abstract algebra - Suppose that $beta$ is a zero of $f(x)=x^4+x+1$ in some field extensions of $E$ of $Z_2$.Write $f(x)$ as a product of linear factors in $E[x]$

Suppose that $\beta$ is a zero of $f(x)=x^4+x+1$ in some field extensions of $E$ of $Z_2$.Write $f(x)$ as a product of linear factors in $E[x]$

Attempt: In $\mathbb Z_2: \beta^4+\beta+1=0$

Going by the long division method, I was able to factorize $f(x)=x^4+x+1 = (x-\beta)(x^3+\beta x^2+ \beta^2x + \beta^3+1)$.

if $g(x)=x^3+\beta x^2+ \beta^2x + \beta^3+1$, then $g(1)=\beta^2 =g(-1)~;~$

Is there any other method than trial and error in this problem. Please note that this problem is listed in the field extension chapter and even before the chapter on algebraic extensions and finite fields

Thank you for your help.

Answer

$\require{cancel}$For a problem with small coefficients like that, trial and error works. We have
$$f(x+y) = (x+y)^4 + x + y + 1 = f(x) + f(y) + 1$$ This implies
$$f(x+1) = f(x) + f(1) + 1 = 0 + 1 + 1 = f(x)$$
So if $x$ is a root of $f$, $x+1$ is another root of $f$. In particular $\beta+1$ is a root of $f$.

Besides $f(\beta^2) = (f(\beta))^2 = 0$, because the Frobenius is a field automorphism (as Jyrki Lahtonen told you in the comments). It follows that the roots are $\beta$, $\beta+1$, $\beta^2$, $\beta^2+1$.