Dùng hệ số bất định ta có dạng \(\left(x^2+ax+2\right)\left(x^2+bx+2\right)=x^4+\left(a+b\right)x^3+\left(2+2+ab\right)x^2+2\left(a+b\right)x+4\Rightarrow\left\{{}\begin{matrix}a+b=3\\4+ab=0\\2\left(a+b\right)=6\end{matrix}\right.\)
\(\Rightarrow a=4,b=-1\). Vậy \(x^4+3x^3+6x+4=\left(x^2+4x+2\right)\left(x^2-x+2\right)=0\)
\(\Leftrightarrow\left(\left(x+2\right)^2-2\right)\left(x^2-x+2\right)=0\)
\(\Leftrightarrow\left(x+2-\sqrt{2}\right)\left(x+2+\sqrt{2}\right)\left(x^2-x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=-2+\sqrt{2}\\x=-2-\sqrt{2}\end{matrix}\right.\)
