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;