Difference between revisions of "2022 AMC 12A Problems/Problem 15"
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==Solution 1 (Vieta's Formulas)== | ==Solution 1 (Vieta's Formulas)== | ||
− | 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>. | ||
By vieta's formulas, we know that: | By vieta's formulas, we know that: | ||
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<math>a+b+c = \frac{-B}{A} = \frac{39}{10}</math>. | <math>a+b+c = \frac{-B}{A} = \frac{39}{10}</math>. | ||
− | We can substitute these into the expression, obtaining <math>V = \frac{6}{10} + 2(\frac{29}{10}) + 4(\frac{39}{10}) + 8 = \boxed{(D) 30}</math> | + | We can substitute these into the expression, obtaining |
+ | |||
+ | <math>V = \frac{6}{10} + 2(\frac{29}{10}) + 4(\frac{39}{10}) + 8 = \boxed{(D) 30}</math> | ||
- phuang1024 | - phuang1024 |
Revision as of 23:28, 11 November 2022
Contents
Problem
The roots of the polynomial 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 ,
,
be the three roots of the polynomial. The lenghtened prism's area is
.
By vieta's formulas, we know that:
.
We can substitute these into the expression, obtaining
- 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 , where
is a factor of the constant
and
is a factor of the leading coefficient
. Therefore,
is
and q is
Doing Synthetic Division, we find that is a root of the cubic:
Then, we have a quadratic Using the Quadratic Formula, we can find the other two roots:
which simplifies to
To find the new volume, we add to each of the roots we found:
Simplifying, we find that the new volume is
-MathWizard09
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.