If one root of $4{\mathrm{x}}^3-24{\mathrm{x}}^2+\mathrm{kx}-15=0$  may be the sum of the other two roots, then k=

Given polynomial is  $4{\mathrm{x}}^3-24{\mathrm{x}}^2+\mathrm{kx}-15=0$
One root is sum of the other two roots
Let $\mathrm{\alpha }=\mathrm{\beta }+\mathrm{\gamma }$
But $\mathrm{\Sigma \alpha }=\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=\frac{-\mathrm{b}}{\mathrm{a}}=\frac{-\left(-24\right)}{4}=6$
$\begin{array}{r}\mathrm{\alpha }+\mathrm{\alpha }=6\\ 2\mathrm{\alpha }=6\Rightarrow \mathrm{\alpha }=3\end{array}$
Since $\text{'}\alpha \text{'}$  is a root of polynomial $\Rightarrow f\left(\alpha \right)=0$
$\begin{array}{l}\therefore \mathrm{f}\left(3\right)=0\\ 4(3{)}^3-24(3{)}^2+\mathrm{k}\left(3\right)-15=0\\ 4\left(27\right)-24\left(9\right)+3\mathrm{k}-15=0\\ 108-216+3\mathrm{k}-15=0\\ \Rightarrow 108-231+3\mathrm{k}=0\\ -123+3\mathrm{k}=0\\ 3\mathrm{k}=123\\ \mathrm{k}=41\end{array}$