# 2021 AMC 12A Problems/Problem 12

The following problem is from both the 2021 AMC 12A #12 and 2021 AMC 10A #14, so both problems redirect to this page.

## 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 formulas, the sum of the six roots is $10$ and the product of the six 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, calculating just the $z^3$ terms, we get $B = -32 - 48 - 8 = \boxed{\textbf{(A) }{-}88}$.

~JHawk0224

## Solution 2

Using the same method as Solution 1, we find that the roots are $2, 2, 2, 2, 1,$ and $1$. 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 $\binom {6}{3} = 20$ products $r_a \cdot r_b \cdot r_c,$ we obtain $$B= - \left(\binom {4}{3} \binom {2}{0} \cdot 2^{3} + \binom {4}{2} \binom{2}{1} \cdot 2^{2} \cdot 1 + \binom {4}{1} \binom {2}{2} \cdot 2 \right) = -\left(32+48+8 \right) = \boxed{\textbf{(A) }{-}88}.$$ ~ike.chen

## Video Solution (🚀 Just 2 min 🚀)

~Education, the Study of Everything

~ pi_is_3.14

## Video Solution by TheBeautyofMath

https://youtu.be/t-EEP2V4nAE?t=1080 (for AMC 10A)

https://youtu.be/ySWSHyY9TwI?t=271 (for AMC 12A)

~IceMatrix