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

$p(x)=3 x^{3}+x^{2}-20 x+12$

$g(x)=3 x-2=3\left(x-\frac{2}{3}\right)$

Putting $x=\frac{2}{3}$ in $p(x)$, we get

$p\left(\frac{2}{3}\right)=3 \times\left(\frac{2}{3}\right)^{3}+\left(\frac{2}{3}\right)^{2}-20 \times \frac{2}{3}+12=\frac{8}{9}+\frac{4}{9}-\frac{40}{3}+12=\frac{8+4-120+108}{9}=\frac{120-120}{9}=0$

Therefore, by factor theorem, (3x − 2) is a factor of p(x).

Hence, g(x) is a factor of p(x).