Problem:
Let g(x) be a polynomial with leading coefficient 1 , whose three roots are the reciprocals of the three roots of f(x)=x3+ax2+bx+c, where 1<a<b<c. What is g(1) in terms of a,b, and c ?
Answer Choices:
A. c1+a+b+c
B. 1+a+b+c
C. c21+a+b+c
D. c2a+b+c
E. a+b+c1+a+b+c
Solution:
Answer (A): Division of f(x) by x3 gives a cubic polynomial in x1, whose roots are the roots of g. Therefore
g(x)=ccx3+bx2+ax+1,
so
g(1)=cc⋅13+b⋅12+a⋅1+1=c1+a+b+c.
OR
Suppose the roots of f are r,s, and t. Then
f(x)=(x−r)(x−s)(x−t)=x3−(r+s+t)x2+(rs+rt+st)x−rst,
so −a=r+s+t,b=rs+rt+st, and −c=rst. Because the roots of g are r1,s1, and t1,
g(x)=(x−r1)(x−s1)(x−t1)
and
g(1)=(1−r1)(1−s1)(1−t1).
Expanding gives
g(1)=1−(r1+s1+t1)+(rs1+rt1+st1)−rst1.
Rewriting with common denominator r st gives
g(1)=rstrst−(rs+rt+st)+(r+s+t)−1.
Therefore
g(1)=−c−c−b−a−1=c1+a+b+c.
The problems on this page are the property of the MAA's American Mathematics Competitions