\(x^2+2x+4x+8\)
\(=x\left(x+2\right)+4\left(x+2\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
Ta có: \(\left(x+2\right)\left(x+4\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2< 0\\x+4>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2>0\\x+4< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< -2\\x>-4\end{matrix}\right.\\\left\{{}\begin{matrix}x>-2\\x< -4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2>x>-4\\-2< x< -4\text{(vô lí)}\end{matrix}\right.\)
Vậy để biểu thức âm thì -2 > x > -4.
\(x^2+2x+4x+8< 0\)
\(\Rightarrow x\left(x+2\right)+4\left(x+2\right)< 0\)
\(\Rightarrow\left(x+4\right)\left(x+2\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+4< 0\Rightarrow x< -4\\x+2>0\Rightarrow x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x+4>0\Rightarrow x>-4\\x+2< 0\Rightarrow x< -2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-4< x< -2\)
