Difference between revisions of "2007 AMC 12A Problems/Problem 18"
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==Solution== | ==Solution== | ||
| − | A fourth degree polynomial has four [[root]]s. Since the coefficients are real, the remaining two roots must be the [[complex conjugate]]s of the two given roots, namely <math>2 | + | A fourth degree polynomial has four [[root]]s. Since the coefficients are real, the remaining two roots must be the [[complex conjugate]]s of the two given roots, namely <math>2-i,-2i</math>. Now we work backwards for the polynomial: |
<div style="text-align:center;"><math>(x-(2+i))(x-(2-i))(x-2i)(x+2i) = 0</math><br /> | <div style="text-align:center;"><math>(x-(2+i))(x-(2-i))(x-2i)(x+2i) = 0</math><br /> | ||
Revision as of 23:24, 25 December 2008
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:
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 | |
