For knots in S-3, it is well-known that the Alexander polynomial of a ribbon knot factorizes as f(t)f(t(-1)) for some polynomial f(t). By contrast, the Alexander polynomial of a ribbon 2-knot in S-4 is not even symmetric in general. Via an alternative notion of ribbon 2-knots, we give a topological condition on a 2-knot that implies the factorization of the Alexander polynomial.