Using factor theorem, show that g(x) is a factor of p(x), when

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

$g(x)=x+3$

Putting $x=-3$ in $p(x)$, we get

$p(-3)=69+11 \times(-3)-(-3)^{2}+(-3)^{3}=69-33-9-27=0$

Therefore, by factor theorem, (x + 3) is a factor of p(x).

Hence, g(x) is a factor of p(x).