2009 AMC 12A Problem 21

Problem 21 of 25HarderAlgebra

Let

p(x) = x^3 + ax^2 + bx + c,

where

a,

b,

and

c

are complex numbers. Suppose that

\begin{gathered} p(2009 + 9002\pi i) \\ = p(2009) \\ = p(9002) = 0. \end{gathered}

What is the number of nonreal zeros of

x^{12} + ax^8 + bx^4 + c?

Show solution

Solution

Since

x^{12} + ax^8 + bx^4 + c = p(x^4),

a value is a zero exactly when

x^4

equals one of the roots of

p,

namely

2009 + 9002\pi i,

2009,

9002.

The equation

x^4 = 2009 + 9002\pi i

has four distinct nonreal roots. Each of

x^4 = 2009

and

x^4 = 9002

has two real roots and two nonreal roots. So the nonreal zeros number

4 + 2 + 2 = 8.

Thus, the correct answer is C.