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| − | ==Problem==
| + | #redirect [[2022 AMC 10A Problems/Problem 16]] |
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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}}
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| − | {{MAA Notice}}
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