2009 AMC 12A Problems/Problem 21
Contents
[hide]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 |
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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