Since f(x) has real coefficients, complex roots occur in conjugate pairs. So, since 4+4i is a root, so is 4-4i.
f(x) = A(x-(-1))(x-4)(x-(4+4i))(x-(4-4i))
= A(x+1)(x-4)[(x-4)-4i][(x-4)+4i]
= A(x2-3x-4)[(x-4)2+16]
= A(x2-3x-4)(x2-8x+32)
Since f(1) = -150, we have A(-6)(25) = -150
A = 1
f(x) = (x2-3x-4)(x2-8x+32)
You can multiply it out, if necessary.