Show that,

(i) Let $p(x)=x^{3}-x^{2}+11 x+69$

We have to show that, $x+3$ is a factor of $p(x)$.

ie. $\quad D(-3)=0$

Now,         $p(-3)=(-3)^{3}-(-3)^{2}+11(-3)+69$

$=-27-9-33+69=-69+69=0$

Hence, $(x+3)$ is a factor of $p(x)$.

(ii) Let $p(x)=2 x^{3}-9 x^{2}+x+12$

We have to show that, $2 x-3$ is a factor of $p(x)$.

i.e.,             $p\left(\frac{3}{2}\right)=0$

Now,           $p\left(\frac{3}{2}\right)=2\left(\frac{3}{2}\right)^{3}-9\left(\frac{3}{2}\right)^{2}+\frac{3}{2}+12$

$=2 \times \frac{27}{8}-9 \times \frac{9}{4}+\frac{3}{2}+12$

$=\frac{27}{4}-\frac{81}{4}+\frac{3}{2}+12$

$=\frac{27-81+6+48}{4}=\frac{81-81}{4}=0$

Hence, $(2 x-3)$ is a factor of $p(x)$.