\(x^4+5x^3-8x-40=0\)
\(\Leftrightarrow x^3\left(x+5\right)-8\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^3-8\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\left(x+5\right)=0\)
Ta có : \(x^2+2x+4=x^2+2x+1+3=\left(x+1\right)^2+3\ge3\)\(\Rightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
\(x^4+5x^3-8x-40=0\)
\(\Leftrightarrow x^4+3x^3-10x^2+2x^3+6x^2-20x+4x^2+12x-40=0\)
\(\Leftrightarrow x^2\left(x^2+3x-10\right)+2x\left(x^2+3x-10\right)+4\left(x^2+3x-10\right)=0\)
\(\Leftrightarrow\left(x^2+3x-10\right)\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow\left(x^2-2x+5x-10\right)\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow\left[x\left(x-2\right)+5\left(x-2\right)\right]\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+5\right)\left(x^2+2x+4\right)=0\)
Dễ thấy: \(x^2+2x+4=x^2+2x+1+3=\left(x+1\right)^2+3>0\) (vô nghiệm)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
