The coefficients of the complex polynomial z^{4} + az^{3} + bz^{2} + cz + d satisfy a^{2}d = c^{2} ≠ 0. Show that the ratio of two of the roots equals the ratio of the other two.

**Solution**

We start with a straight slog. Let the roots be p, q, r, s. We have a^{2}d - c^{2} = (p + q + r + s)^{2}pqrs - (pqr + pqs + prs + qrs)^{2}. The terms like 2p^{2}q^{2}rs all cancel, leaving p^{3}qrs + pq^{3}rs + pqr^{3}s + pqrs^{3} - p^{2}q^{2}r^{2} - p^{2}q^{2}s^{2} - p^{2}r^{2}s^{2} - q^{2}r^{2}s^{2}.

The trick now is to factorise this. We might suspect that pq - rs is a factor. But in that case pr - qs and ps - qr would presumably also be factors. (pq - rs)(pr - qs)(ps - qr) has degree 6, as required. It also has the correct number of terms (8). So we try multiplying it out and find that it is the same.

So (pq - rs)(pr - qs)(ps - qr) = 0. But that means that at least one factor must be zero, which gives the result. Note that the only reason for giving us that a^{2}d is non-zero, is because that implies that none of the roots are zero (their product d is non-zero) and so having got pq = rs, we can divide to get p/r = s/q.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002