2001 AMC 12 Problems/Problem 23
Contents
Problem
A polynomial of degree four with leading coefficient 1 and integer coefficients has two zeros, both of which are integers. Which of the following can also be a zero of the polynomial?
Solution
Let the polynomial be and let the two integer zeros be and . We can then write for some integers and .
If a complex number with is a root of , it must be the root of , and the other root of must be .
We can then write .
We can now examine each of the five given complex numbers, and find the one for which the values and are integers. This is , for which we have and .
(As an example, the polynomial has zeroes , , and .)
Solution 2
By Vieta, we know that the product of all four zeros of the polynomial equals the constant at the end of the polynomial. We also know that the two imaginary roots are a conjugate pair (I.E if one is a+bi, the other is a-bi). So the two imaginary roots must multiply to give you an integer. Taking the 5 answers into hand, we find that is our only integer giving solution.
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 | |
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