Difference between revisions of "2024 AMC 12A Problems/Problem 15"
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==Solution 1== | ==Solution 1== | ||
| − | You can factor <math>(p^2 + 4)(q^2 + 4)(r^2 + 4)</math> as | + | You can factor <math>(p^2 + 4)(q^2 + 4)(r^2 + 4)</math> as <math>(p+2i)(p-2i)(q+2i)(q-2i)(r+2i)(r-2i)</math>. |
For any polynomial <math>f(x)</math>, you can create a new polynomial <math>f(x+2)</math>, which will have roots that instead have the value subtracted. | For any polynomial <math>f(x)</math>, you can create a new polynomial <math>f(x+2)</math>, which will have roots that instead have the value subtracted. | ||
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-ev2028 | -ev2028 | ||
| + | ~Latex by eevee9406 | ||
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| + | ==Solution 2== | ||
| + | Let <math>f(x)=x^3 + 2x^2 - x + 3</math>. Then | ||
| + | <math>(p^2 + 4)(q^2 + 4)(r^2 + 4)=(p+2i)(p-2i)(q+2i)(q-2i)(r+2i)(r-2i)=f(2i)f(-2i)</math>. | ||
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| + | We find that <math>f(2i)=-8i-8-2i+3=-10i-5</math> and <math>f(-2i)=8i-8+2i+3=10i-5</math>, so <math>f(2i)f(-2i)=(-5-10i)(-5+10i)=(-5)^2-(10i)^2=25+100=\boxed{\textbf{(D) }125}</math>. | ||
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| + | ~eevee9406 | ||
| + | |||
==Solution 3== | ==Solution 3== | ||
First, denote that | First, denote that | ||
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<cmath>=(-3)^2+4((-1)^2-2(-3)(-2))+4^2((-2)^2-2(-1))+4^3</cmath> | <cmath>=(-3)^2+4((-1)^2-2(-3)(-2))+4^2((-2)^2-2(-1))+4^3</cmath> | ||
<cmath>=\fbox{(D) 125}</cmath> | <cmath>=\fbox{(D) 125}</cmath> | ||
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==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=A|num-b=14|num-a=16}} | {{AMC12 box|year=2024|ab=A|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:10, 8 November 2024
Problem
The roots of 
are 
and 
What is the value of ![\[(p^2 + 4)(q^2 + 4)(r^2 + 4)?\]](http://latex.artofproblemsolving.com/9/3/8/938ecf6869b053c8335864493848733a3f242e2e.png)
Solution 1
You can factor 
as 
.
For any polynomial 
, you can create a new polynomial 
, which will have roots that instead have the value subtracted.
Substituting 
and 
into 
for the first polynomial, gives you 
and 
as 
for both equations. Multiplying and together gives you 
.
-ev2028 ~Latex by eevee9406
Solution 2
Let 
. Then
.
We find that 
and 
, so 
.
~eevee9406
Solution 3
First, denote that
![\[p+q+r=-2, pq+pr+qr=-1, pqr=-3\]](http://latex.artofproblemsolving.com/5/f/d/5fd7adfa7164625bc4556437c944b4fe9c65594f.png)
Then we expand the expression
![\[(p^2+4)(q^2+4)(r^2+4)\]](http://latex.artofproblemsolving.com/7/9/c/79c9797d0b2895f4ab088d5ff787d893c4eb7e3c.png)
![\[=(pqr)^2+4((pq)^2+(pr)^2+(qr)^2)+4^2(p^2+q^2+r^2)+4^3\]](http://latex.artofproblemsolving.com/3/2/1/3217d6c2547f5ff7bf3821bb3c4d16e5ec4f6d32.png)
![\[=(-3)^2+4((pq+pr+qr)^2-2pqr(p+q+r))+4^2((p+q+r)^2-2(pq+pr+qr))+4^3\]](http://latex.artofproblemsolving.com/9/8/9/9892c5a6f5545bc6c5390b1e8035af2acac54d35.png)
![\[=(-3)^2+4((-1)^2-2(-3)(-2))+4^2((-2)^2-2(-1))+4^3\]](http://latex.artofproblemsolving.com/1/f/4/1f47291e68d0412309c509a07f76311f3ab4fa6f.png)
![\[=\fbox{(D) 125}\]](http://latex.artofproblemsolving.com/7/4/d/74d1dee45a2a8fdd3f738d3c79ba8b068341582c.png)
See also
| 2024 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. 
