Ta có \(P\left(x\right)\equiv\left(ax-3x\right)+\left(b+2\right)[modQ\left(x\right)]\)
Mà: \(P\left(x\right)⋮Q\left(x\right)\)
\(\Rightarrow\left\{{}\begin{matrix}ax-3x=0\\b+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=3\\b=-2\end{matrix}\right.\)
Vậy \(a=3;b=-2\)
Chúc bạn học tốt!