If $x^4 + 4x^3 + 6px^2 + 4qx + r$ is exactly divisible by $x^3 + 3x^2 + 9x + 3$, the value of $(p + q)r$ is: 
$\textbf{(A)}\ - 18 \qquad  \textbf{(B) }\ 12 \qquad  \textbf{(C) }\ 15 \qquad  \textbf{(D) }\ 27 \qquad  \textbf{(E) }\ 45 \qquad$
Let $f(x)=x^3+3x^2+9x+3$ and $g(x)=x^4+4x^3+6px^2+4qx+r$. 
Let 3 roots of $f(x)$ be $r_1, r_2$ and $r_3$. As  $f(x)|g(x)$ , 3 roots of 4 roots of $g(x)$ will be same as roots of $f(x)$. Let  the 4th root of $g(x)$ be $r_4$.  By vieta's formula
In $f(x)$
$r_1+r_2+r_3=-3$
$r_1r_2+r_2r_3+r_1r_3=9$
$r_1r_2r_3=-3$
In $g(x)$
$r_1+r_2+r_3+r_4=-4$
$=>r_4=-1$
$r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4=6p$
$=>r_1r_2+r_1r_3+r_2r_3+r_4(r_1+r_2+r_3)=6p$
$=>9+(-1)(-3)=6p$
$=>p=2$
$r_1r_2r_3+r_1r_3r_4+r_1r_2r_4+r_2r_3r_4=-4q$
$=>-3+r_4(r_1r_3+r_1r_2+r_2r_3)=-4q$
$=>-3+(-1)(9)=-4q$
$=>q=3$
$r_1r_2r_3r_4=r$
$=>(-3)(-1)=r$
$=>r=3$
so $(p+q)r=\fbox{15}$
By ~Ahmed_Ashhab