Difference between revisions of "2021 AMC 12A Problems/Problem 12"

Revision as of 22:22, 11 February 2021

Problem

All the roots of the polynomial $z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16$ are positive integers, possibly repeated. What is the value of $B$ ?

$\textbf{(A) }-88 \qquad \textbf{(B) }-80 \qquad \textbf{(C) }-64 \qquad \textbf{(D) }-41\qquad \textbf{(E) }-40$

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 $(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)$ . Therefore, $B = -32 - 48 - 8 = \boxed{\textbf{(A)} -88}$ . ~JHawk0224

Solution 2:

Using the same method as Solution 1, we find that the roots are $2, 2, 2, 2, 1, 1$ . Note that $B$ is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the $\binom {6}{3}$ products, we obtain $B= - \left(\binom {4}{3} \binom {2}{0} \cdot 2^{3} + \binom {4}{2} \binom{2}{1} \cdot 2^{2} \cdot 1 + \binom {4}{1} \cdot \binom {2}{2} \cdot 2 \right) = -(32+48+8) = \boxed{\textbf{(A)} -88}$ ~ ike.chen

Video Solution by Hawk Math

https://www.youtube.com/watch?v=AjQARBvdZ20

Video Solution by OmegaLern (Using Vieta's Formulas & Combinatorics)

https://youtu.be/5U4MJTo3F5M

~ pi_is_3.14

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.

@@ Line 10: / Line 10: @@
 ==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.
+Using the same method as Solution 1, we find that 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==

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