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

The given equation is $f\left(x\right)=4x^3+16x^2+kx-36=0$ 

Given that product of two roots $\left(\alpha \beta \right)=-6$

And  $\alpha \beta \gamma =\frac{36}{4}$

$\Rightarrow \gamma =\frac{-3}{2}$

Substitute $x=-\frac{3}{2}$  in $f\left(x\right)$

$\Rightarrow f\left(-\frac{3}{2}\right)=4{\left(-\frac{3}{2}\right)}^3+16{\left(-\frac{3}{2}\right)}^2+k\left(-\frac{3}{2}\right)-36=0$

After simplification, we get  $k=-9$