2024 AMC 12A Problems/Problem 15

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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 125

-ev2028

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

See also

2024 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AMC 12 Problems and Solutions

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