Difference between revisions of "2022 AMC 10A Problems/Problem 16"
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==Video Solution== | ==Video Solution== | ||
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| − | + | ~Education, the Study of Everything | |
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==See also== | ==See also== | ||
Revision as of 19:01, 13 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 volume is
.
By vieta's formulas, we know that:
.
We can substitute these into the expression, obtaining
- phuang1024
Solution 2 (Guessing roots)
From the answer choices, we can assume the roots are rational numbers, and therefore this polynomial should be easily factorable.
The coefficients of 
must multiply to 
, so these coefficients must be 
or 
in some order. We can try one at a time, and therefore write the factored form as follows: ![\[(5x-p)(2x-q)(x-r)\]](http://latex.artofproblemsolving.com/0/6/b/06b22feee7c4effbc13bfcd186aa4c2b08c8235c.png)
have to multiply to 6, so they must be either 
or 
in some order. Again, we can try one at a time in different positions and see if they multiply correctly.
We try 
and multiply the 
terms, and sure enough they add up to 
. You can try to add up the 
terms and they add up to 
. Therefore the roots are 
, 
and 
. Now if you add 
to each root, you get the volume is 
~KingRavi
Solution 3 (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:
![\[\begin{array}{c|rrrr}&10&-39&29&-6\\3&&30&-27&6\\\hline\\&10&-9&2&0\\\end{array}.\]](http://latex.artofproblemsolving.com/c/e/0/ce042aa3bfca79495b52a640d9a1a38ff94ee226.png)
Then, we have a quadratic 
Using the Quadratic Formula, we can find the other two roots:
![\[x=\frac{9 \pm \sqrt{(-9)^2-4(10)(2)}}{2 \cdot 10},\]](http://latex.artofproblemsolving.com/2/a/e/2aefd3e7a36bfde2a7fff705ac7f68fe8bdcc79e.png)
which simplifies to 
To find the new volume, we add to each of the roots we found:
![\[(3+2)\cdot(\frac{1}{2}+2)\cdot(\frac{2}{5}+2).\]](http://latex.artofproblemsolving.com/8/e/5/8e50570908f3bad355fc8473bf298d07d29f1247.png)
Simplifying, we find that the new volume is 
-MathWizard09
Solution 4
Let 
, and let 
be the roots of 
. The roots of 
are then 
so the product of the roots of is the area of the desired rectangular prism.
has leading coefficient and constant term 
.
Thus, by Vieta's Formulas, the product of the roots of is 
.
-Orange_Quail_9
Video Solution
~Education, the Study of Everything
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
