Difference between revisions of "2021 AMC 12A Problems/Problem 12"
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<math>\textbf{(A) }-88 \qquad \textbf{(B) }-80 \qquad \textbf{(C) }-64 \qquad \textbf{(D) }-41\qquad \textbf{(E) }-40</math> | <math>\textbf{(A) }-88 \qquad \textbf{(B) }-80 \qquad \textbf{(C) }-64 \qquad \textbf{(D) }-41\qquad \textbf{(E) }-40</math> | ||
− | ==Solution== | + | ==Solution 1:== |
− | 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 | ||
+ | |||
+ | ==Solution 2:== | ||
+ | |||
+ | Using the same method as solution 1, we find the roots are <math>2, 2, 2, 2, 1, 1</math>. Note that <math>B</math> is the negation of the 3rd symmetric sum of the roots; using casework on the number of 1's in each of the <math>\binom {6}{3}</math> products, we obtain <math>B= - \left(\binom {4}{3} \binom {2}{0} \cdot 2^{3} + \binom {4}{2} \binom{2}{1} \cdot 2^{2} \cdot 1 + \binom {4}{1} \cdot \binom {2}{2} \cdot 2 \right) = -(32+48+8) = \boxed{\textbf{(A)} -88}</math> ~ ike.chen. | ||
==Video Solution by Hawk Math== | ==Video Solution by Hawk Math== |
Revision as of 22:21, 11 February 2021
Contents
[hide]Problem
All the roots of the polynomial are positive integers, possibly repeated. What is the value of ?
Solution 1:
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
Solution 2:
Using the same method as solution 1, we find the roots are . 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 OmegaLern (Using Vieta's Formulas & Combinatorics)
~ pi_is_3.14
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 |
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