2007 AMC 12A Problems/Problem 18
Solution
The polynomial has real coefficients, and
What is
Solution
A fourth degree polynomial has four roots. Since the coefficients are real, the remaining two roots must be the complex conjugates of the two given roots, namely . Now we work backwards for the polynomial:
![$(x-(2+i))(x-(2-i))(x-2i)(x+2i) = 0$](http://latex.artofproblemsolving.com/4/a/2/4a2892d359087ed1923e128f7eb8ca5a29178df5.png)
![$x^4 - 4x^3 + 9x^2 - 16x + 20 = 0$](http://latex.artofproblemsolving.com/d/0/a/d0a46838b90acb80b3df33c37320d99d5ad1d974.png)
Thus our answer is .
See also
2007 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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All AMC 12 Problems and Solutions |