We know that if $(x-a)$ is a factor of $p ( x )$, then $p ( a )=0$.

$(i)$ Let $P(x)=3 x^{2}+6 x-24$

If $x-2$ is a factor of $p(x)=3 x^{2}+6 x-24,$ then $p(2)$ should be equal to $0 .$

Now, $p(2)=3(2)^{2}+6(2)-24$

$=3(4)+6(2)-24$

$=12+12-24$

$=0$

$\therefore$ By factor theorem, $(x-2)$ is factor of $3 x^{2}+6 x-24.$

$(ii)$ Let $p(x)=4 x^{2}+x-2$

If $x-2$ is a factor of $p(x)=4 x^{2}+x-2,$ then, $p(2)$ should be equal to $0 .$

Now, $\quad p(2)=4(2)^{2}+2-2$

$=4(4)+2-2$

$=16+2-2$

$=16 \neq 0$

$\therefore x-2$ is not a factor of $4 x^{2}+x-2$