Solutions to Problems; Chapter (^6 351)
= b2 +2d+ac.
The sum of all products of three of the zeros is
(uw + 2v)(cw + z”) + (uw + 2z)(cu + v2)
= ucuw + uw(v2 + z2) + 2c(vw + zu) + 2vz(z + v)
= ubc - c(u + w)(v + z) + uw(v2 + z2) + c2 + c(vw + ZU)
- 2db - Suvzw
= abc + c2 + 2bd - (uz + vw)(uv + wz) + uw(v2 + z2) - 2uvzw
= abc + c2 + 2bd - vz(u2 + w2 + 21120)
= ubc + c2 + 2bd - a2d.
The product of the zeros is
(cu + v2)(cw + z2) = c2uw + c(uz2 + WV”) + v2z2
= c2(uw + v + z) - C”(V + z) - c[c(z + v) - VZ(U + w)] + v2z2
= bc2 - acd + d2.
Hence, the required quartic is
x4 - 2bx3 + (2d + b2 + uc)x2 - (2bd - a2d + c2 + abc)x - (d2 + bc2 - acd) = 0.
Remark. Problem 4860 in the Amer. Math. Monthly 66 (1959), 596; 67
(1960), 598 generalizes this problem to polynomials of degree n in which
symmetric functions of degree q of any p of the zeros are taken. Here, n = 4,
p = 3, q = 2.
4.12. a must be a value assumed by -(t4 - 14t3 + 61t2 - 84t) for four
integer values oft (possibly counting repetitions). Now
t4 - 14t3 + 61t2 - 84t = t(t - 3)(t - 4)(t - 7)
so that the left side assumes the same value for k and 7 - k. Suppose that
a is such that k and 7 - k are zeros for some integer k. Then
t4 - 14t3 + 61t2 - 84t + a = [t” - 7t + k(7 - k)][t” - 7t + (k - 3)(k - 4)].
The discriminant of the second factor is 50 - (2k - 7)2, which is square only
for k = 0, 1, 3, 4, 6, 7. Hence a = 0 or a = 36. Indeed, t4 - 14t3 + 61t2 -
84t + 36 = (t - 1)2(t - 6)“.
4.13. Suppose that p(z) is of degree n 1 2. Then p( z)p(-z) - p(z) is of
degree 2n and the coefficient of t2”-l is 0. Hence the sum of the roots of
the equation is 0. It follows that if there are zeros in one half plane, then