Exercise 2.2 (Revised) - Chapter 2 - Polynomials - Ncert Solutions class 9 - Maths

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Chapter 2 - Polynomials | NCERT Solutions for Class 9 Maths

Ex 2.2 Question 1.

Find the value of the polynomial $5 x-4 x^2+3$ at
(i) $x=0$
(ii) $x=-1$
(iii) $x=2$

Answer.

(i) Let $f(x)=5 x-4 x^2+3$.

We need to substitute 0 in the polynomial $f(x)=5 x-4 x^2+3$ to get $f(0)=5(0)-4(0)^2+3$ $=0-0+3$
$=3$

Therefore, we conclude that at $x=0$, the value of the polynomial $5 x-4 x^2+3$ is 3 .
(ii) Let $f(x)=5 x-4 x^2+3$.

We need to substitute -1 in the polynomial $f(x)=5 x-4 x^2+3$ to get. $f(-1)=5(-1)-4(-1)^2+3$
$
\begin{aligned}
& =-5-4+3 \\
& =-6
\end{aligned}
$

Therefore, we conclude that at $x=-1$, the value of the polynomial $5 x-4 x^2+3$ is -6

(iii) Let $f(x)=5 x-4 x^2+3$.

We need to substitute 0 in the polynomial $f(x)=5 x-4 x^2+3$ to get $f(2)=5(2)-4(2)^2+3$
$
\begin{aligned}
& =10-16+3 \\
& =-3
\end{aligned}
$

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

Ex 2.2 Question 2.

Find $p(0), p(1)$ and ${ }^{p(2)}$ for each of the following polynomials:
(i) $p(y)=y^2-y+1$
(ii) $p(t)=2+t+2 t^2-t^3$
(iii) $p(x)=x^3$
(iv) $p(x)=(x-1)(x+1)$

Answer.

(i) $p(y)=y^2-y+1$

At $^{p(0)}:$
$p(0)=(0)^2-0+1=1$
$\mathrm{At}^{p(1)}$ :
$p(1)=(1)^2-1+1=1-0=1$

At $^{p(2)}:$
$
p(2)=(2)^2-2+1=4-1=3
$
(ii) $p(t)=2+t+2 t^2-t^3$

$\begin{aligned}
& \mathrm{At}^{p(0)} \\
& p(0)=2+(0)+2(0)^2-(0)^3=2 \\
& \mathrm{At}^{p(1)} \\
& p(1)=2+(1)+2(1)^2-(1)^3=2+1+2-1=4 \\
& \mathrm{At}^p(2) \\
& p(2)=2+(2)+2(2)^2-(2)^3=4+8-8=4
\end{aligned}$

(iii) $p(x)=(x)^3$

At $p(0)$ $p(0)=(0)^3=0$

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

At $p(2)$
$p(2)=(2)^3=8$

(vi) $p(x)=(x-1)(x+1)$

At $p(0)$ :
$p(0)=(0-1)(0+1)=(-1)(1)=-1$

At $p(1)$
$p(1)=(1-1)(2+1)=(0)(3)=0$

At $p(2)$ :
$
p(2)=(2-1)(2+1)=(1)(3)=3
$

Ex 2.2 Question 3.

Verify whether the following are zeroes of the polynomial, indicated against them.
(i) $p(x)=3 x+1, \quad x=-\frac{1}{3}$
(ii) $p(x)=5 x-\pi, x=\frac{4}{5}$
(iii) $p(x)=x^2-1, \quad x=-1,1$
(iv) $p(x)=(x+1)(x-2), x=-1,2$
(v) $p(x)=x^2, x=0$
(vi) $p(x)=l x+m, x=-\frac{m}{l}$
(vii) $p(x)=3 x^2-1, \quad x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$
(viii) $p(x)=2 x+1, \quad x=-\frac{1}{2}$

Answer.

(i) $p(x)=3 x+1, \quad x=-\frac{1}{3}$

We need to check whether $p(x)=3 x+1$ at $x=-\frac{1}{3}$ is equal to zero or not.

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

Therefore, we can conclude that $x=-\frac{1}{3}$ is a zero of the polynomial $p(x)=3 x+1$.
(ii) $p(x)=5 x-\pi, x=\frac{4}{5}$

We need to check whether $p(x)=5 x-\pi$ at $x=\frac{4}{5}$ is equal to zero or not.
$
p\left(\frac{4}{5}\right)=5\left(\frac{4}{5}\right)-\pi=4-\pi
$

Therefore, $x=\frac{4}{5}$ is not a zero of the polynomial $p(x)=5 x-\pi$.
(iii) $p(x)=x^2-1, x=-1,1$

We need to check whether $p(x)=x^2-1$ at $x=-1,1$ is equal to zero or not.

At $x=-1$
$
p(-1)=(-1)^2-1=1-1=0
$

At $x=1$

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

Therefore, $x=-1,1$ are the zeros of the polynomial $p(x)=x^2-1$.
(iv) $p(x)=(x+1)(x-2), x=-1,2$

We need to check whether $p(x)=(x+1)(x-2)$ at $x=-1,2$ is equal to zero or not.

At $x=-1$
$
p(-1)=(-1+1)(-1-2)=(0)(-3)=0
$

At $x=2$
$
p(2)=(2+1)(2-2)=(3)(0)=0
$

Therefore, $x=-1,2$ are the zeros of the polynomial $p(x)=(x+1)(x-2)$.
(v) $p(x)=x^2, x=0$

We need to check whether $p(x)=x^2$ at $x=0$ is equal to zero or not.
$
p(0)=(0)^2=0
$

Therefore, we can conclude that $x=0$ is a zero of the polynomial $p(x)=x^2$.

(vi) $p(x)=l x+m, x=-\frac{m}{l}$

We need to check whether $p(x)=$ $+m=0$
$
p\left(-\frac{m}{l}\right)=l\left(-\frac{m}{l}\right)+m=m+m=0
$

Therefore $x=-\frac{m}{l}$ is a zero of the polynomial $p(x)=l x+m$.
(vii)
$
p(x)=3 x^2-1, \quad x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}
$

We need to check whether
$
p(x)=3 x^2-1 \text { at } x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}} \text { is equal to zero or not. }
$
$x=\frac{-1}{\sqrt{3}}$
$
p\left(-\frac{1}{\sqrt{3}}\right)=3\left(-\frac{1}{\sqrt{3}}\right)^2-1=3\left(\frac{1}{3}\right)-1=1-1=0
$
$
\begin{aligned}
& x=\frac{2}{\sqrt{3}} \\
& \text { At } \\
& p\left(\frac{2}{\sqrt{3}}\right)=3\left(\frac{2}{\sqrt{3}}\right)^2-1=3\left(\frac{4}{3}\right)-1=4-1=3
\end{aligned}
$

Therefore, we can conclude that $x=\frac{-1}{\sqrt{3}}$ is a zero of the polynomial $p(x)=3 x^2-1$ but $x=\frac{-1}{\sqrt{3}}$ is not a zero of the polynomial $p(x)=3 x^2-1$.
(viii)
$
p(x)=2 x+1, x=-\frac{1}{2}
$

We need to check whether $p(x)=2 x+1$ at $x=-\frac{1}{2}$ is equal to zero or not.
$
p\left(-\frac{1}{2}\right)=2\left(-\frac{1}{2}\right)+1=-1+1=0
$

Therefore, $x=-\frac{1}{2}$ is a zero of the polynomial $p(x)=2 x+1$

Ex 2.2 Question 4.

Find the zero of the polynomial in each of the following cases:
(i) $p(x)=x+5$
(ii) $p(x)=x-5$
(iii) $p(x)=2 x+5$
(iv) $p(x)=3 x-2$
(v) $p(x)=3 x$
(vi) $p(x)=a x, a \neq 0$
(vii) $p(x)=c x+d, c \neq 0, c, d$ are real numbers.

Answer.

(i) $p(x)=x+5$
$a x+b$, where $a \neq 0$ and $b \neq 0$, and $a$ and $b$ are real numbers, we need to find $p(x)=0$.

On putting $p(x)=x+5$ equal to 0 , we get
$
x+5=0 \quad \Rightarrow x=-5
$

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

(ii) $p(x)=x-5$
$a x+b$, where $a \neq 0$ and $b \neq 0$, and $a$ and $b$ are real numbers, we need to find $p(x)=0$.

On putting $p(x)=x-5$ equal to 0 , we get
$
x-5=0 \quad \Rightarrow x=5
$

Therefore, we conclude that the zero of the polynomial $p(x)=x-5$ is 5 .
(iii) $p(x)=2 x+5$
$a x+b$, where $a \neq 0$ and $b \neq 0$, and $a$ and $b$ are real numbers, we need to find $p(x)=0$.

On putting $p(x)=2 x+5$ equal to 0 , we get
$
2 x+5=0 \quad \Rightarrow x=\frac{-5}{2}
$

Therefore, we conclude that the zero of the polynomial $p(x)=2 x+5$ is $\frac{-5}{2}$.
(iv) $p(x)=3 x-2$
$a x+b$, where $a \neq 0$ and $b \neq 0$, and $a$ and $b$ are real numbers, we need to find $p(x)=0$.

On putting $p(x)=3 x-2$ equal to 0 , we get
$
3 x-2=0 \Rightarrow x=\frac{2}{3}
$

Therefore, we conclude that the zero of the polynomial $p(x)=3 x-2$ is $\frac{2}{3}$.
(v) $p(x)=3 x$
$a x+b$, where $a \neq 0$ and $b \neq 0$, and $a$ and $b$ are real numbers, we need to find $p(x)=0$.

On putting $p(x)=3 x$ equal to 0 , we get
$
3 x=0 \quad \Rightarrow x=0
$

Therefore, we conclude that the zero of the polynomial $p(x)=3 x$ is .
(vi) $p(x)=a x, a \neq 0$
$a x+b$, where $a \neq 0$ and $b \neq 0$, and $a$ and $b$ are real numbers, we need to find $p(x)=0$.

On putting $p(x)=a x$ equal to 0 , we get
$
a x=0 \Rightarrow x=0
$

Therefore, we conclude that the zero of the polynomial $p(x)=a x, a \neq 0$ is0.
(vii) $p(x)=c x+d, c \neq 0, c, d$ are real numbers.
$a x+b$, where $a \neq 0$ and $b \neq 0$, and $a$ and $b$ are real numbers, we need to find $p(x)=0$.

On putting $p(x)=c x+d$ equal to 0 , we get
$
\begin{aligned}
& c x+d=0 \\
& \Rightarrow \quad x=-\frac{d}{c} .
\end{aligned}
$

Therefore, we conclude that the zero of the polynomial $p(x)=c x+d, c \neq 0, c, d$ are real numbers. is $-\frac{d}{c}$.