2021 AMC 10A Problem 14

1. If the six roots are $r_1, r_2, \ldots, r_6$, the polynomial is $(z - r_1)(z - r_2)\cdots(z - r_6)$. Multiplying this out tells us two things:
- the $z^5$ term is $-(r_1 + r_2 + \cdots + r_6)\,z^5$, and it must equal $-10z^5$, so the roots add to $10$;
- the constant term is $(-r_1)(-r_2)\cdots(-r_6)$; the six minus signs cancel, so the roots multiply to $16$.
2. Positive whole numbers whose product is $16$ must each divide $16$, so every root is $1$, $2$, $4$, $8$ or $16$.
3. Could a root be $4$ or more? Then the other roots multiply to $4$ or less, and the smallest possible totals are $4 + 2 + 2 + 1 + 1 + 1 = 11$ and $4 + 4 + 1 + 1 + 1 + 1 = 12$. Both are more than $10$, and using $8$ or $16$ is even worse. So every root is $1$ or $2$.
4. To multiply to $16 = 2 \times 2 \times 2 \times 2$ we need exactly four $2$s, and the other two roots are $1$. Check the sum: $2 + 2 + 2 + 2 + 1 + 1 = 10$.
5. So the polynomial is $(z - 2)^4 (z - 1)^2$, where
$$(z - 2)^4 = z^4 - 8z^3 + 24z^2 - 32z + 16, \qquad (z - 1)^2 = z^2 - 2z + 1.$$
6. $B$ is the number in front of $z^3$. A $z^3$ term comes from a term of the first bracket times a term of the second whose powers add to $3$:
- $-8z^3$ times $1$: gives $-8$;
- $24z^2$ times $-2z$: gives $-48$;
- $-32z$ times $z^2$: gives $-32$.
7. Adding: $B = -8 - 48 - 32 = -88$.

The answer is $\boxed{\textbf{(A)}\ -88}$.

The first and last numbers of a polynomial tell you the sum and the product of its roots. With the extra fact that the roots are positive whole numbers, those two facts are enough to find every root. After that, any coefficient is just careful multiplication; keep track of the minus signs.

$B$ is minus the total of all products of three roots chosen from $2, 2, 2, 2, 1, 1$:
- three $2$s: $4$ ways, each product $8$, total $32$;
- two $2$s and one $1$: $6 \times 2 = 12$ ways, each product $4$, total $48$;
- one $2$ and both $1$s: $4$ ways, each product $2$, total $8$.

The total is $32 + 48 + 8 = 88$, so $B = -88$.

By Vieta, the six roots sum to $10$ and multiply to $16$. Positive integers with product $16 = 2^4$ are all powers of $2$, so the roots come from $\{1, 2, 4, 8, 16\}$.

Distribute four factors of $2$ among six roots so the sum is $10$. Any root of $4$ or more overshoots: the smallest such sums are $4 + 2 + 2 + 1 + 1 + 1 = 11$, $4 + 4 + 1 + 1 + 1 + 1 = 12$, $8 + 2 + 1 + 1 + 1 + 1 = 14$, $16 + 1 + 1 + 1 + 1 + 1 = 21$. So every root is $1$ or $2$. With four $2$s and two $1$s the sum is $8 + 2 = 10$. The polynomial is
$$(z-2)^4 (z-1)^2 = (z^4 - 8z^3 + 24z^2 - 32z + 16)(z^2 - 2z + 1).$$

The $z^3$ coefficient collects the products whose degrees add to $3$:
$$(-8)(1) + (24)(-2) + (-32)(1) = -8 - 48 - 32 = -88.$$

The answer is $\boxed{\textbf{(A)}\ -88}$.