Problem:
A polynomial
P(x)=c2004βx2004+c2003βx2003+β―+c1βx+c0β
has real coefficients with c2004βξ =0 and 2004 distinct complex zeros zkβ=akβ+bkβi, 1β€kβ€2004 with akβ and bkβ real, a1β=b1β=0, and
k=1β2004βakβ=k=1β2004βbkβ
Which of the following quantities can be a nonzero number?
Answer Choices:
A. c0β
B. c2003β
C. b2βb3ββ¦b2004β
D. βk=12004βakβ
E. βk=12004βckβ
Solution:
Since z1β=0, it follows that c0β=P(0)=0. The nonreal zeros of P must occur in conjugate pairs, so βk=12004βbkβ=0 and βk=12004βakβ=0 also. The coefficient c2003β is the sum of the zeros of P, which is
k=1β2004βzkβ=k=1β2004βakβ+ik=1β2004βbkβ=0
Finally, since the degree of P is even, at least one of z2β,β¦,z2004β must be real, so at least one of b2β,β¦,b2004β is 0 and consequently b2βb3ββ¦b2004ββ=0. Thus the quantities in (A), (B), (C), and (D) must all be 0 .
Note that the polynomial
P(x)=x(x2β)(x3β)(x2003β)x+k=2β2003βk)
satis es the given conditions, and βk=12004βckβ=P(1)ξ =0.
The problems on this page are the property of the MAA's American Mathematics Competitions