Question:
Verify that
(i) 1 and 2 are the zeros of the polynomial $p(x)=x^{2}-3 x+2$.
(ii) 2 and $-3$ are the zeros of the polynomial $q(x)=x^{2}+x-6$.
(iii) 0 and 3 are the zeros of the polynomial $r(x)=x^{2}-3 x$.
Solution:
(i) $p(x)=x^{2}-3 x+2=(x-1)(x-2)$
$\Rightarrow p(1)=(1-1) \times(1-2)$
$=0 \times(-1)$
$=0$
Also,
$p(2)=(2-1)(2-2)$
$=(-1) \times 0$
$=0$
Hence, 1 and 2 are the zeroes of the given polynomial.
(ii) $p(x)=x^{2}+x-6$
$\Rightarrow p(2)=2^{2}+2-6$
$=4-4$
$=0$
Also,
$p(-3)=(-3)^{2}+(-3)-6$
$=9-9$
$=0$
Hence, 2 and −3 are the zeroes of the given polynomial.
(iii) $p(x)=x^{2}-3 x$
$\Rightarrow p(0)=0^{2}-3 \times 0$
$\mathrm{Also}$,
$p(3)=3^{2}-3 \times 3$
$=9-9$
$=0$
Hence, 0 and 3 are the zeroes of the given polynomial.