Difference between revisions of "2021 AMC 12A Problems/Problem 12"
MRENTHUSIASM (talk | contribs) m (OmegaLern -> OmegaLearn) |
Fnothing4994 (talk | contribs) m (→Solution 1:) |
||
Line 7: | Line 7: | ||
==Solution 1:== | ==Solution 1:== | ||
− | By Vieta's | + | By Vieta's formulas, the sum of the 6 roots is 10 and the product of the 6 roots is 16. By inspection, we see the roots are 1, 1, 2, 2, 2, and 2, so the function is <math>(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)</math>. Therefore, <math>B = -32 - 48 - 8 = \boxed{\textbf{(A)} -88}</math>. |
~JHawk0224 | ~JHawk0224 | ||
Revision as of 14:50, 28 March 2021
- The following problem is from both the 2021 AMC 12A #12 and 2021 AMC 10A #14, so both problems redirect to this page.
Contents
[hide]Problem
All the roots of the polynomial are positive integers, possibly repeated. What is the value of ?
Solution 1:
By Vieta's formulas, the sum of the 6 roots is 10 and the product of the 6 roots is 16. By inspection, we see the roots are 1, 1, 2, 2, 2, and 2, so the function is . Therefore, . ~JHawk0224
Solution 2:
Using the same method as Solution 1, we find that the roots are and . Note that is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the products we obtain ~ ike.chen
Video Solution by Hawk Math
https://www.youtube.com/watch?v=AjQARBvdZ20
Video Solution by OmegaLearn (Using Vieta's Formulas & Combinatorics)
~ pi_is_3.14
Video Solution by Power Of Logic (Using Vieta's Formulas)
Video Solution by TheBeautyofMath
https://youtu.be/t-EEP2V4nAE?t=1080 (for AMC 10A)
https://youtu.be/ySWSHyY9TwI?t=271 (for AMC 12A)
~IceMatrix
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 |
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.