If $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3+{\mathrm{px}}^2+\mathrm{qx}+30$ is divisible by (x-5) and when it is divided by (x+6), the remainder is ‘-396’, then

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3+{\mathrm{px}}^2+\mathrm{qx}+30\\ \mathrm{f}\left(5\right)=0\\ \Rightarrow 125+25\mathrm{p}+5\mathrm{q}+30=0\\ 25\mathrm{p}+5\mathrm{q}=-155\\ 5\mathrm{p}+\mathrm{q}=-31\to \left(1\right)\end{array}$
Given  $\mathrm{f}\left(-6\right)=-396$
$\begin{array}{l}-216+36\mathrm{p}-6\mathrm{q}+30=-396\\ 36\mathrm{p}-6\mathrm{q}=-210\\ 6\mathrm{p}-\mathrm{q}=-35\to \left(2\right)\end{array}$
From (1) & (2)   
$\begin{array}{r}\mathrm{p}=-6\\ \mathrm{q}=-1\end{array}$