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 x3+2x2x+3x^3+2x^2-x+3 are pp, qq, and rr. What is the value of (p2+4)(q2+4)(r2+4)? (p^2+4)(q^2+4)(r^2+4)?

6464

7575

100100

125125

144144

Answer: D
Concepts:complex numberpolynomial
Difficulty rating: 1710
Solution:

Since P(x)=(xp)(xq)(xr),P(x)=(x-p)(x-q)(x-r), grouping p2+4=(p2i)(p+2i)p^2+4=(p-2i)(p+2i) over all roots gives (p2+4)=P(2i)P(2i). \prod(p^2+4)=P(2i)\,P(-2i). Compute P(2i)=8i82i+3P(2i)=-8i-8-2i+3 =510i=-5-10i and P(2i)=8i8+2i+3P(-2i)=8i-8+2i+3 =5+10i.=-5+10i. Their product is (5)2+102=25+100=125.(-5)^2+10^2=25+100=125. Thus, the correct answer is D.

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