Verify that

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 −3-3 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.