2024 AMC 12A Problem 15

Attempt Problem 15 of the 2024 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 12A solutions, or check the answer key.

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15.

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)?

64

75

100

125

144

p^2+4=(p-2i)(p+2i),

so the product relates to

P(2i)

and

P(-2i),

where

P(x)=x^3+2x^2-x+3

P(2i)

and

P(-2i)

are complex conjugates; multiply them to get a real number

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.

Problem 15 in other years