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2022 AMC 12A Problems/Problem 15

Revision as of 19:40, 11 November 2022 by Phuang1024 (talk | contribs) (Created page with "==Problem== The roots of the polynomial <math>10x^3 - 39x^2 + 29x - 6</math> are the height, length, and width of a rectangular box (right rectangular prism. A new rectangu...")
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Problem

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?

Solution

Let $a$, $b$, $c$ be the three roots of the polynomial. The lenghtened prism's area 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:

$abc = \frac{-D}{A} = \frac{6}{10}$

$ab+ac+bc = \frac{C}{A} = \frac{29}{10}$

$a+b+c = \frac{-B}{A} = \frac{39}{10}$.

We can substitute these into the expression, obtaining $V = \frac{6}{10} + 2(\frac{29}{10}) + 4(\frac{39}{10}) + 8 = (D) 30$

- phuang1024

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