If product of two roots of $4{\mathrm{x}}^3+16{\mathrm{x}}^2+\mathrm{kx}–36=0$ is –6 then k =

$\begin{array}{l}Let\alpha ,\beta ,\gamma arerootsof4x^3+16x^2+kx-36=0\\ Given\beta \gamma =-6\\ \therefore \alpha \beta \gamma =\frac{36}{4}\Rightarrow \alpha (-6)=9\Rightarrow \alpha =\frac{-9}{6}=\frac{-3}{2}\\ \because \alpha isaroot\Rightarrow 4{\left(\frac{-3}{2}\right)}^3+16{\left(\frac{-3}{2}\right)}^2+k\left(\frac{-3}{2}\right)-36=0\\ \Rightarrow 4\left(\frac{-27}{8}\right)+16\left(\frac{9}{4}\right)-\frac{3k}{2}-36=0\\ \Rightarrow \frac{-27+72-3k-72}{2}=0\\ \Rightarrow 3k=-27\Rightarrow k=-9\end{array}$