2023 AMC 10A Problems/Problem 21
Contents
Problem
Let be the unique polynomial of minimal degree with the following properties:
- has a leading coefficient ,
- is a root of ,
- is a root of ,
- is a root of , and
- is a root of .
The roots of are integers, with one exception. The root that is not an integer and can be written as , where and are relatively prime integers. What is ?
Solution 1
From the problem statement, we know , and . Therefore, we know that , , and are roots. So, we can factor as , where is the unknown root. Since , we plug in which gives , therefore . Therefore, our answer is
~aiden22gao
~cosinesine
~walmartbrian
~sravya_m18
~ESAOPS
~Andrew_Lu
Solution 2
We proceed similarly to solution one. We get that . Expanding, we get that . We know that , so the sum of the coefficients of the cubic expression is equal to one. Thus . Solving for , we get that . Therefore, our answer is
~Aopsthedude
Video Solution 1 by OmegaLearn
Video Solution
Video Solution by CosineMethod
https://www.youtube.com/watch?v=HEqewKGKrFE
Video Solution 2
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution 3 by EpicBird08
https://www.youtube.com/watch?v=D4GWjJmpqEU&t=25s
Video Solution 4 by MegaMath
https://www.youtube.com/watch?v=4Hwt3f1bi1c&t=1s
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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.