Question

多项式 $10x^3 - 39x^2 + 29x - 6$ 的三个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：韦达定理多项式体积

难度评级：1420

解答：

设原长方体的三个尺寸为 $h, l$ 和 $w$ 。新长方体的体积为展开得再由韦达定理，且 $(h + 2)(l + 2)(w + 2).$ $hlw + 2(hl + hw + lw)$ $+ 4(l + h + w) + 8.$ $hlw = -\dfrac{D}{A} = \dfrac{3}{5},$ $hl + hw + lw = \dfrac{C}{A} = \dfrac{29}{10},$ $l + h + w = -\dfrac{B}{A} = \dfrac{39}{10}.$

把这些值代入展开式，得到 $\dfrac{3}{5} + 2 \cdot \dfrac{29}{10} + 4 \cdot \dfrac{39}{10} + 8 = 30.$

所以正确答案是 D。

Let $h, l,$ and $w$ be the dimensions of the old box. Then the volume of the new box is $(h + 2)(l + 2)(w + 2).$ Expanding, we get $hlw + 2(hl + hw + lw)$ $+ 4(l + h + w) + 8.$ We can use Vieta's formulas to find the terms in this expression. We get that $hlw = -\dfrac{D}{A} = \dfrac{3}{5},$ $hl + hw + lw = \dfrac{C}{A} = \dfrac{29}{10},$ and $l + h + w = -\dfrac{B}{A} = \dfrac{39}{10}.$

Plugging these values into the expression, we get $\dfrac{3}{5} + 2 \cdot \dfrac{29}{10} + 4 \cdot \dfrac{39}{10} + 8 = 30.$

Thus, D is the correct answer.

2022 AMC 10A 第 16 题

答案：D

难度评级：1420

解答：

其他年份的第 16 题