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

Any help please??

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- December 5th 2012, 10:32 AMTimsBobby2Factor 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?? - December 5th 2012, 11:38 AMDevenoRe: Factor x^16+x as a product of irreducible polynomials
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?