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.

mathmathj28 Jun 8, 2020

#4**+5 **

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

Alan Jun 8, 2020

#5**0 **

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?

Melody
Jun 8, 2020

#6**+4 **

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

\(g(1)=f(1)^{16}=(-\frac{3}{2}+\frac{1}{2}+4)^{16}=3^{16}=43,046,721 \)

\(g(-1)=f(-1)^{16}=(-\frac{3}{2}-\frac{1}{2}+4)^{16}=2^{16}=65,536\)

\(\frac{g(1)+g(-1)}{2}=\frac{43,046,721+65,536}{2}=\frac{43,112,257}{2}=21,556,128.5\)

Thanks again for helping me solve this!

mathmathj28
Jun 8, 2020