If the biquadratic $x^4+4x^3+6p{x}^2+4qx+r$ is divisible by $x^3+3x^2+9x+3$, find the value of  $2(p+q)r$

Since the given bi-quadratic expression

$x^4+4x^3+6p{x}^2+4qx+r$ is divisible by the expression $x^3+3x^2+9x+3$ , so we can write it as 

$\begin{array}{l}{x}^4+4x^3+6p{x}^2+4qx+r\\ =\left(x^3+3x^2+9x+3\right)(x-a)\\ =x^4+(3-\alpha )x^3+3(3+\alpha )x^2+3(1+3\alpha )x-3\alpha \end{array}$

Comparing the co-efficients of x³, x², x and constant terms, we get $(3-\alpha )=4,3(3+\alpha )=6p,3(1+3\alpha )=4q$ and $-3\alpha =r$ Thus, $\alpha =-1,p=1,q=-\frac{3}{2},r=3$ Hence, the value of $2(p+q)r=2\times \left(1-\frac{3}{2}\right)\times 3=-3$