Question

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

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

视频讲解：

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文字解答：

由 Vieta 公式，六个正整数根的和为 $10$ 。积为 $16$ 。 $2$ ， $2$ 。 $2$ $2$ $1$ $10$ 。 $1,1,2,2,2,2.$

由于所有根都是正整数，唯一可能的根为 $B$ $1$ $\begin{aligned} B&=-\left(\binom43 2^3+2\binom42 2^2\right.\\ &\qquad\left.+\binom41 2\right)\\ &=-88. \end{aligned}$

因此多项式为只计算项，可得所以正确答案是 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

答案：A

难度评级：1540

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