Solve the equation $x^4+x^3-16x^2-4x+48=0$ then the product of two of the roots being 6

Let $\alpha , \beta , \gamma , \delta$ be the roots of 

$x^4+x^3-16x^2-4x+48=0$ 

Given $\alpha \beta =6........\left(1\right)$

$$\begin{gathered} S_1=\alpha +\beta +\gamma +\delta =-1 \\ {S}_2=\left(\alpha +\beta \right)\left(\gamma +\delta \right)+\alpha \beta +\gamma \delta =-16 \\ {S}_3=\left(\alpha +\beta \right)\gamma \delta +\left(\gamma +\delta \right)\alpha \beta =4 \\ {S}_4=\alpha \beta \gamma \delta =48 \\ From \left(1\right) \Rightarrow \left(6\right)\gamma \delta =48\Rightarrow \gamma \delta =\frac{48}{6}=8 \\ sub \alpha \beta , \gamma \delta values in S_2 \\ \Rightarrow \left(\alpha +\beta \right), \gamma \delta values in S_2 \\ \Rightarrow \left(\alpha +\beta \right)\left(\gamma +\delta \right)+6+8=-16 \\ \left(\alpha +\beta \right)\left(\gamma +\delta \right)=-30 \\ {S}_1=\left(\alpha +\beta \right)+\left(\gamma +\delta \right)=-1 \\ If \alpha +\beta , \gamma +\delta are roots then \end{gathered}$$

$$\begin{gathered} x^2-\left[\left(\alpha +\beta \right)+\left(\gamma +\delta \right)\right]x+\left(\alpha +\beta \right)\left(\gamma +\delta \right)=0 \\ {x}^2-\left(-1\right)x+\left(-30\right)=0 \\ {x}^2+x-30=0 \\ x=-6 x=5 \\ \alpha +\beta =5 \gamma +\delta =-6 \\ \alpha \beta =5 \gamma \delta =8 \\ If \alpha ,\beta are roots then x^2-\left(\alpha +\beta \right)x+\alpha \beta =0 \\ \Rightarrow x^2-5x+6=0; \left(x-3\right)\left(x-2\right)=0 \\ x=3, or x=2 \\ \alpha =3, \beta =2 \end{gathered}$$

$$\begin{gathered} \gamma , \delta are roots then x^2-\left(\gamma +\delta \right)x+\gamma \delta =0 \\ {x}^2+6x+8=0; \left(x+4\right)\left(x+2\right)=0 \\ x=-4 or x=-2 \Rightarrow \gamma =-4, \delta =-2 \\ \therefore The roots are \left\{3, 2, -4, -2\right\} \end{gathered}$$