  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 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.
