Tìm x biết \(x^3+2x^2+4x+8=0\)
Hướng dẫn giải
\(x^3+2x^2+4x+8=0\)
\(\Leftrightarrow\left(x^3+8\right)+\left(2x^2+4x\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)+2x\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4+2x\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+4\right)=0\Leftrightarrow\orbr{\begin{cases}x+2=0\\x^2+4=0\end{cases}}\)
Do \(x^2+4>0\forall x\Rightarrow x+2=0\Rightarrow x=-2.\)