Thread: Factor x^16+x as a product of irreducible polynomials

Factor x^16+x as a product of irreducible polynomials in Z2[x].

Any help please??

well, ONE factor is obvious: x.

x¹⁶ + x = x(x¹⁵ + 1).

so now we just need to factor x¹⁵ + 1 in Z²[x].

note that x¹⁵ + 1 = (x³ + 1)(x¹² + x⁹ + x⁶ + x³ + 1).

we can factor x³ + 1 as (x + 1)(x² + x + 1) <---two irreducible factors (why?)

so now we just have to factor:x¹² + x⁹ + x⁶ + x³ + 1.

note that: x¹⁵ + 1 also factors as (x⁵ + 1)(x¹⁰ + x⁵ + 1).

since x² + x + 1 is an irreducible factor, it must divide either x⁵ + 1 or x¹⁰ + x⁵ + 1. prove it doesn't divide x⁵ + 1.

we can further factor x⁵ + 1 as (x + 1)(x⁴ + x³ + x² + 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 (xx¹⁰ + x⁵ + 1)/(x² + x + 1).

argue that since GF(16) is every root of x¹⁶ - x, and GF(8) is a subfield of GF(16), there must be an irreducible factor of x¹⁶ - x of degree 8 (we've already found the irreducible factors of orders 1,2 and 4). why does this mean we're done?