2024 AMC 12A Problem 15

Since $P(x)=(x-p)(x-q)(x-r),$ grouping $p^2+4=(p-2i)(p+2i)$ over all roots gives $$\prod(p^2+4)=P(2i)\,P(-2i).$$ Compute $P(2i)=-8i-8-2i+3$ $=-5-10i$ and $P(-2i)=8i-8+2i+3$ $=-5+10i.$ Their product is $(-5)^2+10^2=25+100=125.$ Thus, the correct answer is D.