Question

Verify whether the following are zeroes of the polynomial, indicated against them: $\mathrm{p}\left(\mathrm{x}\right)=(\mathrm{x}+1)(\mathrm{x}–2),\mathrm{x}=-1,2$

Answer 1

We have,

$\mathrm{p}\left(\mathrm{x}\right)=(\mathrm{x}+1)(\mathrm{x}–2),\mathrm{x}=-1,2$

Putting the value $\mathrm{x}=-1$ in $\mathrm{p}\left(\mathrm{x}\right)=(\mathrm{x}+1)(\mathrm{x}–2)$, we get

$$\begin{gathered} \mathrm{p}(-1)=(-1+1)(-1–2) \\ \Rightarrow \mathrm{p}(-1)=\left(0\right)(-3) \\ \Rightarrow \mathrm{p}(-1)=0 \end{gathered}$$

Hence, $\mathrm{x}=-1$ is a zero of the polynomial $\mathrm{p}\left(\mathrm{x}\right)=(\mathrm{x}+1)(\mathrm{x}–2)$.

Now,

Putting the value $\mathrm{x}=2$ in polynomial $\mathrm{p}\left(\mathrm{x}\right)=(\mathrm{x}+1)(\mathrm{x}–2)$, we get

$$\begin{gathered} \mathrm{p}\left(2\right)=(2+1)(2–2) \\ \Rightarrow \mathrm{p}\left(2\right)=\left(3\right)\left(0\right) \\ \Rightarrow \mathrm{p}\left(2\right)=0 \end{gathered}$$

Hence, $\mathrm{x}=2$ is the zero of the polynomial $\mathrm{p}\left(\mathrm{x}\right)=(\mathrm{x}+1)(\mathrm{x}–2)$.