MCQ Questions for Class 9 Maths Polynomials Quiz 4

$\displaystyle x=-4$ is a solution of equation

$\displaystyle { x }^{ 2 }-x-20=0$ ?

0%
Yes
0%
No
0%
Can't say
0%
None

Explanation

Put

$x=-4$ in equation

$\displaystyle { x }^{ 2 }-x-20=0$

$\displaystyle { \left( -4 \right) }^{ 2 }-\left( -4 \right) -20=16+4-20=0$

$\displaystyle \therefore \quad x=-4$ is a solution of equation

$\displaystyle { x }^{ 2 }-x-20=0$ .

Therefore, option

$A$ is correct.

The zero(s) of

$p(x) = 5x - 4$ is/are:

0%
$-4$
0%
$5$
0%
$\dfrac{4}{5}$
0%
$- \dfrac{4}{5}$

Explanation

$p(x) = 5x - 4$
A linear polynomial has at most one zero and to find that we equate the polynomial to zero

$\implies 5x - 4 = 0$

$\implies 5x = 4$

$\therefore x = \dfrac {4}{5}$ .

Therefore, option

$C$ is correct.

Which of the following is INCORRECT?

0%
$p(x) = 5x + 5,$ degree $= 1$
0%
$p(x) = 4x^4 + 4,$ degree $= 4$
0%
$p(x) = x^8 ,$ degree $= 8$
0%
$p(x) = 9,$ degree $= 9$

Explanation

A polynomial is a function of the form $f(x) = a_{n}x^{n} + a_{n−1}x^{n−1} + ... + a_{2}x^{2} + a_1x + a_0$

where $a_{n}, a_{n−1} , ... a_{2} ,a_1 , a_0$ . are contants

and $n$ is a natural number.

The degree of a polynomial is the highest power of $x$ in its expression.

In option A, $p(x)=5x+5$

Highest power of $x$ is $1$

So, degree of $p(x)$ is $1$ .

In option B, $p(x)=4x^{4}+4$

Highest power of $x$ is $4$

So, degree of $p(x)$ is $4$ .

In option C, $p(x)=x^{8}$

Highest power of $x$ is $8$

So, degree of $p(x)$ is $8$ .

Looking at Option $D$

$p(x) = 9=9.x^{0}$

Highest power of $x$ in $p(x)$ is $=0$

So, degree of the polynomial $p(x)$ is $0$ .

Thus, it is incorrect.

Hence, the answer is option D.

$x=9$ a solution of

$\displaystyle { x }^{ 2 }-63-2x=0$ ?

0%
Yes
0%
No
0%
Can't say
0%
None

Explanation

Put

$x=9$ in equation

$\displaystyle { x }^{ 2 }-63-2x=0$ .

$\displaystyle= { \left( 9 \right) }^{ 2 }-63-2\left( 9 \right) =81-63-18$

$\displaystyle = 81-81$

$\displaystyle =0$ .

$\displaystyle \therefore x=9$ is solution of equation

$\displaystyle { x }^{ 2 }-63-2x=0$ .

Therefore, option

$A$ is correct.

$\displaystyle x=\frac { 4 }{ 3 }$ is a root of equation

$\displaystyle { x }^{ 2 }-\frac { 16 }{ 9 } =0$

0%
Yes
0%
No
0%
Can't possible
0%
None

Explanation

$\displaystyle { x }^{ 2 }-\dfrac { 16 }{ 9 } =0,\\\left( x+\dfrac { 4 }{ 3 } \right) \left( x-\dfrac { 4 }{ 3 } \right) =0$
Since,

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

$\displaystyle x=\dfrac { -4 }{ 3 }\ and\ \dfrac { 4 }{ 3 }$ are the roots of

$\displaystyle { x }^{ 2 }-\dfrac { 16 }{ 9 } =0$

$p(x) = 25$ is a _______ polynomial.

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

Explanation

The general form of a constant polynomial is

$p(x)=c$ with constant

$c$ .

Since

$p(x)=25$ is a polynomial with constant term

$25$ and there is no variable in it.

Hence,

$p(x)=25$ is a constant polynomial.

Find the polynomial with degree

$6$ .

0%
$6{ x }^{ 4 }+7{ x }^{ 2 }$
0%
$8{ x }^{ 4 }{ y }^{ 2 }+7{ x }^{ 5 }{ y }^{ 3 }-\dfrac { 3 }{ 5 }$
0%
$8{ x }^{ 2 }{ y }^{ 2 }{ z }^{ 2 }+7{ x }^{ 2 }{ y }^{ 2 }z+3{ x }^{ 4 }$
0%
$8{ x }^{ 6 }{ y }^{ 6 }+{ y }^{ 2 }+7{ x }^{ 3 }$

Explanation

In the case of a polynomial in one variable, the highest power of the variable is called the degree of the polynomial.

In the case of polynomials in more than one variable, the sum of the powers of the variables in each term is taken up and the highest sum so obtained is called the degree of the polynomial.

$(A)6x^{4}+7x^{2}$

The highest power of this expression is 4.Then degree is 4.

$(B)8x^{4}y^{2}+7x^{5}y^{3}-\frac{3}{5}$

The highest power of this expression is 5+3=8. Then degree is 8.

$(C)8x^{2}y^{2}z^{2}+7x^{2}y^{2}z+3x^{4}$

The highest power of this expression is 2+2+2=6. Then degree is 6.

$(D)8x^{6}y^{6}+y^{2}+7x^{3}$

The highest power of this expression is 6+6=12. Then degree is 12.

In all option (C) is the expression with degree 6

Find the solution of

$(2x+6)^2$ .

0%
$x^2+24x+36$
0%
$4x^2+24x+6$
0%
$4x^2+24x+36$
0%
$4x^2-24x+36$

Explanation

$(2x+6)^2 = (2x+6)(2x+6)$
=

$2x.2x+2x.6+6.2x+6.6$
=

$4x^2+12x+12x+36$
=

$4x^2+24x+36$

$107\times 107+93\times93=$ ?

0%
$19578$
0%
$19418$
0%
$20098$
0%
$21908$
0%
None of these

Explanation

$107\times 107+93\times 93={107}^{2}+{93}^{2}$

$={(100+7)}^{2}+{(100-7)}^{2}$

$=2\times [{(100)}^{2}+{7}^{2}]$ ...................... [Ref:

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

$=20098$

$217\times 217+183\times 183=$ ?

0%
$79698$
0%
$80578$
0%
$80698$
0%
$81268$

Explanation

${(217)}^{2}+{(183)}^{2}={(200+17)}^{2}+{(200-17)}^{2}$

$=2\times [{(200)}^{2}+{(17)}^{2}]$ [Ref:

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

$=2[40000+289]$

$=2\times 40289$

$=80578$

Find a zero of the polynomial

$p(x) = 3x + 1$

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

Explanation

Finding a zero of

$p(x)$ , is same as solving the equation

$p(x) = 0$
i.e.

$3x + 1 = 0$

$\Rightarrow 3x = -1$

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

$-\dfrac {1}{3}$ is a zero of the polynomial

$3x + 1$ .

Find zero of the polynomial

$f(x) = x + 2$ :

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

Explanation

Given polynomial is

$f(x)=x+2$

We know that a polynomial is zero polynomial when

$f(x)=0$ .

Then

$x+2=0$

On subtracting

$2$ on both the sides, we get

$x+2-2=-2$

$\Rightarrow x=-2$

So, value of

$x$ is

$-2$ for zero polynomial.

Find the zero of the polynomial

$f(x) = x - 2$ :

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

Explanation

Given polynomial is

$f(x)=x-2$

We know that a polynomial is zero polynomial when

$f(x)=0$

Then

$x-2=0$

Add

$2$ on both sides, we get

$x-2+2=2$

$\Rightarrow x=2$

So, value of

$x$ is

$2$ for zero polynomial.

Find the zero of the polynomial

$f(x) = 2x + 3$ .

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

Explanation

Given polynomial

$f(x)=2x+3$

We know that by putting

$f(x)=0$ and solving for

$x$ we can find the zeroes of a polynomial.

$\therefore 2x+3=0$

Add -3 both side we get

$2x+3-3=-3$

$\Rightarrow 2x=-3$

Divided by 2 both sides we get

$x=-\dfrac{3}{2}$

So value of zero of the polynomial is

$-\dfrac{3}{2}$

The degree of the polynomial

$2x^{2} - 4x^{3} + 3x + 5$ is

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

Explanation

The degree of the polynomial means the highest power of the variable.

Therefore, the degree of the polynomial

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

$3$ .

The degree of the polynomial

$5x^7 - 6x^5 + 7x - 6$ is

0%
$4$
0%
$5$
0%
$6$
0%
$7$

Explanation

The degree of a polynomial is the highest exponent of variable.
Here highest exponent is

$7$ .
So, the degree is

$7$ .
Option

$D$ is correct.

The degree of the polynomial

$x^{2} - 5x^{4} +\dfrac {3}{4}x^{7} - 73x + 5$ is ____

0%
$7$
0%
$\dfrac {3}{4}$
0%
$4$
0%
$-73$

Explanation

Given equation is

$P(x)=x^{2}-5x^{4}+\dfrac{3}{4}x^{7}-73x+5$

We know that, the degree of a polynomial is the highest power of its variables when the polynomial is expressed in its canonical form consisting of a linear combination of monomials.

In the given equation,

$7$ is the highest power

Then, the degree of polynomial

$P(x)$ is

$7$ .

$(a + b)^{2} = (a + b)\times$ ____

0%
$ab$
0%
$2ab$
0%
$(a + b)$
0%
$(a - b)$

Explanation

We know that

$x^{2}=x\times x$

Then

$(a+b)^{2}=(a+b)\times (a+b)$

Hence, the answer is (C)

The root of the polynomial equation

$3x-1=0$ is

0%
$x=-\dfrac{1}{3}$
0%
$x=\dfrac{1}{3}$
0%
$x=1$
0%
$x=3$

Explanation

Given:

$3x-1=0$

$\Rightarrow 3x=1$

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

Hence root of

$3x-1=0$ is

$\dfrac{1}{3}$

The degree of the polynomial

$4x^2-7x^3+6x+1$ is.

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

Explanation

A polynomial is a function of the form

$f(x) = a_{n}x^{n} + a_{n−1}x^{n−1} + ... + a_{2}x^{2} + a_1x + a_0$

where

$a_{n}, a_{n−1} , ... a_{2} ,a_1 , a_0$ . are contants and

$n$ is a natural number.

Let

$p(x) = 4x^{2}-7x^{3}+6x+1$

$\Rightarrow$ The degree of a polynomial is the highest power of

$x$ in its expression.

$\Rightarrow$ Highest power of

$x$ in

$p(x)$ is $$=3

Thus, the degree

$=3$

$p(x) = 4x^{2}-7x^{3}+6x+1$ is a cubic polynomial.

Hence, option

$C$ is correct.

The roots of the polynomial equation

$x^2+2x=0$ are:

0%
$x=0, 2$
0%
$x=1, 2$
0%
$x= 1, -2$
0%
$x=0, -2$

Explanation

The given equation is

$x^2+2x=0$

$x\times(x+2)=0$

$x=0$ or

$x+2=0$

$\therefore x=0$ or

$x=-2$

The degree of

$(6{x}^{7} -7{x}^{3} + 3{x}^{2} + 2x -1)$ is ______.

0%
$7$
0%
$6$
0%
$3$
0%
$5$

Explanation

The degree is the highest power of an exponent in the polynomial.

In the given polynomial,

$(6{x}^{7} -7{x}^{3} + 3{x}^{2} + 2x -1)$

It can be clearly observed that the first term

$6{x}^{7}$ has the highest power of

$x$ i.e

$7$ .

Thus, the degree of the given polynomial is

$7$ .

Hence,

$7$ is the correct answer.

The zero of the polynomial

$P(x)=\sqrt { 5 } x-5$

0%
$\sqrt { 5 }$
0%
$-\sqrt { 5 }$
0%
$\cfrac { \sqrt { 5 } }{ 5 }$
0%
$-5$

Explanation

Given polynomial is

$P(x)=\sqrt { 5 } x-5$ .

To get the zeros of

$P(x)$ , equate it to zero,

$\Rightarrow\sqrt { 5 } x-5=0$

$\Rightarrow\sqrt{5} x=5$

$\Rightarrow x=\sqrt { 5 }$

Hence, the zero of

$P(x)$ is

$\sqrt{5}$ .

The degree of

$(6x^{4} - 7x^{3} + 3x^{2} + 2x - 1)$ is _______.

0%
$4$
0%
$6$
0%
$3$
0%
$5$

Explanation

Given polynomial is

$6x^4-7x^3+3x^2+2x-1$

For the given polynomial, highest power of

$x$ is

$4$ .

And it should be Integer,

Thus degree of polynomial is

$4$ .

Number of zeros of the 'zero polynomial' is equal to?

0%
$0$
0%
$1$
0%
$2$
0%
Infinite

Explanation

A zero polynomial is a polynomial of the form

$P(x)=0$ where all coefficients of the polynomial are equal to zero.

Any value of

$x$ can be a zero of a zero polynomial.

So, number of zeroes of a zero polynomial

$=$

$infinite$ .

$[D]$

The degree of the polynomial

$p(x) =x^7 - 5x^3 - 3x^2 + 2x$ is _____

0%
$7$
0%
$2$
0%
$3$
0%
$1$

Explanation

Given the polynomial is

$p(x) =x^7 - 5x^3 - 3x^2 + 2x$ .

Now the power of the highest degree term is

$7$ so the degree of the polynomial is

$7$ .

Zeros of the polynomial p (x) =

$x^{2}-2x$ is /are

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

Explanation

Zeros of the polynomial

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

$x^2-2x=0$

$\Rightarrow x(x-2)=0$

$\Rightarrow x=0$ or

$x=2$

$\therefore$ Zeros of polynomial

$p(x)$ are

$0$ and

$2$ .

Zero's of the polynomial p(x) =

$x^{2}$ -2x is / are :

0%
0, 2
0%
0, 1
0%
1, 2
0%
2, 3

Explanation

Zeros of the polynomial

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

$x^2-2x=0$

$\Rightarrow x(x-2)=0$

$\Rightarrow x=0$ or

$x=2$

$\therefore$ Zeros of polynomial

$p(x)$ are

$0$ and

$2$

The degree of the polynomial

$x^2-5x^4+\dfrac{3}{4}x^7-73x+5$ is

0%
$7$
0%
$\dfrac{3}{4}$
0%
$4$
0%
$-73$

Explanation

The degree of a polynomial is the highest degree of its monomials.

Degree of

${x}^{2}$ is

$2$

Degree of

$-5{x}^{4}$ is

$4$

Degree of

$\dfrac{3}{4}{x}^{7}$ is

$7$

Degree of

$-73x$ is

$1$

Degree of

$5$ is

$0$

The highest degree of the polynomial is

$7$ .

Sum of zeros of the polynomial

$x^2-6$ is _______

0%
$0$
0%
$-6$
0%
$1$
0%
$6$

Explanation

For

$ax^2+bx+c=0$ where

$a\neq0$ , the sum of roots is given by:

$\dfrac{-b}{a}$ .

Here,

$b=0$ , therefore, sum of roots

$=0$

CBSE Questions for Class 9 Maths Polynomials Quiz 4 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Polynomials Quiz 4 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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