Difference between revisions of "2003 AIME II Problems/Problem 9"
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<math>P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{6}</math> | <math>P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{6}</math> | ||
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+ | == Solution 2 == | ||
== See also == | == See also == |
Revision as of 13:47, 22 December 2019
Contents
Problem
Consider the polynomials and
Given that
and
are the roots of
find
Solution
When we use long division to divide by
, the remainder is
.
So, since is a root,
.
Now this also follows for all roots of
Now
Now by Vieta's we know that ,
so by Newton Sums we can find
So finally
Solution 2
See also
2003 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.