# Ex.2.2 Q3 Polynomials Solution - NCERT Maths Class 9

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## Question

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}

Video Solution
Polynomials
Ex 2.2 | Question 3

## Text 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 $$p(x)$$

Video Solution
Polynomials
Ex 2.2 | Question 3

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