Since f(-2+i) = 0, f(-2-i) = 0.
By the Factor Theorem, x + 6, x - (-2+i). and x - (-2-i) are factors of f(x).
f(x) = (x + 6)[x - (-2+i)][x - (-2-i)]
= (x + 6)[(x+2) - i][(x+2) + i]
= (x + 6)[(x+2)2 - i2]
= (x + 6)(x2 + 4x + 5) = x3 + 4x2 + 5x + 6x2 + 24x + 30
f(x) = x3 + 10x2 + 29x + 30