Common roots

Let us find the roots of the cubic equation. Using rational root theorem , we note that $x=2$ is a root. Then, we have:

$\begin{aligned} x^3-4x^2+8x-8 & = 0 \\ (x-2)\left(x^2 - 2x+4\right) & = 0 \end{aligned}$

We note that $x^2 - 2x+4 = 0$ gives two complex roots, therefore the cubic equation has one real root 2 and two complex roots. For the quadratic equation with real rational coefficients $ax^2+bx+c=0$ to share two roots, the two shared roots must be the two conjugate complex roots. Therefore, $ax^2+bx+c \equiv x^2 - 2x+4 = 0$ and $2b+c = 2(-2)+4 = \boxed{0}$ .