2007 AMC 12A Problem 18 — Solution | LIVE by Po-Shen Loh

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2007 AMC 12A Problem 18

Below is the professionally curated solution for Problem 18 of the 2007 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 12A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:complex number polynomial

Difficulty rating: 1630

18.

The polynomial $f(x)=x^4+ax^3+bx^2+cx+d$ has real coefficients, and $f(2i)=f(2+i)=0.$ What is $a+b+c+d?$

$0$

$1$

$4$

$9$

$16$

Solution:

Since $f$ has real coefficients, the conjugates $-2i$ and $2-i$ are also roots. Thus $f(x)=(x^2+4)(x^2-4x+5)=x^4-4x^3+9x^2-16x+20.$

Then $a+b+c+d=-4+9-16+20=9.$ Equivalently, $a+b+c+d=f(1)-1=(1+4)(1+1)-1=9.$

Thus, the correct answer is D.

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