a: Thay x=-2 vào pt, ta được:
\(-8+4a+2a-4=0\)
=>6a-12=0
hay a=2
Vậy: Pt là \(x^3+2x^2-2x-4=0\)
b: \(x^3+2x^2-2x-4=0\)
\(\Leftrightarrow x^2\left(x+2\right)-2\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2\right)=0\)
hay \(x\in\left\{-2;\sqrt{2};-\sqrt{2}\right\}\)