$x^4+2x^3-3x^2-8x-4=0$
$⇔(x^4+x^3)+(x^3+x^2)-(4x^2+4x)-(4x+3)=0$
$⇔x^3(x+1)+x^2(x+1)-4x(x+1)-4(x+1)=0$
$⇔(x+1)(x^3+x^2-4x-4)=0$
$⇔(x+1).[x^2(x+1)-4(x+1)]=0$
$⇔(x+1)^2.(x^2-4)=0$
$⇔$\(\left[ \begin{array}{l}(x+1)^2=0\\x^2-4=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=1\\x=±2\end{array} \right.\)
Vậy: `S={1;±2}`