Difference between revisions of "2024 AMC 12A Problems/Problem 15"
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| − | The roots of x^3 + 2x^2 | + | ==Problem== |
| + | The roots of <math>x^3 + 2x^2 - x + 3</math> are <math>p, q,</math> and <math>r.</math> What is the value of <cmath>(p^2 + 4)(q^2 + 4)(r^2 + 4)?</cmath><math>\textbf{(A) } 64 \qquad \textbf{(B) } 75 \qquad \textbf{(C) } 100 \qquad \textbf{(D) } 125 \qquad \textbf{(E) } 144</math> | ||
| − | ==Solution== | + | ==Solution 1== |
You can factor (p^2 + 4)(q^2 + 4)(r^2 + 4) as (p − 2i)(p + 2i)(q − 2i)(q + 2i)(r − 2i)(r + 2i). | You can factor (p^2 + 4)(q^2 + 4)(r^2 + 4) as (p − 2i)(p + 2i)(q − 2i)(q + 2i)(r − 2i)(r + 2i). | ||
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-ev2028 | -ev2028 | ||
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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 | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC12 box|year=2024|ab=A|num-b=14|num-a=16}} | ||
| + | {{MAA Notice}} | ||
Revision as of 17:57, 8 November 2024
Contents
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 (p^2 + 4)(q^2 + 4)(r^2 + 4) as (p − 2i)(p + 2i)(q − 2i)(q + 2i)(r − 2i)(r + 2i).
For any polynomial f(x), you can create a new polynomial f(x+2), which will have roots that instead have the value subtracted.
Substituting x-2 and x+2 into x for the first polynomial, gives you 10i-5 and -10i-5 as c for both equations. Multiplying 10i-5 and -10i-5 together gives you 125
-ev2028
Solution 2
Let 
. Then
.
We find that 
and 
, so 
.
~eevee9406
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. 
