The coefficients of the complex polynomial z4 + az3 + bz2 + cz + d satisfy a2d = c2 ≠ 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 a2d - c2 = (p + q + r + s)2pqrs - (pqr + pqs + prs + qrs)2. The terms like 2p2q2rs all cancel, leaving p3qrs + pq3rs + pqr3s + pqrs3 - p2q2r2 - p2q2s2 - p2r2s2 - q2r2s2.
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 a2d 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