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Difference between revisions of "2022 AMC 12A Problems/Problem 15"

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units. What is the volume of the new box?
 
units. What is the volume of the new box?
  
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==Solution==
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==Solution 1==
  
 
Let <math>a</math>, <math>b</math>, <math>c</math> be the three roots of the polynomial. The lenghtened prism's area is <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>.
 
Let <math>a</math>, <math>b</math>, <math>c</math> be the three roots of the polynomial. The lenghtened prism's area is <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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- phuang1024
 
- phuang1024
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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)
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2022|ab=A|num-b=14|num-a=16}}
 
{{AMC12 box|year=2022|ab=A|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:31, 11 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?

Solution 1

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

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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