$4x^2 + 3x$
0%
$4x^2 + 3$

Explanation

Consider option A
let

$p(x)=4x^2-3x$ . Then

$p(-3)=4(9)+9=45$ Hence

$p(-3)\neq 27$ .

Consider option B
let

$p(x)=4x-3x=x$ . Then

$p(-3)=-3$ Hence

$p(-3)\neq 27$ .

Consider option C
let

$p(x)=4x^2+3x$ . Then

$p(-3)=4(9)-9=27$ Hence

$p(-3)=27$ .

Consider option D
let

$p(x)=4x^2+3$ . Then

$p(-3)=4(9)+3=39$ Hence

$p(-3)\neq 27$ .

Hence the correct option is option C.

$p(x) = 5x^2 - 4x + 2$ , then

$p(3) =$

0%
$35$
0%
$30$
0%
$45$
0%
$33$

Explanation

The given polynomial is

$p(x)=5x^2-4x+2$ we substitute

$x=3$ in the polynomial:

$p(3)=5(3)^2-4(3)+2=5\times 9-12+2=45-12+2=47-12=35$

Hence,

$p(3)=35$ .

$p(1) = 8$ , then

$p(x) =$

0%
$3x + 6$
0%
$3x + 3$
0%
$3x + 4$
0%
$3x + 5$

Explanation

Let

$p(x)=3x+5$ and substitute

$x=1$ as shown below:

$p(1)=\left( 3\times 1 \right) +5\\=3+5\\=8$

Hence,

$p(x)=3x+5$ .

The highest power of the variable in a polynomial is called its _______.

0%
degree
0%
constant
0%
zero
0%
co-efficient

Explanation

The polynomial

$x^2+x+a$ is in the standard form.

The power is simply number in the exponent. In the polynomial,

$x^2+x+a$ , the power of the first term is

$2$ . Since the polynomial has the largest exponent that is

$2$ , which is the degree of the polynomial.

Hence, the highest power of the variable in a polynomial is called its degree.

$5x + 3$ is a polynomial in

$x$ of degree

0%
$2$
0%
$1$
0%
$3$
0%
$5$

Explanation

The given polynomial is

$5x+3$

The power of

$x$ in the first term is

$1$ ,and the power of

$x$ in the second term is

$0$

Since in the polynomial the largest exponent of

$x$ is

$1$ , it is the degree of the polynomial.

Hence, the degree of the polynomial is

$1$

$p(a) = 3a^2 + 4a - 4$ is a polynomial in

$a$ of degree

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

Explanation

The polynomial

$ax^2+bx+c$ is in standard form.On comparing

$3a^2+4a-4$ with the standard polynomial we see, the power of the first term is

$2$ . Since the polynomial has the largest exponent, that is

$2$ which is the degree of the polynomial.

Hence, the degree of the polynomial is

$2$ .

Which of the following is NOT a constant polynomial?

0%
$p(x) = 15$
0%
$p(x) = 1$
0%
$p(x) = x$
0%
$p(x) = 20$

Explanation

$\textbf{Step-1: Apply the concept of polynomial.}$

$\text{Since}$

$p(x)=x$

$\text{is a polynomial with variable}$

$x$

$\text{and there is no constant term in it.}$

$\text{In the other given options, we can see that the polynomials have}$

$\text{constant terms, none of them has any variable term.}$

$\text{So,}$

$p(x)=x$

$\text{is not a constant polynomial.}$

$\textbf{Hence, correct option is C}$

Which of the following polynomial defines constant polynomials?

0%
$p(x) = ax^3 + bx^2 + cx + d$
0%
$p(x) = ax^2 + bx + c$
0%
$p(x) = c$
0%
$p(x) = ax + b$

Explanation

Since a polynomial

$p(x)=c$ is a constant polynomial with constant term

$c$ ,

A polynomial

$ax^2+bx+c$ is a quadratic polynomial with variables

$x,y$ and constant

$c$ and

A polynomial

$ax+b$ is a linear polynomial with variable

$x$ and constant

$b$

Hence,

$p(x)=c$ is a constant polynomial.

$p(y) = 5y^3 - 2y^2 + y + 10$ is a polynomial in

$y$ of degree

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

Explanation

$5y^3-2y^2+y+10$ we see the power of the first term is

$3$ , the power of the second term is

$2$ and the power of the third term is

$1$ . Since the polynomial has the largest exponent, that is

$3$ which is the degree of the polynomial.

Hence, the degree of the polynomial is

$3$ .

Which of the following is a constant polynomial?

0%
$p(x) = \dfrac{15}{2}$
0%
$p(x) = x$
0%
$p(x) = x^2$
0%
$p(x) = x^3$

Explanation

$\textbf{Step-1: Apply the concept of polynomial.}$

$\text{We know that, a constant polynomial is a polynomial}$

$\text{that has only constant term and no variables.}$

$\text{Since,}$

$p(x)=\dfrac { 15 }{ 2 },$

$\text{is a polynomial with constant term}$

$\dfrac { 15 }{ 2 }$

$\text{having no variable, it is a constant polynomial.}$

$\text{So,}$

$p(x)=\dfrac { 15 }{ 2 },$

$\text{is a constant polynomial.}$

$\textbf{Hence, correct option is A}$

Which of the following is a constant polynomial?

0%
$p(x) = 7 + 3x$
0%
$p(x) = 7$
0%
$p(x) = 7x + 7$
0%
$p(x) = 4x + 3$

Explanation

Since

$p(x)=7$ is a polynomial with constant term

$7$ and there is no variable in it.

Hence,

$p(x)=7$ is a constant polynomial.

$p(x) = c$ , where

$c$ is a real number.

$p(x)$ is a

0%
linear polynomial
0%
quadratic polynomial
0%
cubic polynomial
0%
constant polynomial

Explanation

Since

$p(x)=c$ is a polynomial with only constant term

$c$ and no variable term present in the it.

Hence,

$p(x)$ is a constant polynomial.

$p(x) = 6x^2 - 2x^6$ is a polynomial in

$x$ of degree

0%
$6$
0%
$3$
0%
$1$
0%
none of these

Explanation

$p(x) = 6x^2 - 2x^6$ .

Here, the highest degree of the variable

$x$ is

$6$ .

$\therefore p(x)$ is a polynomial of degree

$6$

$p(5) = 5$ then, which polynomial from the following, it corresponds to?

0%
$p(x) = x^2 - 5x + 5$
0%
$p(x) = x^2 - 5x$
0%
$p(x) = x^2 - x + 5$
0%
$p(x) = x^2$

Explanation

Let

$p(x)=x^2-5x+5$ and substitute

$x=5$ as shown below:

$p(5)=(5)^{ 2 }-\left( 5\times 5 \right) +5=25-25+5=5$

Hence,

$p(x)=x^2-5x+5$ .

What is the degree of the polynomial

$p(x) = 5x^3 - 8x^2 + 4x?$

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

Explanation

The polynomial

$ax^3+ bx^2+cx+d=0$ is in standard form.

On comparing

$5x^3-8x^2+4x$ with the standard form, the power of the first term is

$3$ , the power of the second term is

$2$ and the power of the third term is

$1$ . Since the polynomial has the largest exponent, that is

$3$ which is the degree of the polynomial.

Hence, the degree of the polynomial is

$3$

What is the degree of the polynomial

$p(x) = 8x^8 + 9x^9 + 10x^0$ ?

0%
$8$
0%
$9$
0%
$10$
0%
$0$

Explanation

After writing the

$9x^9+8x^8+10x^0$ we can see that the power of the first term is

$9$ , the power of the second term is

$8$ and the power of the third term is

$0$ . Since the polynomial has the largest exponent, that is

$9$ which is the degree of the polynomial.

Hence, the degree of the polynomial is

$9$ .

The constant polynomial whose coefficients are all equal to

$0$ is called ________ polynomial.

0%
Zero
0%
Linear
0%
Quadratic
0%
Cubic

Explanation

For example:-

Consider the polynomial,

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

$a=b=c= 0$ then the expression becomes zero polynomial.

Therefore, zero polynomial can be written as

$p(x) = 0$ .

Hence, the constant polynomial whose coefficients are all equal to

$0$ is called a zero polynomial.

A polynomial whose coefficients are all equal to _______ is called zero polynomial.

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

Explanation

Consider the polynomial,

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

$a=b=c= 0$ then the expression becomes zero polynomial.

Therefore, zero polynomial can be written as

$p(x) = 0$ .

Hence, the constant polynomial whose coefficients are all equal to

$0$ is called a zero polynomial.

Which of the following is not a constant polynomial?

0%
$p(x) = 3^3$
0%
$p(x) = 2^3$
0%
$p(x) = x^3$
0%
$p(x) = 4^3$

Explanation

$\textbf{Step-1: Apply the concept of polynomial.}$

$\text{We know that cubic power of a constant is also a constant.}$

$\text{Therefore,}$

$p(x)=x^3$

$\text{is a polynomial with variable}$

$x$

$\text{and there is no constant term in it.}$

$\text{So,}$

$p(x)=x^3$

$\text{is not a constant polynomial.}$

$\textbf{Hence, correct option is C}$

Which of the following polynomials, has

$p(6) = 36$ value?

0%
$p(x) = x$
0%
$p(x) = x^2$
0%
$p(x) = x^3$
0%
$p(x) = x^0$

Explanation

Let

$p(x)=x^2$ and substitute

$x=6$ as shown below:

$p(6)=(6)^{ 2 }=36$

Hence,

$p(x)=x^2$ .

For

$p(x) = 3x^2 - 5x, p(6) =$

0%
$36$
0%
$72$
0%
$78$
0%
$77$

Explanation

The polynomial given to us is

$p(x)=3x^2-5x$ after substituting the value of

$x=6$ in the polynomial:

$p(6)=3(6)^2-(5\times 6)=(3\times 36)-30=108-30=78$

Hence,

$p(6)=78$ .

Zero polynomial can be written as ________.

0%
$p(x) = x$
0%
$p(x) = 1$
0%
$p(x) = 0$
0%
$p(x) = x^2$

Explanation

Consider the polynomial,

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

$a=b=c= 0$ then the expression becomes zero polynomial.

Therefore, the constant polynomial whose coefficients are all equal to

$0$ is called a zero polynomial.

Hence, zero polynomial can be written as

$p(x) = 0$ .

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

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Practice Class 9 Maths Quiz Questions and Answers

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

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Practice Class 9 Maths Quiz Questions and Answers

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