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:

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

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:

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

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

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

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. $p(x) = 3x^2 - 1, \ x = -\frac{1}{\sqrt3}, \frac{2}{\sqrt3}$

Answer:

Given polynomial is $p(x) = 3x^2-1$ and $x = -\frac{1}{\sqrt3}, \frac{2}{\sqrt3}$

First, check with x = -$\frac{1}{\sqrt3}$, its value is:

$p\left ( -\frac{1}{\sqrt3} \right )= 3 \times \left ( -\frac{1}{\sqrt3} \right )^2-1 = 1-1 =0$

And at $x = \frac{2}{\sqrt3}$

$p\left ( \frac{2}{\sqrt3} \right )= 3 \times \left ( \frac{2}{\sqrt3} \right )^2-1 = 4-1 =3\neq 0$

Therefore, $x = -\frac{1}{\sqrt3}$ is a zeros of polynomial $p(x) = 3x^2-1$ whereas $x = \frac{2}{\sqrt3}$ 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

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

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: $p(x) = ax, \ a\neq 0$

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: $p(x) = cx + d, c\neq 0, c,d$ 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 $p(x) = cx+d$.