a question about Pure maths

By remainder theorem :
f(k) = (k-a)(k-b)(k-4)-4 = -2
(k-a)(k-b)(k-4) = 2
Since a and b are integers , k is an integer , so
k-a , k-b , k-4 are integers ;
Since a<b<4 , so k-a > k-b > k-4 .
Only one possible :
k-a = 1 > k-b = - 1 > k-4 = - 2
k = 2 , b = 3 , a = 1
f(x) = (x-1)(x-3)(x-4)-4 divided by x-2
the remainder = f(2) = (2-1)(2-3)(2-4)-4 = - 2 is correct.