# Exercise 2.2 Polynomials NCERT Solutions Class 9

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## Chapter 2 Ex.2.2 Question 1

Find the value of the polynomial $$5x - 4{x^2} + 3$$ at  

(i) \begin{align}x=0 \end{align}

(ii) \begin{align}x=-1 \end{align}

(iii) \begin{align} x=2\end{align}

### Solution

Steps:

Let, \begin{align}p(x) = 5x - 4{x^2} + 3\end{align}

\begin{align} \rm(i)\;\;\;p(0) &=5(0)-4(0)^{2}+3 \\ &=3 \end{align}

\begin{align} \rm(ii) \;\;\;p(-1) &=5(-1)-4(-1)^{2}+3 \\ &=-5-4+3 \\ &=-6 \end{align}

\begin{align} \rm(iii)\;\; \; p(2) &=5(2)-4(2)^{2}+3 \\ &=10-16+3 \\ &=-3 \end{align}

## Chapter 2 Ex.2.2 Question 2

Find $$p(0), p(1)$$ and $$p(2)$$ for each of the following polynomials:

(i) \begin{align}p(y)=y^{2}-y+1\end{align}

(ii) \begin{align}p(t)=2+t+2 t^{2}-t^{3}\end{align}

(iii) \begin{align}p(x)=x^{3}\end{align}

(iv) \begin{align}p(x)=(x-1)(x+1)\end{align}

### Solution

Steps:

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

\begin{align}{p(0)}&={(0)^{2}-(0)+1=1} \\ {p(1)}&={(1)^{2}-(1)+1=1} \\ {p(2)}&={(2)^{2}-2+1=3}\end{align}

(ii) $$p(t) =2+t+2\left(t^{2}\right)-t^{3}$$

\begin{align}p(0) &=2+0+2(0)^{2}-(0)^{3} \\ &=2+0+0-0=2 \\ p(1) &=2+1+2(1)^{2}-(1)^{3} \\ &=2+1+2-1=4 \\ p(2) &=2+2+2(2)^{2}-(2)^{3} \\ &=2+2+8-8=4 \end{align}

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

\begin{align}p(0) &=(0)^{3}=0 \\ p(1) &=(1)^{3}=1 \\ p(2) &=(2)^{3}=8 \end{align}

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

\begin{align} p(x) &=x^{2}-1 \\ p(0) &=(0)^{2}-1=-1 \\ p(1) &=(1)^{2}-1=0 \\ p(2) &=(2)^{2}-1=3 \end{align}

## Chapter 2 Ex.2.2 Question 3

Verify whether the following are zeroes of the polynomial, indicated against them.

(i) \begin{align}p(x)=3 x+1, x = -\frac{1}{3}\end{align}

(ii) \begin{align}p(x)=5 x - \pi, x = \frac{4}{5}\end{align}

(iii) \begin{align}p(x)= x^{2}-1, x = 1,-1 \end{align}

(iv) \begin{align}p(x)= (x+1)(x-2), x = -1,2 \end{align}

(v) \begin{align}p(x)= x^{2}, x = 0\end{align}

(vi)\begin{align}p(x)=l x+m, x=\frac{-m}{l}\end{align}

(vii) \begin{align}p(x)=3 x^{2}-1, x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\end{align}

(viii) \begin{align}p(x)=2 x+1, x=\frac{1}{2}\end{align}

### Solution

Reasoning:

In general, we say that a zero of a polynomial $$p(x)$$ is a number c such that $$p(c) = 0.$$

Steps:

(i) \begin{align}p(x)=3 x+1, x = -\frac{1}{3}\end{align}

\begin{align}& p(x)=3 x+1, x=\frac{-1}{3} \\&\!{p\left(\frac{-1}{3}\right)\!=3\!\times\!\left(\frac{-1}{3}\right)\!+\!1\!=-\!1\!+\!1\!=\!0}\end{align}

\begin{align}\therefore \frac{-1}{3}\end{align} is a zero of $$p(x)$$

(ii) \begin{align}p(x)=5 x - \pi, x = \frac{4}{5}\end{align}

\begin{align} &p(x) = 5x - \pi ,x = \frac{4}{5} \\&p\left( {\frac{4}{5}} \right) = 5 \times \frac{4}{5} - \pi = 4 - \pi \not= 0\end{align}

\begin{align}\therefore\, \frac{4}{5} \end{align} is not a zero of $$p(x).$$

(iii) \begin{align}p(x)= x^{2}-1, x = 1,-1 \end{align}

\begin{align}p(x)&={x^{2}-1, x=1,-1} \\ p(1) &=1^{2}-1=0 \\ p(-1) &=(-1)^{2}-1=1-1=0 \end{align}

$$\therefore$$ 1 and $$-1$$ are zeroes of $$p(x)$$

(iv) \begin{align}p(x)= (x+1)(x-2), x = -1,2 \end{align}

\begin{align} p(x)&=(x+1)(x-2), x=-1,2 \\p(-1)&\!=\!(-1\!+\!1)\!(-1\!-2)\!=\!0 \!\times\!(\!-3)\!=\!0 \\ p(2)&\!=\!(2\!-\!1)(2\!-\!2)\!=\!1\!\times\!0\!=\!0 \end{align}

$$\therefore-1$$and $$2$$ are zeroes of $$p(x)$$

(v) \begin{align}p(x)= x^{2}, x = 0\end{align}

\begin{align} p(x)&=x^{2}, x=0 \\ p(0)&=0^{2}=0 \end{align}

$$\therefore$$ $$0$$ is a zero of $$p(x)$$

(vi)\begin{align}p(x)=l x+m, x=\frac{-m}{l}\end{align}

\begin{align}& {p(x)=l x+m, x=\frac{-m}{l}} \\ &p\left(\frac{-m}{l}\right) =l \times \frac{-m}{l}+m \\ &\qquad\qquad\;=-m+m=0 \end{align}

\begin{align}\therefore \frac{-m}{l}\end{align} is a zero of $$p(x)$$

(vii) \begin{align}p(x)=3 x^{2}-1, x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\end{align}

\begin{align} p(x)&{=3 x^{2}-1, x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}} \\ p\left(\frac{-1}{\sqrt{3}}\right) &=3 \times\left(\frac{-1}{\sqrt{3}}\right)^{2}-1 \\ &=3 \times \frac{1}{3}-1=1-1=0 \end{align}

\begin{align}\therefore\frac{-1}{\sqrt{3}}\end{align} is a zero of $$p(x)$$

\begin{align} p\left(\frac{2}{\sqrt{3}}\right) &=3 \times\left(\frac{2}{\sqrt{3}}\right)^{2}-1 \\ &=3 \times \frac{4}{3}-1 \\ &=4-1=3 \neq 0\end{align}

\begin{align}\therefore \frac{2}{\sqrt{3}}\end{align}is not a zero of $$p(x)$$

(viii) \begin{align}p(x)=2 x+1, x=\frac{1}{2}\end{align}

\begin{align}p(x)&=2 x+1, x=\frac{1}{2} \\ p\left(\frac{1}{2}\right)&=2 \times \frac{1}{2}+1 \\ \quad&=1+1=2 \neq 0 \end{align}

\begin{align}\therefore \frac{1}{2}\end{align} is not a zero of $$p(x)$$

## Chapter 2 Ex.2.2 Question 4

Find the zero of the polynomials in each of the following cases:

(i) \begin{align} p(x)=x+5 \end{align}

(ii) \begin{align}p(x)=x-5\end{align}

(iii) \begin{align}p(x)=2 x+5\end{align}

(iv)\begin{align}p(x)=3 x-2\end{align}

(v) \begin{align}p(x)=3 x\end{align}

(vi) \begin{align}p(x)=a x, a \neq 0\end{align}

(vii) \begin{align}p(x)=c x+d, c \neq 0, c, d \end{align} are real numbers.

### Solution

Reasoning:

In general, we say that a zero of a polynomial $$p(x)$$ is a number $$c$$ such that$$p(c) = 0.$$

Steps:

(i) \begin{align} p(x)=x+5 \end{align}

\begin{align}p(x)&=x+5 \\ \qquad p(x)&=0 \rightarrow x+5=0 \\ \Rightarrow x&=-5 \end{align}

$$\therefore-5$$ is the zero of $$p(x)$$

(ii) \begin{align}p(x)=x-5\end{align}

\begin{align}{p(x)}&={x-5} \\ p(x)&=0 \rightarrow x-5=0 \\ \Rightarrow x&=5 \end{align}

$$\therefore 5$$ is the zero of $$p(x)$$

(iii) \begin{align}p(x)=2 x+5\end{align}

\begin{align}p(x)&=2 x+5 \\p(x)&=0 \rightarrow 2 x+5=0 \\ {\Rightarrow 2 x}&={-5} \\ {\Rightarrow x}&={\frac{-5}{2}} \\ \end{align}

\begin{align}\therefore \frac{-5}{2}\end{align} is the zero of $$p(x)$$

(iv)\begin{align}p(x)=3 x-2\end{align}

\begin{align}p(x)& =3 x-2 \\ {p(x)}&={0 \rightarrow 3 x-2=0} \\ {\Rightarrow 3 x}&={2} \\ {\Rightarrow x}&={\frac{2}{3}} \end{align}

\begin{align}\therefore \frac{2}{3}\end{align} is the zero of $$p(x)$$

(v) \begin{align}p(x)=3 x\end{align}

\begin{align} p(x)&=3 x \\ p(x)&=0 \rightarrow 3 x =0 \\ x&=0 \\ \end{align}

$$\therefore 0$$ is the zero of $$p(x)$$

(vi) \begin{align}p(x)=a x, a \neq 0\end{align}

\begin{align}p(x)&=a x, a \neq 0 \\ {p(x)}&={0 \rightarrow a x=0} \\ {\quad x}&={0} \end{align}

$$\therefore 0$$ is the zero of $$p(x)$$

(vii) \begin{align}p(x)=c x+d, c \neq 0, c, d \end{align} are real numbers.

\begin{align}{p(x)}&=c x+d, c \neq 0, c, d \quad \\&\text { are real numbers. } \\\\ {p(x)}&={0 \rightarrow c x+d=0} \\ {\Rightarrow c x}&={-d} \\ {\Rightarrow x}&={\frac{-d}{c}}\end{align}

\begin{align}\therefore \frac{-d}{c}\end{align} is the zero of $$p(x)$$

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