Factor x^16+x as a product of irreducible polynomials in Z2[x].
Any help please??
well, ONE factor is obvious: x.
x^{16} + x = x(x^{15} + 1).
so now we just need to factor x^{15} + 1 in Z^{2}[x].
note that x^{15} + 1 = (x^{3} + 1)(x^{12} + x^{9} + x^{6} + x^{3} + 1).
we can factor x^{3} + 1 as (x + 1)(x^{2} + x + 1) <---two irreducible factors (why?)
so now we just have to factor:x^{12} + x^{9} + x^{6} + x^{3} + 1.
note that: x^{15} + 1 also factors as (x^{5} + 1)(x^{10} + x^{5} + 1).
since x^{2} + x + 1 is an irreducible factor, it must divide either x^{5} + 1 or x^{10} + x^{5} + 1. prove it doesn't divide x^{5} + 1.
we can further factor x^{5} + 1 as (x + 1)(x^{4} + x^{3} + x^{2} + x + 1).
prove that the quartic is irreducible (hint: it suffices to show it has no linear or quadratic factors).
use long division to determine the quotient (xx^{10} + x^{5} + 1)/(x^{2} + x + 1).
argue that since GF(16) is every root of x^{16} - x, and GF(8) is a subfield of GF(16), there must be an irreducible factor of x^{16} - x of degree 8 (we've already found the irreducible factors of orders 1,2 and 4). why does this mean we're done?