The polynomial x^(101) + Ax + B is divisible by \(x^2 - 1\) for some real numbers A and B. Find A+B.
f(x)=x^(101) + Ax + B
x^2-1=(x-1)(x+1) so 1 and -1 are roots of f(x)
so
f(1)=0 and f(-1)=0
solve simultaneously.
Easy peazy.