Using factor theorem, show that $$g(x)$$ is a factor of $$p(x)$$, when $$p(x)=3x^3+x^2-20x+12$$, $$g(x)=3x-2$$.
By using factor theorem
$$(3x-2)=0$$
$$3x=2$$
$$x=2/3$$
now, let $$f(x)=3x^3+x^2-20x+12$$
$$f(2/3)=3(2/3)^3+(2/3)^2-20(2/3)+12$$
$$f(2/3)=3(8/27)+(4/9)-(40/3)+12$$
$$f(2/3)=(8/9)+(4/9)-(40/3)+12$$
$$f(2/3)=(12/9)-(40/3)+12$$
$$f(2/3)=(4/3)-(40/3)+12$$
$$f(2/3)=12-(36/3)$$
$$f(2/3)=12-12$$
$$f(2/3)=0$$
$$(3x-2)$$ is a factor of $$3x^3+x^2-20x+12$$