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350 Answers to Exercises and Solutions to Problems

Case4.~=3,u+v=3,sothata=9-a/3~u=27/4.Thecubicis

(x - 3)(x2 - (^32) - g/4) = x3 - 6x2 + 27x/4 + 2714.
4.9. Let yk = zk - 1 for each k. Then the yk are the zeros of
(Y + 1)” + a(y + 1)+l+ a”-‘(y + 1) + 1 = yn + (n + a)y”-1 + *

• [n + a(n - 1) + a”-‘]y + (2 + a + an-l)
  and
  n - 3 = c(xk + 2)(xk - l)--’ = X(1+ 3y,-‘) = n + 3Cyil.
  Hence
  -1 = By,-l = -[n + Q(fa - 1) + a”-I](2 + Q + a”--1)-l
  an-’ + a + 2 = a”-’ + a(n - 1) + n
  =R a(2 - n) = n - 272=2 or a= -1.
  The case Q = -1 yields the polynomial x” - x”-l - x + 1 =
  (x - 1)(x”” - 1). B u t in this case, one of the zeros is 1 and the left
  side of the given equation is undefined. Hence a # -1. The case n = 2
  yields the polynomial x2 + 2ux + 1, whose zeros can be verified to satisfy
  the condition, provided a # -1.
  4.10. u, v, w are zeros of t3 - pt - q, where
  p = -(uv + VW + wu) = (1/2)(U2 + v2 + w2)
  and q = UVW. The result follows from adding the equation u”+~ = p&l +
  &L” to the corresponding equations for v and w.
  4.11. Let u = r+s,v=rs,w=p+q,z=pq.Thenu+w=u,vz=d,
  UW+V+Z = 6, uz+vw = c and the zeros of the required quartic are pu+v,
  qu f v, rw + z and SW + z. Note that
  (pu+v)+(qu+v)=uw+2v=b+v-z
  (rw+z)+(sw+z)=uw+2z=b+z-v
  (pu + v)(qu + v) = (uz + vw)u + v2 = cu + v2
  (rw + z)(sw + z) = (uz + vw)w + z2 = cw + z2.
  The sum of the zeros is 2(uw + v + z) = 2b. The sum of the products of
  pairs of zeros is
  (uw+2v)(uw+2z)+c(u+w)+v2+z2 =(uw+v+z)2+2vz+c(u+w)