2019 AMC 10A Problems/Problem 24
Contents
Problem
Let , , and be the distinct roots of the polynomial . It is given that there exist real numbers , , and such that for all . What is ?
Solution
Solution 1
Multiplying both sides by yields As this is a polynomial identity, and it is true for infinitely many , it must be true for all (since a polynomial with infinitely many roots must in fact be the constant polynomial ). This means we can plug in to find that . Similarly, we can find and . Summing them up, we get that By Vieta's Formulas, we know that and . Thus the answer is .
Note: this process of substituting in the 'forbidden' values in the original identity is a standard technique for partial fraction decomposition, as taught in calculus classes.
See Also
Video Solution: https://www.youtube.com/watch?v=GI5d2ZN8gXY&t=53s
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.