2009 AMC 12A Problems/Problem 21
Problem
Let , where , , and are complex numbers. Suppose that
What is the number of nonreal zeros of ?
Solution
From the three zeroes, we have .
Then .
Let's do each factor case by case:
- : Clearly, all the fourth roots are going to be complex.
- : The real roots are , and there are two complex roots.
- : The real roots are , and there are two complex roots.
So the answer is .
Alternative Thinking
Consider the graph of . It is similar to a parabola, but with a wider "base". First examine and . Clearly they are just being translated down some large amount. This will result in the -axis being crossed twice, indicating 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 or nonreal roots for each of the graphs. For the graph of , it's not even possible to graph it on the Cartesian plane, so all roots will be nonreal. This is total nonreal roots .
See also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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All AMC 12 Problems and Solutions |
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