`(3x-2)(x+1)^2(3x+8)=-16`
Đặt `3x + 3=t`, ta có:
`\frac{t^2}{9}( t - 5)(t+5) = -16`
`<=> t^2( t^2 - 25 ) = -144`
`<=> t^4 - 25t^2 + 144=0`
`<=> t^4 - 4t^3 + 4t^3 -16t^2 -9t^2 + 36t - 36t + 144=0`
`<=> t^3( t - 4) + 4t^2( t - 4 ) -9t( t - 4) -36( t - 4 ) = 0`
`<=> ( t - 4 )( t^3 + 4t^2 -9t - 36) = 0`
`<=> ( t - 4 )( t + 4 )( t^2 - 9 ) =0`
`<=> ( t - 4 )( t + 4 )( t - 3 )( t + 3 ) = 0`
`<=>` \(\left[ \begin{array}{l}t=4\\t=-4\\t=3\\t=-3\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}3x + 3 = 4\\3x + 3 = -4\\3x + 3 = 3\\3x + 3 = -3\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=-\dfrac{7}{3}\\x = 0 \\x = -2\end{array} \right.\)
Vậy `S = { \frac{1}{3}; \frac{-7}{3}; 0;-2}`
