Question

What is the remainder when f(x)=x^2220 + x^2215 + x^2210 + … + x^10 + x^5 + x^0 is divided by x^4 + x^3 + x^2 + x + 1?

Note that:

x^0 = 1

Answer 1

f(x) = (x^2225 - 1)/(x^5 - 1)

q(x) = x^4 + x^3 + x^2 + x + 1 =
(x^5 - 1)/(x-1)

If we write:

f(x) = A(x) q(x) + r(x)

with A(x) and r(x) polynomials, then r(x) is the remainder, which will be a polynomial of third degree. We can compute this as follows. If we take x to be a zero of q(x), then we have:

f(x) = r(x)

q(x) has 4 zeroes, if we compute f(x) at these points that will fix the thord degree polynomial r(x).

The zeroes of q(x), xi satisfy xi^5 = 1, and they are not equal to 1.

f(xi) can be computed by computing the limit of x to xi of f(x) where we write f(x) in the above form (which strictly speaking is't valid there, but it does have a removable singularity at x = xi).

So, we have:

f(xi) = 2225 xi^2224/(5 xi^4) =

445 xi^2220 = 445

It then follows that r(x)is the constant function equal to 445.