2009 AMC 12A Problems/Problem 21

Revision as of 00:04, 14 February 2009 by Xantos C. Guin (talk | contribs) (8 is choice C NOT choice D. Same mistake I made during the test. :))

Problem

Let $p(x) = x^3 + ax^2 + bx + c$, where $a$, $b$, and $c$ are complex numbers. Suppose that

\[p(2009 + 9002\pi i) = p(2009) = p(9002) = 0\]

What is the number of nonreal zeros of $x^{12} + ax^8 + bx^4 + c$?

$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 12$

Solution

From the three zeroes, we have $p(x) = (x - (2009 + 9002\pi i))(x - 2009)(x - 9002)$.

Then $p(x^4) = (x^4 - (2009 + 9002\pi i))(x^4 - 2009)(x^4 - 9002)$.

Let's do each factor case by case:

  • $x^4 - (2009 + 9002\pi i) = 0$: Clearly, all the fourth roots are going to be complex.
  • $x^4 - 2009 = 0$: The real roots are $\pm \sqrt [4]{2009}$, there are two complex roots.
  • $x^4 - 9002 = 0$: Same.

So the answer is $4 + 2 + 2 = 8\ \mathbf{(C)}$.

See also

2009 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AMC 12 Problems and Solutions
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