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

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