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# Ex.2.2 Q4 Polynomials Solution - NCERT Maths Class 9

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## Question

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.

Video Solution
Polynomials
Ex 2.2 | Question 4

## 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)=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 $$p(x)$$

Video Solution
Polynomials
Ex 2.2 | Question 4

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