The polynomial I'm working with is $f(x) = x^5 + x + 1$ in $Z2[x]$. I've tried finding its roots but $f(0) = 1$ and $f(1) = 1 (mod 2).$ I know that $f(x)$ is still reducible even though it doesn't have any roots, but I'm not sure how to verify this and then I'm uncertain how to then find its factors.
I assume that the factors $g(x)$ and $h(x)$ will need to be either of degree 2 and 3 or of degree 1 and 4.
One of the factors is quadratic, one cubic. We may demand the quadratic $x^2 + a x + 1,$ where either $a=0$ or $a=1.$ Well, if $a=0,$ we get a root, since $1^2 + 1 = 0 $ here.
Anyway, you do not do these all at once, you try to get one new coefficient at each step. So far the quadratic must be $x^2 + x + 1.$ At the moment, the cubic is some $x^3 + bx^2 + cx + 1.$ The product is $$ x^5 + (b+1)x^4 + (b+c+1)x^3 + (b+c+1)x^2 + (c+1)x + 1 $$ while we want $x^5 + x + 1.$ The $x^4$ demands $b=1$ while the $x$ demands $c=0.$ We arrive at $$ (x^2 + x + 1)(x^3 + x^2 + 1). $$ Looks good so far. Check it carefully! Check for lack of roots in this field as well.