Ta có : \(x^3+x^2+4=0\)
\(\Leftrightarrow x^3+2x^2-x^2-2x+2x+4=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-x+2\right)=0\)
Vì \(x^2-x+2=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\)
Nên \(x+2=0\Leftrightarrow x=-2\)
Vậy \(x=-2\)
Ta có: \(x^3+x^2+4=0\)
\(\Leftrightarrow x^3+2x^2-x^2-2x+2x+4=0\)
\(\Leftrightarrow x^2\left(x+2\right)-x\left(x+2\right)+2\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-x+2\right)=0\)mà \(x^2-x+2>0\forall x\)
nên x+2=0
hay x=-2
Vậy: S={-2}