Ta có:

`x^3 - 3x^2 - 2x + 5`

`= x^3 + x^2 - 4x^2 - 4x + 2x + 2 + 3`

`= x^2(x+1) - 4(x+1) + 2(x+1) + 3`

`= (x+1)(x^2 - 4 + 2) + 3`

Với `x in ZZ`, `(x + 1)(x^2-4 + 2) vdots x + 1 => 3 vdots x + 1`

`=> x + 1 in Ư(3) = {-3;-1;1;3}`

`=> x in {-4;-2;0;2}`

Vậy `x in {-4;-2;0;2}`

`\ttcolor{cyan}{#PUyen}`