MCQ Questions for Class 9 Maths Polynomials Quiz 5

What is the degree of the polynomials

$\sqrt 7$ ?

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

The expression with only constant is

0%
$y + 3$
0%
$3x$
0%
$5(21 - 7)$
0%
$5 - 5n$

The zero of polynomial

$9x+2$ is

0%
$\cfrac{-2}{9}$
0%
$\cfrac{1}{9}$
0%
$\cfrac{2}{9}$
0%
$\cfrac{-1}{9}$

Explanation

The polynomial given is

$9x+2$

to find the zero of polynomial we need to equate the polynomial to zero

$\Rightarrow 9x+2=0$

$\Rightarrow 9x=-2$

$\Rightarrow x=\dfrac {-2}9$

the zero of the polynomial is

$\dfrac{-2}{9}$

State true or false:
If 5 is constant and y is variable, then 5y and 5+y are variables.

0%
True
0%
False

The degree of the polynomial

$3x^3y-5xy^4-2x+1$ is

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

Explanation

The degree of the polynomial

$3x^3y-5xy^4-2x+1$ is of

$-5xy^4\implies1+4=5$ [as it is the highest]

A cubic polynomial has at most

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

Explanation

The third degree polynomial function has maximum of

$3$ zeros

Example:

$x^3-9x=0$

$x(x^2-9)=0$

$x=0, x=±3$

Therefore, there are

$3$ zeroes in cubic polynomial.

If $p(x) = x^2 -2 \sqrt {2}x + 1$ , then $p(2\sqrt 2)$ is equal to

0%
$0$
0%
$1$
0%
$4\sqrt 2$
0%
$8\sqrt 2 + 1$

Explanation

$p(x)=x^{ 2 }-2\sqrt { 2 } x+1$

$p(2\sqrt { 2 } )=+2\sqrt { 2 } ^{ 2 }-2\sqrt { 2 }\times2\sqrt { 2 } +1$

$p(2\sqrt { 2 } )=8-8+1$

$p(2\sqrt { 2 } )=1$
Hence, option B.

Zero of the polynomial

$p( x) = 3x +5$ is :

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

Explanation

Zero of the polynomial

$p(x)$ is the value of variable

$x$ at which

$p(x)=0$ .

$\Rightarrow 3x+5=0$

$\Rightarrow 3x=-5$

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

Therefore, option

$D$ is correct.

Find the value of the polynomial

$(3x^{ 3 })\times (4x^{ 2 })+7{ x }^{ 5}$ when

$x=3$ .

0%
$4617$
0%
$4613$
0%
$4611$
0%
$4619$

Explanation

$3{ x }^{ 3 }(4{ x }^{ 2 })+7{ x }^{ 5 }=12{ x }^{ 5 }+7{ x }^{ 5 }=19{ x }^{ 5 }$ .

To find the value of the expression for

$x = 3$ , we have to substitute

$x = 3$ .
So,

$19(3)^{ 5 }=19(243)=4617$

$p(y) = (y + 2) (y - 2)$ . Find the value of

$p(1)$ .

0%
$2$
0%
$0$
0%
$-3$
0%
$3$

Explanation

$p(y) = (y+2)(y-2)$

Putting the value of

$y=1$

$p(y) = (1+2)(1-2) = 3\times-1 = -3$

$\textbf{Alternate:}$

$p(y) = (y+2)(y-2) = y^2 - 4$ [using

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

$p(1) = 1^2-4 = -3$

$p(x)=x^{2}-4x+3,$ . Evaluate

$p(2)-p(-1)+p\left ( \dfrac{1}{2} \right )$ .

0%
$\dfrac{-5}{4}$
0%
$\dfrac{ 5}{4}$
0%
$\dfrac{-31}{4}$
0%
$\dfrac{ 31}{4}$

Explanation

After putting

$x=2$ in equation, we get

$-1$ .
After putting

$x=-1$ , we get

$8$
After putting

$x=\dfrac12$ , we get

$\dfrac54$
So,

$p(2)-p(-1)+p\left(\dfrac12\right) = -1-8+\dfrac{5}{4}$

$= -9+\dfrac{5}{4} = \dfrac{-36+5}{4} = \dfrac{-31}{4}$

Degree of the polynomial $4x^4 + 5x + 7$ , is

0%
4
0%
5
0%
2
0%
7

Explanation

Given polynomial is

$4x^4 + 5x + 7$
The highest power of

$x$ having a non zero coefficient is 4.
Hence, the degree is 4.

The value of the polynomial $5x^3- 4x^2 + 3$ when $x = -1$ is

0%
$-6$
0%
$6$
0%
$2$
0%
$-2$

Explanation

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

$x= -1$ , then replace

$x$ with

$-1$ , we get

$5\times(-1)^{ 3 }-4(-1)^{ 2 }+3$

$=-5-4+3=-6$

Zero of the zero polynomial is:

0%
$0$
0%
$1$
0%
Any real number
0%
Not defined

Explanation

Zeroes of a zero polynomial can be any real number.

That is, since a zero polynomial is defined as

$p(x)=0$ for all

$x$ in real numbers,

we can say that, its zero can be any real number.

Therefore, option

$C$ is correct.

Find

$p(0)$ for the following polynomial :

$p(x) =10x -4x^2-3$

0%
$3$
0%
$-3$
0%
$1$
0%
$0$

Explanation

$p(x)=10x-4x^{ 2 }-3$

$\Rightarrow p(0)=10(0)-4(0)^{ 2 }-3$

$\Rightarrow p(0)=0-0-3$

$\Rightarrow p(0)=-3$

If $p(x) = x + 3$ , then $p(3) + p(-3)$ , is equal to

0%
$3$
0%
$2$
0%
$0$
0%
$6$

Explanation

$p(x) = x + 3$

For

$x = 3$ , we get

$p(3) = 3 + 3 = 6$

For

$x = -3$

$p(-3) = -3 + 3 = 0$

So,

$p(3) + p(-3) = 6 + 0 = 6$

What are the zero(s) of the polynomial

$p(x)= 4x^2 -49$ ?

0%
$\frac{7}{2},\frac{7}{2}$
0%
$+\frac{7}{2},-\frac{7}{2}$
0%
$\frac{2}{7},-\frac{2}{7}$
0%
$0,+\frac{7}{2},-\frac{2}{7}$

Simplify:

$(l + m)^2- 4lm$

0%
$(l -m)^2$
0%
$4l^2m^2$
0%
$(l+2m)^2$
0%
$(l-2m)^2$

Explanation

$(l+m)^2-4lm$

$=l^2+2lm+m^2-4lm$

$=l^2-2lm+m^2$

$=(l-m)^2$

$f(x) =2x^{2} - x + 1$ and

$g(x) = x^{3} - 3x + 1$ , then the value of

$f(1) + g(-1)$ is

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

Explanation

$f(x)= { 2x }^{ 2 }-x+1\\ \therefore f(1)=2\times { 1 }^{ 2 }-1+1\\ =2\\ g(x)={ x }^{ 3 }-3x+1\\ \therefore g(-1)=(-1)^{ 3 }-3\times (-1)+1\\ =-1+3+1\\ =3\\ \therefore f(1)+g(-1)=2+3\\ =5$

8 is a polynomial of degree :

0%
1
0%
$\dfrac{1}{2}$
0%
8
0%
0

Explanation

Degree of a polynomial is the highest power of the variable in the polynomial.

Degree of

$8$ is

$0$ since

$8x^0$

$=8$ .

If the two zeroes of a quadratic polynomial $ax^{2} + bx + c$ are negative, which of the following statements holds true?

0%
a and c are of opposite signs.
0%
a, b and c have the same sign.
0%
a and b are of opposite signs.
0%
b and c are of opposite signs.

Zero of a polynomial is always zero.

0%
True
0%
False
0%
Cannot be determined
0%
None of the above

Explanation

We know that the zero of the polynomial is defined as any real value of

$x$ , for which the value of the polynomial becomes zero.

For example:

Let

$f(x)=− 4x + 7$ be the polynomial.

To find the zero of the polynomial, we substitute

$f(x)=0$ as shown below:

$f(x)=−4x+7\\ \Rightarrow 0=−4x+7\\ \Rightarrow 4x=7$

$\Rightarrow x=\dfrac { 7 }{ 4 }$ .

Therefore,

$\dfrac {7}{4}$ is the zero of the polynomial

$f(x)=− 4x + 7$ , which contradicts the statement that zero of a polynomial is always zero.

Hence, the zero of a polynomial is not always zero.

Therefore, the given statement is false.

That is, option

$B$ is correct.

$f(x) = 2x^3 - 13x^2 +17 x+12$ , then find out the value of

$f(-2)$

0%
$-90$
0%
$-85$
0%
$-70$
0%
$-55$

Explanation

$f(x) = 2x^3 - 13x^2 + 17x+12$

$f(-2)=2(-2)^3-13(-2)^2+17(-2)+12$

$=-16 - 52 - 34 +12 =-90$

The zero polynomial has :

0%
One zero
0%
two zeroes
0%
infinite zeroes
0%
No zero

Explanation

Zeroes of the polynomial mean to find those values of

$x$ for which the value of the polynomial becomes zero.

i.e.

$p(x)=0$

Now, we have a zero polynomial

$p(x)=0$ .

This is the zero polynomial and will be true for all values of x belonging to complex numbers.

Hence, infinite no. of solutions.

Evaluate (using formulae):

$\displaystyle \frac{2.43 \times 2.43 - 2 \times 2.43 \times 1.67 +1.67 \times 1.67}{2.43 - 1.67}$

0%
$0.76$
0%
$0.55$
0%
$1$
0%
$1.4$

Explanation

Given:

$\displaystyle \frac {2.43 \times 2.43 - 2 \times 2.43 \times 1.67 + 1.67 \times 1.67}{2.43-1.67}$

$=\displaystyle \frac {(2.43)^2 - 2 \times 2.43 \times 1.67 + (1.67)^2 }{0.76}$

$=\displaystyle \frac {(2.43-1.67)^2}{0.76}$ [

$\because a^2-2ab+b^2=(a-b)^2$ ]

$=\displaystyle \frac {(0.76)^2}{0.76}$

$=0.76$

Use the identity

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

$9.8\times 10.2$ .

0%
$98.96$
0%
$99.86$
0%
$98.86$
0%
$99.96$

Explanation

We know, $9.8 \times 10.2 = (10 -0.2) \times (10+0.2)$ .

Applying the formula $(a+b)(a-b) = { a }^{ 2 }-{ b }^{ 2 }$ , where $a = 10 , b = 0.2$ ,

we get,
$9.8 \times 10.2 = (10 -0.2) \times (10+0.2)=10^2-(0.2)^2 \\ =100-0.04=99.96 .$

Therefore, option $D$ is correct.

Use the identity

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

$4.6\times 5.4$ .

0%
$24.84$
0%
$25.24$
0%
$24.94$
0%
$25.84$

Explanation

We are given the formula

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

We know,

$4.6 \times 5.4 = (5-0.4)(5+0.4)$ ,

where

$a=5$ and

$b=0.4$ .

Now, solve as shown below:

$4.6 \times 5.4 = (5-0.4)(5+0.4)$

$={ 5 }^{ 2 }-{ 0.4 }^{ 2 }=25-0.16=24.84$ .

Hence,

$4.6 \times 5.4 =24.84$ .

Therefore, option

$A$ is correct.

Simplify :

$\displaystyle \left(\frac{3}{2}x - 0.45y\right)^2$

0%
$2.25x^2-1.35xy+0.2015y^2$
0%
$2.15x^2-1.35xy+0.2025y^2$
0%
$2.25x^2-1.35xy+0.2025y^2$
0%
$2.25x^2-1.25xy+0.2025y^2$

Explanation

$\displaystyle \left(\frac{3}{2}x - 0.45y\right)^2$

$=\displaystyle \left(1.5x - 0.45y\right)^2$

$=\displaystyle (1.5x)^2 + (0.45y)^2- 2(1.5x)(0.45y)^2$

$= 2.25x^2 +0.2025 y^2 -1.35xy$

Use the identity

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

$8.3\times 7.7$ .

0%
$63.91$
0%
$63.81$
0%
$64.91$
0%
$64.21$

Explanation

We are given the formula

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

We know,

$8.3\times7.7=(8+0.3)(8-0.3)$ ,

where

$a=8$ and

$b=0.3$ .

Now, solve as shown below:

$8.3\times7.7$

$=(8+0.3)(8-0.3)=8^{ 2 }-(0.3)^{ 2 }\\ \Rightarrow 8.3\times 7.7=64-0.09\\ \Rightarrow 8.3\times 7.7=63.91.$

Hence,

$8.3\times 7.7=63.91$ .

Therefore, option

$A$ is correct.

Use the identity

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

$103\times 97$ .

0%
$8881$
0%
$9991$
0%
$9981$
0%
$9891$

Explanation

We know, $103 \times 97 = (100 + 3) \times (100 - 3)$ .

Applying the formula $(a+b)(a-b) = { a }^{ 2 }-{ b }^{ 2 }$ , where $a = 100 , b = 3$ ,

we get,
$103 \times 97 = (100 + 3) \times (100 - 3) = { 100 }^{ 2 }-{ 3 }^{ 2 } = 10000 - 9 = 9991$ .

Therefore, option $B$ is correct.

CBSE Questions for Class 9 Maths Polynomials Quiz 5 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Polynomials Quiz 5 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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