\(\left(x-4\right)\left(x^2+16\right)\left(x^2-16\right)\left(x+1\right)=0\)
\(\Rightarrow\left\{\begin{matrix}x-4=0\\x^2+16=0\\x^2-16=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=4\\x^2=-16\left(loại\right)\\x^2=16\\x=-1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=4\\x=\pm4\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{4;-4;-1\right\}\)