a)\(G\left(x\right)=\left(x+2\right)\left(x-1\right)\left(x+\frac{1}{3}\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=-2\\x=1\\x=\frac{-1}{3}\end{matrix}\right.\)
b)\(H\left(x\right)=3x^3+3x^2+x^2+x+x+1=0\)
\(H\left(x\right)=\left(x+1\right)\left(3x^2+x+1\right)=0\)
Vì \(3x^2+x+1=0\) vô nghiệm nên x=-1.