a, $x^2-2016x-2017=0$
$⇔x(x-2017)+(x-2017)=0$
$⇔(x-2017)(x+1)=0$
$⇔$\(\left[ \begin{array}{l}x-2017=0\\x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2017\\x=-1\end{array} \right.\)
Vậy $S=${$2017;-1$}
b, $x(x-1)(x+1)(x+2)-24$
$=(x^2+x-2)(x^2+x)-24$
Đặt: $a=x^2+x-1$
$=(a-1)(a+1)-24$
$=a^2-1-24$
$=(a-5)(a+5)$
$=(x^2+x-6)(x^2+x+4)$