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2022 AMC 12A Problems/Problem 15

Revision as of 22:28, 11 November 2022 by Phuang1024 (talk | contribs) (→‎Solution 1 (Vieta's Formulas): Formatting.)

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 1 (Vieta's Formulas)

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

$V = \frac{6}{10} + 2(\frac{29}{10}) + 4(\frac{39}{10}) + 8 = \boxed{(D) 30}$

- phuang1024

Solution 2 (Rational Root Theorem bash)

We can find the roots of the cubic using the Rational Root Theorem, which tells us that the rational roots of the cubic must be in the form $\frac{p}{q}$, where $p$ is a factor of the constant $(-6)$ and $q$ is a factor of the leading coefficient $(10)$. Therefore, $p$ is $\pm (1, 2, 3, 6)$ and q is $\pm (1, 2, 5, 10).$

Doing Synthetic Division, we find that $3$ is a root of the cubic: \[\begin{array}{c|rrrr}&10&-39&29&-6\\3&&30&-27&6\\\hline\\&10&-9&2&0\\\end{array}.\]

Then, we have a quadratic $10x^2-9x+2.$ Using the Quadratic Formula, we can find the other two roots: \[x=\frac{9 \pm \sqrt{(-9)^2-4(10)(2)}}{2 \cdot 10},\] which simplifies to $x=\frac{1}{2}, \frac{2}{5}.$

To find the new volume, we add $2$ to each of the roots we found: \[(3+2)\cdot(\frac{1}{2}+2)\cdot(\frac{2}{5}+2).\] Simplifying, we find that the new volume is $\boxed{30}.$

-MathWizard09

Video Solution

https://youtu.be/08YkinzFcCc

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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