Problem:
Let p(x)=x3+ax2+bx+c, where a,b, and c are complex numbers. Suppose that
p(2009+9002Ï€i)=p(2009)=p(9002)=0.
What is the number of nonreal zeros of x12+ax8+bx4+c?
Answer Choices:
A. 4
B. 6
C. 8
D. 10
E. 12
Solution:
Because x12+ax8+bx4+c=p(x4), the value of this polynomial is 0 if and only if
x4=2009+9002Ï€i or x4=2009 or x4=9002
The first of these three equations has four distinct nonreal solutions, and the second and third each have two distinct nonreal solutions. Thus p(x4)=x12+ ax8+bx4+c has 8​ distinct nonreal zeros.
The problems on this page are the property of the MAA's American Mathematics Competitions