Difference between revisions of "2021 AMC 12A Problems/Problem 12"
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By Vieta's formulae, 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> | By Vieta's formulae, 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 | ||
+ | |||
+ | ==Video Solution by Hawk Math== | ||
+ | https://www.youtube.com/watch?v=AjQARBvdZ20 | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=11|num-a=13}} | {{AMC12 box|year=2021|ab=A|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:54, 11 February 2021
Problem
All the roots of the polynomial are positive integers, possibly repeated. What is the value of ?
Solution
By Vieta's formulae, 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
Video Solution by Hawk Math
https://www.youtube.com/watch?v=AjQARBvdZ20
See also
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.