Question

多项式 $10x^3-39x^2+29x-6$ 的根分别是一个长方体的高、长、宽。将原长方体的每条棱都增加 $2$ 个单位，形成一个新的长方体。新长方体的体积是多少？

The roots of the polynomial $10x^3-39x^2+29x-6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?

$\dfrac{24}{5}$

$\dfrac{42}{5}$

$\dfrac{81}{5}$

$30$

$48$

答案：D

知识点：韦达定理体积

难度评级：1630

解答：

设根为 $r,s,t$ 。由韦达定理， $r+s+t=\dfrac{39}{10}$ ， $rs+rt+st=\dfrac{29}{10}$ ，且 $rst=\dfrac{6}{10}=\dfrac35$ 。

新体积为 $\begin{gathered} (r+2)(s+2)(t+2) \\ =rst+2(rs+rt+st) \\ \quad {}+4(r+s+t)+8 \\ =\frac35+\frac{58}{10} \\ \quad {}+\frac{156}{10}+8 \\ =30. \end{gathered}$

因此，正确答案是 D。

Let the roots be $r,s,t.$ By Vieta's formulas, $r+s+t=\dfrac{39}{10},$ $rs+rt+st=\dfrac{29}{10},$ and $rst=\dfrac{6}{10}=\dfrac35.$

The new volume is $\begin{gathered} (r+2)(s+2)(t+2) \\ =rst+2(rs+rt+st) \\ \quad {}+4(r+s+t)+8 \\ =\frac35+\frac{58}{10} \\ \quad {}+\frac{156}{10}+8 \\ =30. \end{gathered}$

Thus, the correct answer is D.

2022 AMC 12A 第 15 题

答案：D

难度评级：1630

解答：

其他年份的第 15 题