In this case we need to test the remainder r = -1.

R	= f(r) = f(-1) = (-1)3 + 2(-1)2 - 5(-1) - 6 = -1 + 2 + 5 - 6 = 0

Therefore, since f(-1) = 0, we conclude that (x + 1) is a factor of f(x).