2017 AMC 12A Problem 23

Attempt Problem 23 of the 2017 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 2017 AMC 12A solutions, or check the answer key.

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23.

For certain real numbers a,a, b,b, and c,c, the polynomial g(x)=x3+ax2+x+10g(x)=x^3+ax^2+x+10 has three distinct roots, and each root of g(x)g(x) is also a root of the polynomial f(x)=x4+x3+bx2+100x+c. \begin{aligned} &f(x)=x^4+x^3+bx^2 \\ &\quad {}+100x+c. \end{aligned} What is f(1)?f(1)?

9009-9009

8008-8008

7007-7007

6006-6006

5005-5005

Answer: C
Concepts:polynomialfactoring
Difficulty rating: 2380
Solution:

Since gg has three distinct roots all shared by the quartic f,f, we can write f(x)=(xq)g(x)f(x)=(x-q)g(x) for some remaining root q.q. Expanding, f(x)=x4+(aq)x3+(1qa)x2+(10q)x10q. \begin{aligned} &f(x)=x^4+(a-q)x^3 \\ &\quad {}+(1-qa)x^2 \\ &\quad {}+(10-q)x-10q. \end{aligned}

Matching the xx coefficient, 10q=100,10-q=100, so q=90.q=-90. Matching the x3x^3 coefficient, aq=1,a-q=1, so a=89.a=-89.

Then g(1)=1+a+1+10g(1)=1+a+1+10 =1289=77=12-89=-77 and 1q=91,1-q=91, so f(1)=(1q)g(1)=91(77)=7007. \begin{aligned} &f(1)=(1-q)g(1) \\ &=91\cdot(-77)=-7007. \end{aligned}

Thus, the correct answer is C.

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