If the product of two roots of $x^4+x^3-16x^2-4x+48=$0 is 6, find its roots.

Let $\alpha ,\beta ,\gamma ,\delta$ be the roots of $x^4+x^3-16x^2-4x+48=0$, where $\alpha \beta =6$ Clearly, $\alpha +\beta +\gamma +\delta =-1,\alpha \beta \gamma \delta =48$ Then $\gamma \delta =8$

Let $\alpha +\beta =p,\gamma +\delta =q$

Thus 

$\begin{array}{l}{x}^4+x^3-16x^2-4x+48\\ =\left(x^2-px+6\right)\left(x^2-qx+8\right)\\ =x^4-(p+q)x^3+(pq+14)x^2\\ -(8p+6q)x+48\end{array}$

Comparing the co-efficients, we get,

$\begin{array}{l}p+q=-1,pq+14=-16\\ \Rightarrow p+q=-1,pq=-30\end{array}$

$\Rightarrow p=5$ and q=-6

Hence, the roots are

$\begin{array}{l}{x}^2-5x+6=0,x^2+6x+8=0\\ \Rightarrow x=2,3,-2,-4\end{array}$