Question

多项式 $\begin{aligned} &z^6-10z^5+Az^4+Bz^3\\ &\quad+Cz^2+Dz+16 \end{aligned}$ 的所有根都是正整数，且可以重复。求 $B$ 的值。

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

$-88$

$-80$

$-64$

$-41$

$-40$

答案：A

知识点：韦达定理多项式

难度评级：1540

小提示：

用 Vieta 公式确定六个正整数根的和与积。

Use Vieta to determine the sum and product of the six positive integer roots

大提示：

找出积为 $16$ 、和为 $10$ 的唯一六个正整数。

Find the only six positive integers with product $16$ and sum $10$

视频讲解：

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文字解答：

由 Vieta 公式，六个根的和为 $10$ ，积为 $16$ 。因为积是 $2$ 的幂，所以每个正整数根都是 $2$ 的幂。把四个因子 $2$ 分配给六个根，当其中四个根为 $2$ 、两个根为 $1$ 时和最小，而这个和恰好已经等于 $10$ 。因此这些根是 $1,1,2,2,2,2\text{。}$

系数 $B$ 是所有三个根乘积之和的相反数。按取零个、一个还是两个等于 $1$ 的根来分类，得 $\begin{aligned} B&=-\left(\binom43 2^3+2\binom42 2^2\right.\\ &\qquad\left.+\binom41 2\right)\\ &=-88 \end{aligned}\text{。}$

所以正确答案是 A。

By Vieta’s formulas, the six roots have sum $10$ and product $16.$ Because the product is a power of $2,$ every positive integer root is a power of $2.$ Distributing the four factors of $2$ among six roots gives the least possible sum when four roots are $2$ and two roots are $1$ ; that sum is already $10.$ Hence the roots are $1,1,2,2,2,2.$

The coefficient $B$ is the negative of the sum of all products of three roots. Choosing zero, one, or two of the two roots equal to $1$ gives $\begin{aligned} B&=-\left(\binom43 2^3+2\binom42 2^2\right.\\ &\qquad\left.+\binom41 2\right)\\ &=-88. \end{aligned}$

Thus, A is the correct answer.

2021 AMC 10A Spring 第 14 题