The remainder theorem says that if \(x-c\) is a factor of a polynomial \(p(x)\), then the remainder upon dividing \(\dfrac{p(x)}{x-c}\) is 0 so that there exists a lower degree polynomial \(q(x)\) as the quotient:
\(\dfrac{p(x)}{x-c}=q(x)\)
Using the fact that \(x-2\) is a factor, you can find a quadratic \(q(x)\) which is easy to factorize further.
Synthetic division yields
\(q(x)=4x^2+4x-8\)
which can be factored further as
\(4(x^2+x-2)=4(x+2)(x-1)\)
So,
\(f(x)=4x^3-4x^2-16x+16=4(x-2)(x+2)(x-1)\)
The roots are then \(x=-2,1,2\).
