Ta có:
\(3n^3-2n+17=3n^3+6n^2-6n^2-12n+10n+20-3\)
\(=3n^2\left(n+2\right)-6n\left(n+2\right)+10\left(n+2\right)-3\)
\(=\left(3n^2-6n+10\right)\left(n+2\right)-3\)
Để \(3n^3-2n+17⋮\left(n+2\right)\)
Hay \(\left(3n^2-6n+10\right)\left(n+2\right)-3⋮\left(n+2\right)\)
Mà \(\left(3n^2-6n+10\right)\left(n+2\right)⋮\left(n+2\right)\)
Suy ra \(3⋮\left(n+2\right)\)
Vì \(n\in Z\), suy ra \(n+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
TH1: \(n+2=\pm1\Leftrightarrow\left[{}\begin{matrix}n+2=1\\n+2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=-1\\n=-3\end{matrix}\right.\)
TH2:\(n+2=\pm3\Leftrightarrow\left[{}\begin{matrix}n+2=3\\n+2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=1\\n=-5\end{matrix}\right.\)
Vậy \(n\in\left\{-5;-3;-1;1\right\}\)