\(x^3+x^2+4=0\Leftrightarrow x^3+2x^2-x^2-2x+2x+4=0\Leftrightarrow x^2\left(x+2\right)-x\left(x+2\right)+2\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-x+2\right)=0\)
vì x^2 -x +2 >0 nên \(x+2=0\Rightarrow x=-2\)
Vậy nghiệm phương trình là x=-2
x^3+x^2+4=0
x^3+2x^2-x^2-2x+2x+4=0
x^2(x+2)-x(x+2)+2(x+2)=0
(x+2)(x^2-x+2)=0
\(\Rightarrow\orbr{\begin{cases}x+2=0\\x^2-x+2=0\left(voli\right)\end{cases}\Rightarrow\hept{x=-2}}\)
Vay pt co tap hop nghiem la : \(S=\left\{-2\right\}\)