2003 AIME II Problems/Problem 9
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
Let then by Vieta's Formula we have
By Newton's Sums we have
Applying the formula couples of times yields .
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 | ||
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