(x2 + 4x + 8)2 + 3x(x2 + 4x + 8) + 2x2 = 0
Đặt x2 + 4x + 8 = a
<=> a2 + 3xa + 2x2 = 0
<=> a2 + 2ax + ax + 2x2 = 0
<=> (a + x)(a + 2x) = 0
<=> (x2 + 4x + 8 + x)(x2 + 4x + 8 + 2x) = 0
<=> (x2 + 5x + 8)(x2 + 6x + 8) = 0
<=> x2 + 4x + 2x + 8 = 0 (vì x2 + 5x + 8 = (x2 + 5x + 6,25) + 1,75 = (x + 2,5)^2 + 1,75 > 0)
<=> (x + 4)(x + 2) = 0
<=> \(\orbr{\begin{cases}x+4=0\\x+2=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-4\\x=-2\end{cases}}\)
Vậy S = {-4; -2}