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:
Note that f(x1) has the same roots as g(x), if it is multiplied by some monomial so that the leading term is x3 they will be equal. We have
f(x1)=x31+x2a+xb+c
So, we can see that
g(x)=cx3⋅f(x1)
Therefore
g(1)=c1⋅f(1)=(A) c1+a+b+c
OR
Let the three roots of f(x) be d,e, and f.
We know that
abc=−(d+e+f)=de+ef+df=−def
and using Vieta's that
g(1)=1−d1−e1−f1+de1+ef1+df1−def1
This is equal to
defdef−de−df−ef+d+e+f−1=(A) c1+a+b+c
OR
Because the problem doesn't specify what the coefficients of the polynomial actually are, we can just plug in any arbitrary polynomial that satisfies the constraints. Let's take
f(x)=(x+5)3=x3+15x2+75x+125
f(x) has a triple root of x=−5. Then, g(x) has a triple root of −51, and it's monic, so
g(x)=(x+51)3=125125x3+75x2+15x+1
We can see that this is (A) c1+a+b+c.
The problems on this page are the property of the MAA's American Mathematics Competitions