2009 AMC 12A Problems/Problem 21

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 $p(x^4) = (x^4 - (2009 + 9002\pi i))(x^4 - 2009)(x^4 - 9002)=x^{12}+ax^8+bx^4+c$.

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 $\pm \sqrt [4]{2009}$, and there are two complex roots.
  • $x^4 - 9002 = 0$: The real roots are $\pm \sqrt [4]{9002}$, and there are two complex roots.

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

Alternative Thinking

Consider the graph of $x^4$. It is similar to a parabola, but with a wider "base". First examine $x^4-2009$ and $x^4-9002$. Clearly they are just being translated down some large amount. This will result in the $x$-axis being crossed twice, indicating $2$ real zeroes. From the Fundamental Theorem of Algebra we know that a polynomial must have exactly as many roots as its highest degree, so we are left with $4-2$ or $2$ nonreal roots for each of the graphs. For the graph of $x^4-(2009+9002\pi i)$, it's not even possible to graph it on the Cartesian plane, so all $4$ roots will be nonreal. This is $2+2+4 = 8$ total nonreal roots $\Rightarrow \boxed{\text{C}}$.

See also

2009 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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