Let f(x) be a quadratic polynomial such that f(-4)=-22, f(-1)=2, and f(2)=-1. Let g(x)=f(x)^16. Find the sum of the coefficients of the terms ing(x) that have even degree. (For example, the sum of the coefficients of the terms in -5x^3+4+11x-5 that have even degree is (4)+(-5)=-1.)
I figured out that f(x)=-1.5x^2+0.5x+4, but I'm not sure how to solve the problem from there.
The sum of the coefficients is clearly (2^16 + (-1)^16)/2.= (2^16 + 1)/2.
Consider g(1) = f(1)^16 This will consist of the sum of the even degree coefficients and the sum of odd degree coefficients.. Now consider g(-1) = f(-1)^16. This will consist of the sum of the even degree coefficients and the negative of the sum of the odd degree coefficients. Add g(1) and g(-1). This will eliminate the odd degree coefficients, but will leave twice the sum of the even degree coefficients. Hence to get the sum of the even degree coefficients you need:
(g(1) + g(-1))/2 or (f(1)^16 + f(-1)^16)/2
Thanks very much Alan,
It took me a while to work out what you were saying but I have it all sorted now.
I never would have thought to do that!
Have you worked it out mathmathj28?
If so what answer did you finally get?
Thank you so much for the help Alan!! I didn't think of doing that before but I understand why that is helpful now.
If my work is correct, the sum of the coefficients of the terms in g(x) that have even degree is 21,556,128.5
Thanks again for helping me solve this!