2024 AMC 12A Problems/Problem 15

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 $f(2i)f(-2i)=(-5-10i)(-5+10i)=(-5)^2-(10i)^2=25+100=\boxed{\textbf{(D) }125}$ .

Select D

~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

Solution 6

Suppose $y = x^2 + 4$

then $x = \sqrt{y - 4}$ . Substitute $x = \sqrt{y - 4}$ into $x^3 + 2x^2 - x + 3 = 0$

$(\sqrt{y - 4})^3 + 2(\sqrt{y - 4})^2 - \sqrt{y - 4} + 3 = 0 \implies (y - 5)^2(y - 4) = (-2y + 5)^2$ whose constant is 125

according to Vieta's theorem, $y_1y_2y_3 = 125$

$y_1y_2y_3 = 125 \implies (x_1^2 + 4)(x_2^2 + 4)(x_3^2 + 4) = 125 \implies (p^2 + 4)(q^2 + 4)(r^2 + 4) = 125$ . It is same for $x = -\sqrt{y - 4}$

Select D

~JiYang

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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