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Difference between revisions of "2024 AMC 12A Problems/Problem 15"

(→‎Solution 3)
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<cmath>=\fbox{(D) 125}</cmath>
 
<cmath>=\fbox{(D) 125}</cmath>
 
~lptoggled
 
~lptoggled
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 +
==Solution 4 (Reduction of power)==
 +
 +
The motivation for this solution is the observation that <math>(p+c)(q+c)(r+c)</math> is easy to compute for any constant c, since <math>(p+c)(q+c)(r+c)=-f(-c)</math>, where <math>f</math> is the polynomial given in the problem. We attempt to transform the expression involving <math>p^2, q^2, r^2</math> into one involving <math>p, q, r</math>.
 +
 +
Since <math>p^3+px^2-p+3=0</math>, we get <math>p^2=\frac{p-3}{p+1}</math>. Similarly <math>q^2=\frac{q-3}{q+1}</math> and <math>r^2=\frac{r-3}{r+1}</math>.
 +
 +
Hence,
 +
<cmath>(p^2 + 4)(q^2 + 4)(r^2 + 4)</cmath>
 +
<cmath>=(\frac{p-3}{p+1}+4)(\frac{q-3}{q+1}+4)(\frac{r-3}{r+1}+4)</cmath>
 +
<cmath>=(\frac{5p+1}{p+1})(\frac{5q+1}{q+1})(\frac{5r+1}{r+1})</cmath>
 +
<cmath>=125(\frac{p+\frac{1}{5}}{p+1})(\frac{q+\frac{1}{5}}{q+1})(\frac{r+\frac{1}{5}}{r+1})</cmath>
 +
<cmath>=125\frac{-f(-frac{1}{5})}{-f(-1)}</cmath>
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<cmath>=125</cmath>
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 +
Therefore our answer is <math>\boxed{\textbf{(D) }125}</math>.
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 +
~tsun26
  
 
==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 20:23, 8 November 2024

Problem

The roots of $x^3 + 2x^2 - x + 3$ are $p, q,$ and $r.$ What is the value of \[(p^2 + 4)(q^2 + 4)(r^2 + 4)?\]$\textbf{(A) } 64 \qquad \textbf{(B) } 75 \qquad \textbf{(C) } 100 \qquad \textbf{(D) } 125 \qquad \textbf{(E) } 144$

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 $\boxed{\textbf{(D) }125}$.

-ev2028

~Latex by eevee9406

Solution 2

Let $f(x)=x^3 + 2x^2 - x + 3$. Then $(p^2 + 4)(q^2 + 4)(r^2 + 4)=(p+2i)(p-2i)(q+2i)(q-2i)(r+2i)(r-2i)=f(2i)f(-2i)$.


We find that $f(2i)=-8i-8-2i+3=-10i-5$ and $f(-2i)=8i-8+2i+3=10i-5$, so $f(2i)f(-2i)=(-5-10i)(-5+10i)=(-5)^2-(10i)^2=25+100=\boxed{\textbf{(D) }125}$.

~eevee9406

Solution 3

First, denote that \[p+q+r=-2, pq+pr+qr=-1, pqr=-3\] Then we expand the expression \[(p^2+4)(q^2+4)(r^2+4)\] \[=(pqr)^2+4((pq)^2+(pr)^2+(qr)^2)+4^2(p^2+q^2+r^2)+4^3\] \[=(-3)^2+4((pq+pr+qr)^2-2pqr(p+q+r))+4^2((p+q+r)^2-2(pq+pr+qr))+4^3\] \[=(-3)^2+4((-1)^2-2(-3)(-2))+4^2((-2)^2-2(-1))+4^3\] \[=\fbox{(D) 125}\] ~lptoggled

Solution 4 (Reduction of power)

The motivation for this solution is the observation that $(p+c)(q+c)(r+c)$ is easy to compute for any constant c, since $(p+c)(q+c)(r+c)=-f(-c)$, where $f$ is the polynomial given in the problem. We attempt to transform the expression involving $p^2, q^2, r^2$ into one involving $p, q, r$.

Since $p^3+px^2-p+3=0$, we get $p^2=\frac{p-3}{p+1}$. Similarly $q^2=\frac{q-3}{q+1}$ and $r^2=\frac{r-3}{r+1}$.

Hence, \[(p^2 + 4)(q^2 + 4)(r^2 + 4)\] \[=(\frac{p-3}{p+1}+4)(\frac{q-3}{q+1}+4)(\frac{r-3}{r+1}+4)\] \[=(\frac{5p+1}{p+1})(\frac{5q+1}{q+1})(\frac{5r+1}{r+1})\] \[=125(\frac{p+\frac{1}{5}}{p+1})(\frac{q+\frac{1}{5}}{q+1})(\frac{r+\frac{1}{5}}{r+1})\] \[=125\frac{-f(-frac{1}{5})}{-f(-1)}\] \[=125\]

Therefore our answer is $\boxed{\textbf{(D) }125}$.

~tsun26

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

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