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 | + | 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
Contents
Problem
The roots of are and What is the value of
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 Then we expand the expression ~lptoggled
Solution 4 (Reduction of power)
The motivation for this solution is the observation that is easy to compute for any constant c, since (*), where is the polynomial given in the problem. The idea is to transform the expression involving into one involving .
Since is a root of , which gives us that . Then Since and are also roots of , 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 since for all .
~tsun26 ~KSH31415 (final step and clarification)
Solution 5 (Cheesing it out)
Expanding the expression gives us
Notice that everything other than is a multiple of . Solving for using vieta's formulas, we get . Since is , the answer should be as well. The only answer that is is .
~callyaops
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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