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

Revision as of 18:46, 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

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.

@@ Line 33: / Line 33: @@
 <cmath>=(-3)^2+4((-1)^2-2(-3)(-2))+4^2((-2)^2-2(-1))+4^3</cmath>
 <cmath>=\fbox{(D) 125}</cmath>
+~lptoggled
 ==See also==
 {{AMC12 box|year=2024|ab=A|num-b=14|num-a=16}}
 {{MAA Notice}}

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