\(x^3+x^2+4=0\)
\(\Leftrightarrow\left(x^3-4x\right)+\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow x\left(x^2-4\right)+\left(x+2\right)^2=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)+\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-x+2\right)=0\)
\(\Leftrightarrow x+2=0\) (Vì: \(x^2-x+2>0\) )
\(\Leftrightarrow x=-2\)
\(x^3+x^2+4=0\)
\(\Leftrightarrow x^3+2x^2-x^2-2x+2x+4=0\)
\(\Leftrightarrow x^2\left(x+2\right)-x\left(x+2\right)+2\left(x+2\right)\)
\(\Leftrightarrow\left(x^2-x+2\right)\left(x+2\right)\)
Vì \(x^2-x+2=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}>0\)
\(\Rightarrow x+2=0\)
\(\Rightarrow x=-2\)
Vậy phương tình có nghiệm là \(-2\)