Tìm x để x² + x + 1 chia hết cho x – 1.

Ta có:

A = x² + x + 1 = x² − x + 2x − 2 + 3

= x(x − 1) + 2(x − 1) + 3

= (x + 2)(x − 1) + 3.

Vì (x + 2)(x − 1) ⋮ (x − 1) nên để x² + x + 1 ⋮ x − 1 thì 3 ⋮ (x − 1).

Vậy x − 1 $\in$ Ư(3) = {±1; ±3} => x = {−2; 0; 2; 4}.