2017 AMC 10A Problem 24

Attempt Problem 24 of the 2017 AMC 10A 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 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

24.

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 \\ &\quad {}+ bx^2 + 100x + c. \end{aligned} What is f(1)?f(1)?
−9009-9009
−8008-8008
−7007-7007
−6006-6006
−5005-5005
Answer: C
Concepts:polynomialfactoring
Small hint:
Because the roots of gg are roots of ff, factor ff by gg
Big hint:
Compare the x3x^3 and xx coefficients to find the extra root and aa
Solution:
Because the three distinct roots of the cubic gg are also roots of the monic quartic f,f, we can write f(x)=g(x)(x−r)f(x)=g(x)(x-r) for some real number r.r.
The coefficient of xx in this product is 10−r,10-r, so 10−r=10010-r=100 and r=−90.r=-90. The coefficient of x3x^3 is a−r,a-r, so a−r=1a-r=1 and a=−89.a=-89.
Therefore, f(1)=g(1)(1−r)=(1−89+1+10)(91)=(−77)(91)=−7007.\begin{aligned} f(1)&=g(1)(1-r)\\ &=(1-89+1+10)(91)\\ &=(-77)(91)\\ &=-7007. \end{aligned}
Thus, C is the correct answer.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.