`n^3+6n^2-7n+4`
`=n^3-2n^2+8n^2-16n+9n-18+22`
`=(n^3-2n^2)+(8n^2-16n)+(9n-18)+22`
`= n^2(n-2)+8n(n-2)+9(n-2)+22`
`=(n-2)(n^2+8n+9)+22`
Có: `(n-2)⋮(n-2)⇒(n-2)(n^2+8n+9)⋮(n-2)`
Để `[(n-2)(n^2+8n+9)+22]⋮(n-2)⇔22⋮(n-2)`
Vì `n` là số nguyên nên:
`⇒(n-2)∈Ư(22)={1; -1; 2; -2; 11; -11; 22; -22}.`
`⇒n∈{3; 1; 4; 0; 13; -9; 24; -20}.`
Vậy `n∈{3; 1; 4; 0; 13; -9; 24; -20}⇒n^3+6n^2-7n+4⋮(n-2).`