Найдите все целые k при которых (6k^2+k-27) / (3k+2) - целое число.

(6k² + k - 27) / (3k+2) = 2k - 1 - 25 / (3k + 2) ∈ Z если

3k + 2 = ±1; ±5; ±25

1) 3k+2 = 1

3k = - 1

k ∉Z

2) 3k+2 = - 1

3k = - 3

k = - 1

3) 3k + 2 = 5

3k = 3

k = 1

4) 3k + 2 = - 5

3k = - 7

k∉Z

5) 3k + 2 = 25

3k = 23

k∉Z

6) 3k + 2 = - 25

3k = - 27

k = - 9

Ответ: - 9; - 1; 1;