Clearly $p^{\text{'}}\left(x\right)=\lambda (x-1)(x-2)$ where $\lambda >0$

Integrating both sides

$p\left(x\right)=\lambda \left[\frac{x^3}{3}-\frac{3x^2}{2}+2x\right]+C$

Given

$p\left(1\right)=8\Rightarrow \lambda \left(\frac{1}{3}-\frac{3}{2}+2\right)+C=8 \Rightarrow \frac{5\lambda }{6}+C=8 ...\left(1\right)$

$p\left(2\right)=4\Rightarrow \lambda \left(\frac{8}{3}-6+4\right)+\mathrm{C}=4 \Rightarrow \frac{2}{3}\lambda +\mathrm{C}=4 ...\left(2\right)$

By $\left(1\right)$ and $\left(2\right)$ $\Rightarrow C=-12$

$\Rightarrow p\left(0\right)=-12$