If p(x) be a polynomial of degree three that has

Let $p^{\prime}(x)=\lambda(x-1)(x-2)$ where $\lambda>0$

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

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

$\Rightarrow \frac{5 \lambda}{6}+C=8$  ...........(1)

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

$\Rightarrow \frac{2}{3} \lambda+C=4$ .................(2)

From (1) and (2), we get

$C=-12$ and $\lambda=24$

$\Rightarrow p(0)=0+C=-12$