smc

Let $f(x)=x^3+3x^2+9x+3$ and $g(x)=x^4+4x^3+6p{x}^2+4qx+r$. 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 Formulas In $f(x)$ $r_1+r_2+r_3=-3$ $r_1{r}_2+r_2{r}_3+r_1{r}_3=9$ $r_1{r}_2{r}_3=-3$ In $g(x)$ $r_1+r_2+r_3+r_4=-4$ $=>r_4=-1$ $r_1{r}_2+r_1{r}_3+r_1{r}_4+r_2{r}_3+r_2{r}_4+r_3{r}_4=6p$ $=>r_1{r}_2+r_1{r}_3+r_2{r}_3+r_4(r_1+r_2+r_3)=6p$ $=>9+(-1)(-3)=6p$ $=>p=2$ $r_1{r}_2{r}_3+r_1{r}_3{r}_4+r_1{r}_2{r}_4+r_2{r}_3{r}_4=-4q$ $=>-3+r_4(r_1{r}_3+r_1{r}_2+r_2{r}_3)=-4q$ $=>-3+(-1)(9)=-4q$ $=>q=3$ $r_1{r}_2{r}_3{r}_4=r$ $=>(-3)(-1)=r$ $=>r=3$ so $(p+q)r=\boxed{15}$ By