The following problem is from both the 2022 AMC 10A #16 and 2022 AMC 12A #15, so both problems redirect to this page.

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 (Incorrect Guess)

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: $$ (Error compiling LaTeX. Unknown error_msg)\begin{alignat*}{8} a+b+c &= -\frac{B}{A} 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: $(3+2)\cdot\left(\frac{1}{2}+2\right)\cdot\left(\frac{2}{5}+2\right).$ Simplifying, we find that the new volume is $\boxed{\textbf{(D) } 30}.$

-MathWizard09

Solution 4

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

Solution 5

Let $P(x) = 10x^3 - 39x^2 + 29x - 6$ . This can be factored m as $P(x) = 10(x-a)(x-b)(x-c)$ , where $a$ , $b$ , and $c$ are the roots of $P(x)$ . We want $V = (a+2)(b+2)(c+2)$ .

"Luckily" $P(-2) = 10(-2-a)(-2-b)(-2-c) = -10V$ . $P(-2) = -300$ , giving $V = \boxed{\textbf{(D) } 30}$ .

-Oxymoronic15

(It's not just lucky. If $P(x)$ has roots $x \in \{r_i\}$ , $Q(y) = P(x-2)$ has roots $y \in \{r_i+2\}$ . By Vieta, the product $V$ of the roots is the negation of the constant term divided by the leading coefficient $10$ , which is $-Q(0) / 10$ , which is $-P(0-2) / 10$ . -oinava )

Solution 6 (Desperate Final Effort - Estimation Guess)

By Vieta's, we can see that $ABC = \frac{6}{10}$ . Using this, we can see that if each side $ABC$ is the same length, the length is between $0.8$ ( $0.512$ ) and $0.9$ ( $0.729$ ). Adding $2$ to these numbers would give us three numbers that are close to $3$ . Rounding up, we will just assume they are all three. If we multiply all of them, it gives us $27$ . The closest answer choice is $\boxed{\textbf{(D) } 30},$ as all of the other choices are far from this number (the second closest answer choice being $11$ away).

~orenbad

$(2+0.6^{1/3})^3 \approx 23$ is a lower bound for the answer (if the roots are more spread out then adding to a smaller root stretches the product more than adding 2 to a larger root shrinks the product), but a different $P(x)$ with the same product of roots could have roots that lead to a much larger answer (but not exactly 48, it turns out). Going by this bound alone, only answers A, B, and C can be eliminated, leaving a guess between D and E.

-oinava

Video Solution (Quick and Simple)

https://youtu.be/kvNM_USBoyE

~Education, the Study of Everything

Video Solution

https://www.youtube.com/watch?v=08YkinzFcCc

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

2022 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2022 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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2022 AMC 10A Problems/Problem 16

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