Suppose two cubic polynomials f(x) and g(x) satisfy the following: f(2)=g(4), f(4)=g(8), f(8)=g(16), f(16)=g(32)+64. Find the value of g(128)-f(64).
Let h(x) = g(2x)-f(x)
Then h(2)=0, h(4)=0, h(8)=0, h(16)=64. And we are tasked with finding h(64).
From the three zeros h(x) = A*(x-2)*(x-4)*(x-8). Then evaluating at x=16 yields 64=A*14*12*8. Then A=1/21.
h(x) = (1/21)*(x-2)*(x-4)*(x-8). Then h(64) = (1/21)*(62)*(60)*(56) = 9920.