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 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 $2, 2, 2, 2, 1,$ 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 by Hawk Math

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

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

https://youtu.be/5U4MJTo3F5M

~ pi_is_3.14


Video Solution by Power Of Logic (Using Vieta's Formulas)

https://youtu.be/rl6QtVnIbdU

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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS