\(n^2-2n-22\) là bội \(n+3\)
\(\Rightarrow n^2-2n-22⋮n+3\)
\(\Rightarrow n^2+3n-5n-22⋮n+3\)
\(\Rightarrow n\left(n+3\right)-5n-22⋮n+3\)
Ta có: \(n\left(n+3\right)⋮n+3\) nên để \(n^2-2n-22⋮n+3\)
thì \(-5n-22⋮n-3\)\(\Rightarrow-5\left(n-3\right)-7⋮n-3\)
Mà \(-5\left(n-3\right)⋮n-3\) suy ra \(-7⋮n-3\)
\(\Rightarrow n-3\inƯ\left(-7\right)=\left\{1;-1;7;-7\right\}\)
\(\Rightarrow n\in\left\{4;2;10;-4\right\}\)