Problem:
For certain real numbers a,b, and c, the polynomial
g(x)=x3+ax2+x+10
has three distinct roots, and each root of g(x) is also a root of the polynomial
f(x)=x4+x3+bx2+100x+c
What is f(1)?
Answer Choices:
A. −9009
B. −8008
C. −7007
D. −6006
E. −5005
Solution:
Let q be the additional root of f(x). Then
f(x)=(x−q)(x3+ax2+x+10)=x4+(a−q)x3+(1−qa)x2+(10−q)x−10q.
Thus 100=10−q, so q=−90 and c=−10q=900. Also 1=a−q= a+90, so a=−89. It follows, using the factored form of f shown above, that f(1)=(1−(−90))⋅(1−89+1+10)=91⋅(−77)=(C)−7007.
The problems on this page are the property of the MAA's American Mathematics Competitions