AMC10 2022 A

AMC10 2022 A · Q16

AMC10 2022 A · Q16. It mainly tests Vieta / quadratic relationships (basic), Word problems (algebra).

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?

多项式 $10x^3 - 39x^2 + 29x - 6$ 的根是长方体盒子的长、高、宽。将原盒子每个边都延长 $2$ 单位，形成一个新的长方体盒子。新盒子的体积是多少？

(A) frac{24}{5} frac{24}{5}

(B) frac{42}{5} frac{42}{5}

(D) 30 30

(E) 48 48

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let $a$, $b$, $c$ be the three roots of the polynomial. The lengthened prism's volume is \[V = (a+2)(b+2)(c+2) = abc+2ac+2ab+2bc+4a+4b+4c+8 = abc + 2(ab+ac+bc) + 4(a+b+c) + 8.\] By Vieta's formulas, we know that a cubic polynomial $Ax^3+Bx^2+Cx+D$ with roots $a$, $b$, $c$ satisfies: \begin{alignat*}{8} a+b+c &= -\frac{B}{A} &&= \frac{39}{10}, \\ ab+ac+bc &= \hspace{2mm}\frac{C}{A} &&= \frac{29}{10}, \\ abc &= -\frac{D}{A} &&= \frac{6}{10}. \end{alignat*} We can substitute these into the expression, obtaining \[V = \frac{6}{10} + 2\left(\frac{29}{10}\right) + 4\left(\frac{39}{10}\right) + 8 = \boxed{\textbf{(D) } 30}.\]

设 $a$、$b$、$c$ 为多项式的三个根。延长后的棱柱体积为 \[V = (a+2)(b+2)(c+2) = abc+2ac+2ab+2bc+4a+4b+4c+8 = abc + 2(ab+ac+bc) + 4(a+b+c) + 8.\] 根据 Vieta 公式，对于三元多项式 $Ax^3+Bx^2+Cx+D$ 其根 $a$、$b$、$c$ 满足： \begin{alignat*}{8} a+b+c &= -\frac{B}{A} &&= \frac{39}{10}, \\ ab+ac+bc &= \hspace{2mm}\frac{C}{A} &&= \frac{29}{10}, \\ abc &= -\frac{D}{A} &&= \frac{6}{10}. \end{alignat*} 将这些代入表达式，得到 \[V = \frac{6}{10} + 2\left(\frac{29}{10}\right) + 4\left(\frac{39}{10}\right) + 8 = \boxed{\textbf{(D) } 30}.\]

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