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Difference between revisions of "2022 AMC 10A Problems/Problem 16"

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#redirect [[2022 AMC 12A Problems/Problem 15]]
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==Problem==
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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 rectangular box is formed by lengthening each edge of the original box by 2
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units. What is the volume of the new box?
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<math>\textbf{(A) } \frac{24}{5} \qquad \textbf{(B) } \frac{42}{5} \qquad \textbf{(C) } \frac{81}{5} \qquad \textbf{(D) } 30 \qquad \textbf{(E) } 48</math>
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==Solution 1 (Vieta's Formulas)==
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Let <math>a</math>, <math>b</math>, <math>c</math> be the three roots of the polynomial. The lenghtened prism's area is
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<math>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</math>.
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By vieta's formulas, we know that:
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<math>abc = \frac{-D}{A} = \frac{6}{10}</math>
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<math>ab+ac+bc = \frac{C}{A} = \frac{29}{10}</math>
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<math>a+b+c = \frac{-B}{A} = \frac{39}{10}</math>.
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We can substitute these into the expression, obtaining
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<math>V = \frac{6}{10} + 2(\frac{29}{10}) + 4(\frac{39}{10}) + 8 = \boxed{\textbf{(D) } 30}</math>
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- phuang1024
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== Solution 2 (Rational Root Theorem bash) ==
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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 <math>\frac{p}{q}</math>, where <math>p</math> is a factor of the constant <math>(-6)</math> and <math>q</math> is a factor of the leading coefficient <math>(10)</math>. Therefore, <math>p</math> is <math>\pm (1, 2, 3, 6)</math> and q is <math>\pm (1, 2, 5, 10).</math>
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Doing Synthetic Division, we find that <math>3</math> is a root of the cubic:
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<cmath>\begin{array}{c|rrrr}&10&-39&29&-6\\3&&30&-27&6\\\hline\\&10&-9&2&0\\\end{array}.</cmath>
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Then, we have a quadratic <math>10x^2-9x+2.</math> Using the Quadratic Formula, we can find the other two roots:
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<cmath>x=\frac{9 \pm \sqrt{(-9)^2-4(10)(2)}}{2 \cdot 10},</cmath>
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which simplifies to <math>x=\frac{1}{2}, \frac{2}{5}.</math>
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To find the new volume, we add <math>2</math> to each of the roots we found:
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<cmath>(3+2)\cdot(\frac{1}{2}+2)\cdot(\frac{2}{5}+2).</cmath>
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Simplifying, we find that the new volume is <math>\boxed{\textbf{(D) } 30}.</math>
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-MathWizard09
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== Solution 3 ==
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Let <math>P(x) = 10x^3 - 39x^2 + 29x - 6</math>, and let <math>a, b, c</math> be the roots of <math>P(x)</math>. The roots of <math>P(x-2)</math> are then <math>a + 2, b + 2, c + 2,</math> so the product of the roots of <math>P(x-2)</math> is the area of the desired rectangular prism.
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<math>P(x-2)</math> has leading coefficient <math>10</math> and constant term <math>P(0-2) = P(-2) = 10(-2)^3 - 39(-2)^2 + 29(-2) - 6 = -300</math>.
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Thus, by Vieta's Formulas, the product of the roots of <math>P(x-2)</math> is <math>\frac{-(-300)}{10} = \boxed{\textbf{(D) } 30}</math>.
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-Orange_Quail_9
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==Video Solution==
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https://youtu.be/08YkinzFcCc
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==See also==
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{{AMC10 box|year=2022|ab=A|num-b=15|num-a=17}}
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{{AMC12 box|year=2022|ab=A|num-b=14|num-a=16}}
 +
{{MAA Notice}}

Revision as of 00:35, 12 November 2022

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?

$\textbf{(A) } \frac{24}{5} \qquad \textbf{(B) } \frac{42}{5} \qquad \textbf{(C) } \frac{81}{5} \qquad \textbf{(D) } 30 \qquad \textbf{(E) } 48$

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

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

- 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 $\boxed{\textbf{(D) } 30}.$

-MathWizard09

Solution 3

Let $P(x) = 10x^3 - 39x^2 + 29x - 6$, and let $a, b, c$ be the roots of $P(x)$. The roots of $P(x-2)$ are then $a + 2, b + 2, c + 2,$ so the product of the roots of $P(x-2)$ is the area of the desired rectangular prism.

$P(x-2)$ has leading coefficient $10$ and constant term $P(0-2) = P(-2) = 10(-2)^3 - 39(-2)^2 + 29(-2) - 6 = -300$.

Thus, by Vieta's Formulas, the product of the roots of $P(x-2)$ is $\frac{-(-300)}{10} = \boxed{\textbf{(D) } 30}$.

-Orange_Quail_9

Video Solution

https://youtu.be/08YkinzFcCc

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

See also

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 10 Problems and Solutions
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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