MCQ Questions for Class 9 Maths Polynomials Quiz 10

A real number

$\alpha$ is said to be a zero of the polynomial

$p(x)$ , if ___________.

0%
$p(\alpha) = 0$
0%
$p(\alpha) = 1$
0%
$p(\alpha) = x$
0%
$p(\alpha) = \alpha$

Explanation

Any real number

$\alpha$ is said to be a zero of polynomial

$p\left( x \right)$ if

$p\left( \alpha \right) =0.$

Zero can be seen as the value of

$x$ for which the polynomial takes the value as zero.

So, if

$x^2-2x+1$ is a polynomial, we know at

$x=1.$

In this case, the value of

$x^2-2x+1=0,$ hence

$x=1$ can be said as zero of

$x^2-2x+1$

Hence, the answer is

$p\left( \alpha \right) =0.$

That is, option

$A$ is correct.

A real number '

$\alpha$ ' is said to be a ................. of the polynomial

$p(x)$ , if

$p(\alpha) = 0$ .

0%
linear
0%
zero
0%
constant
0%
cubic

Explanation

If we replace the value of

$x$ by a real number, say

$\alpha$ in the polynomial

$p(x)$ , such that

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

$p(\alpha)=0$ , then we say that, the real number '

$\alpha$ ' is a zero of the polynomial

$p(x)$ .

E.g.: Suppose

$p(x)=x-1$ .

Here, the real number

$1$ is a zero of the given polynomial as

$p(1)=1-1=0$ , i.e.

$p(1)=0$ .

Hence, we can say that, a real number

$\alpha$ is said to be a zero of the polynomial

$p(x)$ , if

$p(\alpha)=0$ .

That is, option

$B$ is correct.

Which of the following does NOT represent a zero polynomial?

0%
$p(x) = 0$
0%
$p(x) = 0.x^0$
0%
$p(x) = x^0$
0%
$p(x) = x^0 - 1$

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.

Zero polynomial can be written as

$p(x) = 0$ .

Now, the polynomial

$p(x)=x^0$ or

$p(x)=1$ is a constant polynomial but the coefficient is not equal to

$0$ and therefore it is not of the form

$p(x) = 0$ .

Hence, the polynomial

$p(x)=x^0$ is not a zero polynomial

The zero polynomial is the _________ identity of the additive group of polynomials.

0%
multiplicative
0%
additive
0%
multiplicative inverse
0%
additive inverse

Explanation

Zero polynomial has all coefficients

$=0$

$p(x)$ is a zero polynomial then

$p(x)=0$

It is the additive identity of additive group of polynomials.

Hence,

$Op-B$ is correct.

Using binomial identities, expand the following expression

$(t+3)^2$

0%
$t^2-6t-9$
0%
$t^2+6t-9$
0%
$t^2+6t+9$
0%
$t^2-6t+9$

Explanation

We need to expand

$(t+3)^2$
We know that binomial identities for

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

$a = t, b = 3$

$=$

$t^2+2\times 3t+3^2$

$=$

$t^2+6t+9$

What is the product of

$(x-2x)^2$ ?

0%
$2x^2$
0%
$x-2x^2$
0%
$x^2$
0%
$8x^2$

Explanation

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

$=$

$x.x-x.2x-2x.x+2x.2x$

$=$

$x^2-2x^2-2x^2+4x^2$

$=x^2$

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

$f(2a) = 35$ , then calculate the value of

$a$ .

0%
$-6$
0%
$-3$
0%
$1$
0%
$2$
0%
$6$

Explanation

Given,

$f\left( x \right) ={ x }^{ 2 }-1$ and

$f\left( 2a \right) =35$
Putting

$2a$ in place of

$x$

$\Rightarrow f\left( 2a \right) ={ \left( 2a \right) }^{ 2 }-1=35$

$\Rightarrow { \left( 2a \right) }^{ 2 }-1=35$

$\Rightarrow 4{ a }^{ 2 }=36$

$\Rightarrow { a }^{ 2 }=9$

$\Rightarrow a=\sqrt { 9 } =\pm 3$
Value of

$a$ is

$+3$ or

$-3$
So, correct option is B.

$x^{3}(x^{2} - 5) = -4x$
If

$x > 0$ , what is one possible solution to the equation above?

0%
$1$ or $2$
0%
only $1$
0%
only $2$
0%
none of these

Explanation

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

$\Rightarrow x^5-5x^3+4x=0$

$\Rightarrow x(x^4-5x^2+4)=0$

$\Rightarrow x=0$ or

$x^4-5x^2+4=0$

But given

$x>0$

$\therefore x^4-5x^2+4=0$

$x^4-4x^2-x^2+4=0$

$\Rightarrow (x^2-4)(x^2-1)=0$

$x^2-4=0$ or

$x^2-1=0$

$x^2-4=0 \implies x = \pm 2$

$x^2-1=0 \implies x = \pm 1$
Since it is given

$x > 0$

$x=1$ or

$x=2$

$x\Box y=(x+y)^2-(x-y)^2$ . Then

$\sqrt 4\Box \sqrt 5=$ .

0%
$0$
0%
$8\sqrt5$
0%
$10\sqrt5$
0%
$15$
0%
$20$

Find the value of

$(2x+5)(2x-5)$ .

0%
$4x-25$
0%
$4x^2-25$
0%
$2x^2-25$
0%
$4x^2-5$

Explanation

We know that binomial identities for

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

$\therefore (2x+5)(2x-5) = (2x)^2-5^2$

$=$

$4x^2-25$

Simplify the following expression:

$(s-r)^2-(s+r)^2$

0%
$2sr$
0%
$2s^2+2r^2$
0%
$-4sr$
0%
$2s^2-2r^2$

Explanation

We know that binomial identities for

$(x+y)^2=x^2+2xy+y^2$
and

$(x-y)^2=x^2-2xy+y^2$
So,

$(s-r)^2-(s+r)^2$

$=$

$s^2-2sr+r^2-s^2-2sr-r^2$

$=$

$-4sr$

$(3+\sqrt{2})$ is a root of

${x}^{3}-11{x}^{2}+37x-35=0$ , find the rational root.

0%
$-3$
0%
$5$
0%
$2$
0%
None of these

Reduce the following expression using binomial identities:

$(3x+5)(3x+10)$ .

0%
$9x^2+45x+50$
0%
$9x^2-45x+50$
0%
$9x^2+45x-50$
0%
$9x^2+45-50$

Explanation

We know that binomial identities for

$(x+y)(x+z)=x^2+(y+z)x+yz$
So,

$x = 3x, y = 5, z = 10$

$=$

$(3x)^2+(5+10)3x+5\times 10$

$=$

$9x^2+45x+50$

The value of

$(7.2)^2$ is (use an identity to expand):

0%
$49.9$
0%
$14.4$
0%
$51.84$
0%
$49.04$

Explanation

We will use,

$(a+b)^2$ =

$a^2+2ab+b^2$ to find the required result.

Substitute

$7$ for

$a$ and

$0.2$ for

$B$ in the above identity,

$(7+0.2)^2=(7)^2+2(7)(0.2)+(0.2)^2$

$=49+2.8+0.04$

$=51.84$

Hence, option

$C$ is correct.

The expansion of

$(2x - 3y)^2$ is:

0%
$2x^2+3y^2+6xy$
0%
$4x^2+9y^2-12xy$
0%
$2x^2+3y^2-6xy$
0%
$4x^2+9y^2+12xy$

Explanation

Given,

$(2x-3y)^2$ .

We know,

$(a-b)^2$

$=$

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

Then,

$(2x-3y)^2$

$=$

$(2x)^2-2 (2x)(3y)+(3y)^2$

$=$

$4x^2-12xy+9y^2$ .

Therefore, option

$B$ is correct.

The degree of zero polynomial is zero.

0%
True
0%
False

Find the value of

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

$x = 1$ .

0%
$6$
0%
$7$
0%
$8$
0%
$9$

Explanation

Given polynomial is

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

We need to find value of polynomial at

$x=1$ .

Then put the value of

$x=1$ in the given polynomial, we get

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

$\Rightarrow p(x)= 4\times 1-3\times 1+7$

$\Rightarrow p(x)= 4-3+7$

$\Rightarrow p(x)= 11-3=8$

So, value of polynomial is

$8$ .

Write whether the following statement is True or False.
A binomial may have degree

$3$ .

0%
True
0%
False

Explanation

Since the degree of the polynomial is always a whole number and

$3$ is also a whole number.

Hence, a binomial may have degree

$3$ .

eg.

$ax^3+ b$ .

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

0%
$4ab$
0%
$2ab$
0%
$a^{2} + 2ab + b^{2}$
0%
$2(a^{2} + b^{2})$

Explanation

Given,

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

We know,

$(a+b)^{2}$

$=\left ( a^{2}+2ab+b^{2} \right )$

and

$(a-b)^{2}$

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

Then,

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

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

$=a^{2}+2ab+b^{2}-a^{2}+2ab-b^{2}\\=4ab$ .

Therefore, the required answer is

$4ab$ .

Hence, option

$A$ is correct.

The number of real zeroes of polynomial

$P(x) = x^3 - x$ is ..............

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

Explanation

Given

$P(x)=x^3-x$

To find the zeros of

$P(x)$ , let

$P(x)=0$

$\implies x^3-x=0$

$\implies x(x^2-1)=0$

$\implies x(x-1)(x+1)=0$

$\implies x=0$ and

$x-1=0$ and

$x+1=0$

$\implies x=0,1,-1$ are the zeros of

$P(x)$

Thus, the number of zeros of

$P(x)$ is

$3$ .

Hence, the answer is

$3$ .

The zero of the polynomial

$2x-5$ is:

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

Explanation

Given $p(x)=2x-5$

To find the zero, we put, $p(x)=0$

$\Rightarrow 2x-5 =0$

$\Rightarrow x=\dfrac52$

Hence, $x=\dfrac52$ is the zero of the polynomial $p(x)$

Hence, option $A$ is correct.

What is the degree of the given monomial

$7y$ ?

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

Explanation

We know that the degree is the term with the greatest exponent and,

To find the degree of a monomial with more than one variable for the same term, just add the exponents for each variable to get the degree.

Here, the given monomial

$7y$ has one variable

$y$ where the power of

$y$ is

$1$ . Therefore degree of the monomial

$7y$ is:

$1$ (The exponent of the variable)

Hence, the degree of the monomial

$7y$ is

$1$ .

Find the degree of the given algebraic expression:

$3x-15$

0%
$1$
0%
$3$
0%
$2$
0%
$-15$

Explanation

The given algebraic expression

$3x-15$ has two terms. The first one is

$3x$ and the second is

$-15$ .

The exponent of the first term is

$1$ because

$3x = 3x^1$

The exponent of the second term is

$0$ because

$15 = 15x^0$

Since the highest exponent is

$1$ , therefore, the degree of

$3x-15$ is

$1$ .

Hence, the degree of the algebraic expression

$3x-15$ is

$1$ .

The repeated root of

$x^3+4x^2+5x+2=0$ is

0%
-5/3
0%
-2
0%
-1
0%
1

Explanation

$x^{3}+4{x^{2}}+5{x}+2=(x+1)(x^{2}+3{x}+2)$

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

$x^{3}+4{x^{2}}+5{x}+2=(x+1)^{2}(x+2)$

The repeated root is

$-1$

Find the degree of the given algebraic expression

$ax^2+bx+c$ .

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

Explanation

We know that the degree is the term with the greatest exponent.

Since the highest exponent is

$2$ , therefore, the degree of

$ax^2+bx+c$ is

$2$ .

Hence, the degree of the algebraic expression

$ax^2+bx+c$ is

$2$ .

The value of

$\displaystyle\left (5^{\tfrac {1}{2}} + 3^{\tfrac {1}{2}}\right ) \left (5^{\tfrac {1}{2}} - 3^{\tfrac {1}{2}}\right )$ is:

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

Explanation

Given,

$\displaystyle\left (5^{\tfrac {1}{2}} + 3^{\tfrac {1}{2}}\right ) \left (5^{\tfrac {1}{2}} - 3^{\tfrac {1}{2}}\right )$ .

We know,

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

Then,

$\displaystyle\left (5^{\tfrac {1}{2}} + 3^{\tfrac {1}{2}}\right ) \left (5^{\tfrac {1}{2}} - 3^{\tfrac {1}{2}}\right )$

$=\displaystyle 5^{\tfrac {1}{2}\times2} - 3^{\tfrac {1}{2}\times2}$

$=\displaystyle 5^{1} - 3^{1}$

$=\displaystyle 5- 3$

$=\displaystyle 2$ .

Hence, option

$A$ is correct.

Find the product of

$\left(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\right)$ and

$\left(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\right)$ .

0%
$\dfrac{1}{9}y^4-\dfrac{1}{4}x^4$
0%
$\dfrac{1}{4}x^4-\dfrac{1}{9}y^4$
0%
$4x^2-9y^4$
0%
$\dfrac{1}{4}x^4+\dfrac{1}{9}y^4$

Explanation

Given, product of

$\Bigl(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\Bigr)$ and

$\Bigl(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\Bigr)$

i.e.

$\Bigl(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\Bigr)\times \Bigl(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\Bigr)$ .

We know, the identity

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

Then,

$\Bigl(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\Bigr)\times \Bigl(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\Bigr)$

$=\Bigl(\dfrac{1}{2}x^2\Bigr)^2-\Bigr(\dfrac{1}{3}y^2\Bigl)^2$

$=\dfrac{1}{4}x^4-\dfrac{1}{9}y^4$ .

Therefore, option

$B$ is correct.

The polynomial having atmost 3 zero

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

Explanation

A polynomial of degree

$3$ is called a cubic polynomial.

For example,

$x^3 - 1,$ and

$4a^3 - 100a^2 + a - 6$ are cubic polynomials with atmost

$3$ zeroes (having degree

$3$ ).

Find the value of

$f(x) = x^2+\dfrac{1}{x^2}$ at

$x= 1, 2$ .

0%
$f(1)=2, f(2)=\dfrac{17}{4}$
0%
$f(2)=2, f(1)=\dfrac{17}{4}$
0%
$f(1)=-2, f(2)=\dfrac{17}{4}$
0%
$f(2)=2, f(1)=-\dfrac{17}{4}$

Explanation

Given:

$f(x)=x^2 + \dfrac{1}{x^2}$

$x=1 ;$

$f(1)=1^{2}+\dfrac {1}{1^{2}}$

$\Rightarrow f(1)=2$

$x=2 ;$

$f(2)=2^{2}+\dfrac {1}{2^{2}}$

$\Rightarrow f(2)=\dfrac {17}{4}$

Check whether

$3$ is zero of polynomial

$2x^{3} - x^{2} -13x -6$ .

0%
True
0%
False

Explanation

Substitute

$x=3$ in

$p(x) = 2x^3-x^2-13x-6$

$p(3) = 0$ then 3 is zero of the polynomial

$p(x)$

$\Rightarrow 2x^3-x^2-13x-6$

$\Rightarrow 2\times(3^3)-(3^2)-13\times3-6$

$\Rightarrow 54-9-39-6$

$\Rightarrow 0$

Hence,

$3$ is zero of this polynomial

CBSE Questions for Class 9 Maths Polynomials Quiz 10 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Polynomials Quiz 10 - MCQExams.com

0:0:2

Practice Class 9 Maths Quiz Questions and Answers

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