1. Let the cubic polynomial be $p(x)=ax^3+bx^2+cx+d$ Since it has a local maximum at $x=1$ and a local minimum at $x=2$, we must have $p'(1)=0,\, p'(2)=0$

2. Differentiate $p'(x)=3ax^2+2bx+c$ Since $x=1,2$ are the critical points, the derivative must be of the form $p'(x)=k(x-1)(x-2)$ for some constant $k$.

3. Integrate to get $p(x)$ First expand: $p'(x)=k(x^2-3x+2)$ Integrating, $p(x)=k\left(\frac{x^3}{3}-\frac{3x^2}{2}+2x\right)+C$ where $C$ is a constant.

4. Use the given values We are given: $p(1)=8, \qquad p(2)=4$

   Compute $p(1)$:

   =k\left(\frac{2-9+12}{6}\right)+C =\frac{5k}{6}+C$$ So, $$\frac{5k}{6}+C=8 \quad ...(1)$$ Compute $p(2)$: $$p(2)=k\left(\frac83-6+4\right)+C =k\left(\frac{8-18+12}{3}\right)+C =\frac{2k}{3}+C$$ So, $$\frac{2k}{3}+C=4 \quad ...(2)$$

5. Solve for $k$ and $C$ Subtract (2) from (1): $\frac{5k}{6}-\frac{2k}{3}=8-4$ $\frac{5k}{6}-\frac{4k}{6}=4$ $\frac{k}{6}=4$ $k=24$

   Put into (2): $\frac{2(24)}{3}+C=4$ $16+C=4$ $C=-12$

6. Find $p(0)$ $p(0)=C=-12$

7. Check max/min condition $p'(x)=24(x-1)(x-2)$ For $x<1$, $p'(x)>0$; for $1<x<2$, $p'(x)<0$; for $x>2$, $p'(x)>0$. Hence $x=1$ is indeed a local maximum and $x=2$ is a local minimum.

Therefore, $\boxed{p(0)=-12}$ So the correct option is C.