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

(Solution)
 
(17 intermediate revisions by 7 users not shown)
Line 1: Line 1:
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)?
+
==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 <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 f(x), you can create a new polynomial f(x+2), 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.
  
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
+
Substituting <math>x-2</math> and <math>x+2</math> into <math>x</math> for the first polynomial, gives you <math>10i-5</math> and <math>-10i-5</math> as <math>c</math> for both equations. Multiplying <math>10i-5</math> and <math>-10i-5</math> together gives you <math>\boxed{\textbf{(D) }125}</math>.
  
 
-ev2028
 
-ev2028
 +
 +
~Latex by eevee9406
 +
 +
==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>.
 +
 +
 +
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>.
 +
 +
~eevee9406
 +
 +
==Solution 3==
 +
First, denote that
 +
<cmath>p+q+r=-2,
 +
pq+pr+qr=-1,
 +
pqr=-3</cmath>
 +
Then we expand the expression
 +
<cmath>(p^2+4)(q^2+4)(r^2+4)</cmath>
 +
<cmath>=(pqr)^2+4((pq)^2+(pr)^2+(qr)^2)+4^2(p^2+q^2+r^2)+4^3</cmath>
 +
<cmath>=(-3)^2+4((pq+pr+qr)^2-2pqr(p+q+r))+4^2((p+q+r)^2-2(pq+pr+qr))+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>
 +
~lptoggled
 +
 +
==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. The idea is to transform the expression involving <math>p^2, q^2, r^2</math> into one involving <math>p, q, r</math>.
 +
 +
Since <math>p</math> is a root of <math>f</math>,
 +
<cmath>p^3+2p^2-p+3=0\implies p^3+2p^2=p^2(p+2)=p-3,</cmath>
 +
which gives us that <math>p^2=\frac{p-3}{p+2}</math>. Then
 +
<cmath>p^2+4=\frac{p-3}{p+2} + 4 = \frac{5p+5}{p+2}=5\cdot \frac{p+1}{p+2}.</cmath>
 +
Since <math>q</math> and <math>r</math> are also roots of <math>f</math>, the same analysis holds, so
 +
 +
\begin{align*}
 +
(p^2 + 4)(q^2 + 4)(r^2 + 4)&= \left(5\cdot \frac{p+1}{p+2}\right)\left(5\cdot \frac{q+1}{q+2}\right)\left(5\cdot \frac{r+1}{r+2}\right)\\
 +
&= 125 \frac{(p+1)(q+1)(r+1)}{(p+2)(q+2)(r+2)}\\
 +
&= 125 \frac{-f(-1)}{-f(-2)} \\
 +
&= 125\cdot 1\\
 +
&=\boxed{\textbf{(D) }125}.
 +
\end{align*}
 +
 +
(*) This is because
 +
<cmath>(p+c)(r+c)(q+c)=(-1)^3(-c-p)(-c-r)(-c-q)=-f(-c),</cmath>
 +
since
 +
<cmath>f(x)=(x-p)(x-q)(x-r)</cmath>
 +
for all <math>x</math>.
 +
 +
~tsun26
 +
~KSH31415 (final step and clarification)
 +
 +
==Solution 5 (Cheesing it out)==
 +
Expanding the expression
 +
<cmath>(p^2 + 4)(q^2 + 4)(r^2 + 4)</cmath>
 +
gives us
 +
<cmath>(pqr)^2+4p^2q^2+4p^2r^2+4q^2r^2+16p^2+16q^2+16r^2+64</cmath>
 +
 +
Notice that everything other than <math>(pqr)^2</math> is a multiple of <math>4</math>. Solving for <math>(pqr)^2</math> using vieta's formulas, we get <math>9</math>. Since <math>9</math> is <math>1\pmod4</math>, the answer should be as well. The only answer that is <math>1\pmod4</math> is <math>\boxed{\textbf{(D) }125}</math>.
 +
 +
~callyaops
 +
 +
==See also==
 +
{{AMC12 box|year=2024|ab=A|num-b=14|num-a=16}}
 +
{{MAA Notice}}

Latest revision as of 23:46, 9 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. The idea is to transform the expression involving $p^2, q^2, r^2$ into one involving $p, q, r$.

Since $p$ is a root of $f$, \[p^3+2p^2-p+3=0\implies p^3+2p^2=p^2(p+2)=p-3,\] which gives us that $p^2=\frac{p-3}{p+2}$. Then \[p^2+4=\frac{p-3}{p+2} + 4 = \frac{5p+5}{p+2}=5\cdot \frac{p+1}{p+2}.\] Since $q$ and $r$ are also roots of $f$, the same analysis holds, so

\begin{align*} (p^2 + 4)(q^2 + 4)(r^2 + 4)&= \left(5\cdot \frac{p+1}{p+2}\right)\left(5\cdot \frac{q+1}{q+2}\right)\left(5\cdot \frac{r+1}{r+2}\right)\\ &= 125 \frac{(p+1)(q+1)(r+1)}{(p+2)(q+2)(r+2)}\\ &= 125 \frac{-f(-1)}{-f(-2)} \\ &= 125\cdot 1\\ &=\boxed{\textbf{(D) }125}. \end{align*}

(*) This is because \[(p+c)(r+c)(q+c)=(-1)^3(-c-p)(-c-r)(-c-q)=-f(-c),\] since \[f(x)=(x-p)(x-q)(x-r)\] for all $x$.

~tsun26 ~KSH31415 (final step and clarification)

Solution 5 (Cheesing it out)

Expanding the expression \[(p^2 + 4)(q^2 + 4)(r^2 + 4)\] gives us \[(pqr)^2+4p^2q^2+4p^2r^2+4q^2r^2+16p^2+16q^2+16r^2+64\]

Notice that everything other than $(pqr)^2$ is a multiple of $4$. Solving for $(pqr)^2$ using vieta's formulas, we get $9$. Since $9$ is $1\pmod4$, the answer should be as well. The only answer that is $1\pmod4$ is $\boxed{\textbf{(D) }125}$.

~callyaops

See also

2024 AMC 12A (ProblemsAnswer KeyResources)
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. AMC logo.png