Question

Verify that
(i) 1 and 2 are the zeros of the polynomial p(x) = x² − 3x + 2.
(ii) 2 and −3 are the zeros of the polynomial q(x) = x² + x − 6.
(iii) 0 and 3 are the zeros of the polynomial r(x) = x² − 3x.

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)p\left(x\right)=x^2-3x+2=\left(x-1\right)\left(x-2\right) \\ \Rightarrow p\left(1\right)=\left(1-1\right)\times \left(1-2\right) \end{gathered}$$
$$\begin{gathered} =0\times \left(-1\right) \\ =0 \end{gathered}$$
Also,
$p\left(2\right)=\left(2-1\right)\left(2-2\right)$
$$\begin{gathered} =\left(-1\right)\times 0 \\ =0 \end{gathered}$$

Hence, 1 and 2 are the zeroes of the given polynomial.

$$\begin{gathered} \left(\mathrm{ii}\right)p\left(x\right)=x^2+x-6 \\ \Rightarrow p\left(2\right)=2^2+2-6 \end{gathered}$$
$$\begin{gathered} =4-4 \\ =0 \end{gathered}$$
Also,
$$\begin{gathered} p\left(-3\right)={\left(-3\right)}^2+\left(-3\right)-6 \\ =9-9 \\ =0 \end{gathered}$$

Hence, 2 and $-3$ are the zeroes of the given polynomial.

$$\begin{gathered} \left(\mathrm{iii}\right)p\left(x\right)=x^2-3x \\ \Rightarrow p\left(0\right)=0^2-3\times 0 \end{gathered}$$

Also,
$$\begin{gathered} p\left(3\right)=3^2-3\times 3 \\ =9-9 \\ =0 \end{gathered}$$

Hence, 0 and 3 are the zeroes of the given polynomial.