Short-Answer Questions

$x^{2}+6 x-\left(a^{2}+2 a-8\right)=0$

$\Rightarrow x^{2}+6 x-(a+4)(a-2)=0$

$\Rightarrow x^{2}+[(a+4)-(a-2)] x-(a+4)(a-2)=0$

$\Rightarrow x^{2}+(a+4) x-(a-2) x-(a+4)(a-2)=0$

$\Rightarrow x[x+(a+4)]-(a-2)[x+(a+4)]=0$

$\Rightarrow[x+(a+4)][x-(a-2)]=0$

$\Rightarrow x+(a+4)=0$ or $x-(a-2)=0$

$\Rightarrow x=-(a+4)$ or $x=(a-2)$

Hence, $-(a+4)$ and $(a-2)$ are the roots of the given equation.