2007 AMC 12A Problems/Problem 18
Problem
The polynomial has real coefficients, and
What is
Solution
A fourth degree polynomial has four roots. Since the coefficients are real(meaning that complex roots come in conjugate pairs), 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 |
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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