NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2 - Polynomials

Q1 (i) Find the value of the polynomial $5x-4x^2+3$ at $x=0$

Answer:

Given $5x-4x^2+3$ and $x=0$

Therefore, the value is:

$\Rightarrow 5(0)-4(0{)}^2+3=0-0+3=3$

Therefore, value of polynomial $5x-4x^2+3$ at x = 0 is 3.

Q1 (ii) Find the value of the polynomial $5x-4x^2+3$ at $x=-1$

Answer:

Given $5x-4x^2+3$ and $x=-1$

Therefore, the value is:

$\Rightarrow 5(-1)-4(-1{)}^2+3=-5-4+3=-6$

Therefore, value of polynomial $5x-4x^2+3$ at x = -1 is -6.

Q1 (iii) Find the value of the polynomial $5x-4x^2+3$ at $x=2$

Answer:

Given $5x-4x^2+3$ and $x=2$

Therefore, the value is:

$\Rightarrow 5(2)-4(2{)}^2+3=10-16+3=-3$

Therefore, value of polynomial $5x-4x^2+3$ at x = 2 is -3.

Q2 (i) Find p(0) , p(1) and p(2) for each of the following polynomials: $p(y)=y^2-y+1$

Answer:

Given a polynomial is $p(y)=y^2-y+1$

Now, according to the question:

$p(0)=(0{)}^2-0+1=1$

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

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

Therefore, values of p(0), p(1) and p(2) are 1, 1 and 3, respectively.

Q2 (ii) Find p(0) , p(1) and p(2) for each of the following polynomials: $p(t)=2+t+2t^2-t^3$

Answer:

Given a polynomial is $p(t)=2+t+2t^2-t^3$

Now, according to the question:

$p(0)=2+0+2(0{)}^2-(0{)}^3=2$

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

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

Therefore, values of p(0), p(1) and p(2) are 2, 4 and 4, respectively.

Q2 (iii) Find p(0), p(1) and p(2) for each of the following polynomials: $p(x)=x^3$

Answer:

Given a polynomial is $p(x)=x^3$

Now, according to the question:

$p(0)=(0{)}^3=0$

$p(1)=(1{)}^3=1$

$p(2)=(2{)}^3=8$

Therefore, values of p(0), p(1) and p(2) are 0, 1 and 8, respectively.

Q2 (iv) Find p(0), p(1) and p(2) for each of the following polynomials: $p(x)=(x-1)(x+1)$

Answer:

Given a polynomial is $p(x)=(x-1)(x+1)=x^2-1$

Now, according to the question:

$p(0)=(0{)}^2-1=-1$

$p(1)=(1{)}^2-1=0$

$p(2)=(2{)}^2-1=3$

Therefore, values of p(0), p(1) and p(2) are -1, 0 and 3, respectively.

Q3 (i) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x)=3x+1,x=-\frac{1}{3}$

Answer:

Given polynomial is $p(x)=3x+1$ and $x=-\frac{1}{3}$.

Therefore, the value is:

$p(-\frac{1}{3})=3\times (-\frac{1}{3})+1=-1+1=0$

Therefore, yes $x=-\frac{1}{3}$ is a zero of the polynomial $p(x)=3x+1$, because we get 0 as an answer after putting the value.

Q3 (ii) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x)=5x-\pi ,x=\frac{4}{5}$

Answer:

Given polynomial is $p(x)=5x-\pi$ and $x=\frac{4}{5}$

Therefore, the value is:

$p\left(\frac{4}{5}\right)=5\times \left(\frac{4}{5}\right)-\pi =4-\pi \ne 0$

Therefore, no $x=\frac{4}{5}$ is not a zero of a polynomial $p(x)=5x-\pi$, because we get a number other than 0 as an answer after putting the value.

Q3 (iii) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x)=x^2-1,x=1,-1$

Answer:

Given polynomial is $p(x)=x^2-1$ and x = 1, -1

First, check with x = 1, therefore we get:

$p(1)=(1{)}^2-1=1-1=0$

And at x = -1

$p(-1)=(-1{)}^2-1=1-1=0$

Therefore, yes, x = 1, -1 are zeros of the polynomial $p(x)=x^2-1$, because we get 0 as an answer after putting the value.

Q3 (iv) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x)=(x+1)(x-2),x=-1,2$

Answer:

Given polynomial is $p(x)=(x+1)(x-2)$ and x = -1, 2

First, check with x = -1, therefore we get:

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

And at x = 2

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

Therefore, yes, x = 2, -1 are zeros of the polynomial $p(x)=(x+1)(x-2)$, because we get 0 as an answer after putting the value.

Q3 (v) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x)=x^2.x=0$

Answer:

Given polynomial is $p(x)=x^2$

Now, at x = 0 its value is

$p(0)=(0{)}^2=0$

Therefore, yes x = 0 is a zeros of polynomial $p(x)=(x+1)(x-2)$

Q3 (vi) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x)=lx+m,x=-\frac{m}{l}$

Answer:

Given polynomial is $p(x)=lx+m$ and $x=-\frac{m}{l}$ it's value is

$p(-\frac{m}{l})=l\times (-\frac{m}{l})+m=-m+m=0$

Therefore, yes, $x=-\frac{m}{l}$ is a zero of the polynomial $p(x)=lx+m$, because we get 0 as an answer after putting the value.

Q3 (vii) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x)=3x^2-1,x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}$

Answer:

Given polynomial is $p(x)=3x^2-1$ and $x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}$

First, check with x = -$\frac{1}{\sqrt{3}}$, its value is:

$p(-\frac{1}{\sqrt{3}})=3\times {(-\frac{1}{\sqrt{3}})}^2-1=1-1=0$

And at $x=\frac{2}{\sqrt{3}}$

$p\left(\frac{2}{\sqrt{3}}\right)=3\times {\left(\frac{2}{\sqrt{3}}\right)}^2-1=4-1=3\ne 0$

Therefore, $x=-\frac{1}{\sqrt{3}}$ is a zeros of polynomial $p(x)=3x^2-1$ whereas $x=\frac{2}{\sqrt{3}}$ is not a zeros of polynomial $p(x)=3x^2-1$

Q3 (viii) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x)=2x+1,x=\frac{1}{2}$

Answer:

Given polynomial is $p(x)=2x+1$ and $x=\frac{1}{2}$

Therefore, its value is

$p\left(\frac{1}{2}\right)=2\times \left(\frac{1}{2}\right)+1=1+1=2\ne 0$

Therefore, $x=\frac{1}{2}$ is not a zeros of polynomial $p(x)=2x+1$

Q4 (i) Find the zero of the polynomial in each of the following cases: $p(x)=x+5$

Answer:

Given polynomial is $p(x)=x+5$

The zero of a polynomial is the value of the variable at which the polynomial's value is obtained as 0.

Now, $p(x)=0$

$\Rightarrow x+5=0$

$\Rightarrow x=-5$

Therefore, $x=-5$ is the zero of polynomial $p(x)=x+5$

Q4 (ii) Find the zero of the polynomial in each of the following cases: $p(x)=x-5$

Answer:

Given polynomial is $p(x)=x-5$

The zero of a polynomial is the value of the variable at which the polynomial's value is obtained as 0.

Now, $p(x)=0$

$\Rightarrow x-5=0$

$\Rightarrow x=5$

Therefore, $x=5$ is a zero of polynomial $p(x)=x-5$

Q4 (iii) Find the zero of the polynomial in each of the following cases: $p(x)=2x+5$

Answer:

Given polynomial is $p(x)=2x+5$

The zero of a polynomial is the value of the variable at which the polynomial's value is obtained as 0.

Now, $p(x)=0$

$\Rightarrow 2x+5=0$

$\Rightarrow x=-\frac{5}{2}$

Therefore, $x=-\frac{5}{2}$ is a zero of polynomial $p(x)=2x+5$

Q4 (iv) Find the zero of the polynomial in each of the following cases: $p(x)=3x-2$

Answer:

Given polynomial is $p(x)=3x-2$

The zero of a polynomial is the value of the variable at which the polynomial's value is obtained as 0.

Now, $p(x)=0$

$\Rightarrow 3x-2=0$

$\Rightarrow x=\frac{2}{3}$

Therefore, $x=\frac{2}{3}$ is a zero of polynomial $p(x)=3x-2$

Q4 (v) Find the zero of the polynomial in each of the following cases: $p(x)=3x$

Answer:

Given polynomial is $p(x)=3x$

The zero of a polynomial is the value of the variable at which the polynomial's value is obtained as 0.

Now, $p(x)=0$

$\Rightarrow 3x=0$

$\Rightarrow x=0$

Therefore, $x=0$ is a zero of polynomial $p(x)=3x$

Q4 (vi) Find the zero of the polynomial in each of the following cases: $p(x)=ax,a\ne 0$

Answer:

Given polynomial is $p(x)=ax$

The zero of a polynomial is the value of the variable at which the polynomial's value is obtained as 0.

Now, $p(x)=0$

$\Rightarrow ax=0$

$\Rightarrow x=0$

Therefore, $x=0$ is a zero of polynomial $p(x)=ax$

Q4 (vii) Find the zero of the polynomial in each of the following cases: $p(x)=cx+d,c\ne 0,c,d$ are real numbers

Answer:

Given polynomial is $p(x)=cx+d$

The zero of a polynomial is the value of the variable at which the polynomial's value is obtained as 0.

Now, $p(x)=0$

$\Rightarrow cx+d=0$

$\Rightarrow x=-\frac{d}{c}$

Therefore, $x=-\frac{d}{c}$ is a zero of polynomial $p(x)=cx+d$.