Factor $f(x)=x^2+x+1$ in $z_2[x]/(x^2+x+1)$ into the form if possible
$$ f(x)=c_0(x-c_1)(x-c_2)$$
where $c_{i} \in z_2[x]/(x^2+x+1)$ where $i\in[0,2]$
Attempt the elemts of $z_2[x]/(x^2+x+1)$ should be $$ [0],[1],[x],[x+1]$$ obviously $[x]$ is a root and it is also the case for $[x+1]$ since $$ f([x+1])=[(x+1)^2+(x+1)+1]=[x(x+1)+1(x+1)+(x+1)+1]=[x^2+x+1]$$ Now found 2 roots of the possible 4 which is easy in this case. but trying to put in factor form $$ f(x)=c_0(x-x)(x-(x+1))$$ Not sure that is correct. because of $x-x$. Would like to stay away from $i$
You must introduce elements in $\mathbf F_4=\mathbf Z_2[x]/(x^2+x+1)$. So denote $\omega$ the class of $x$ modulo $x^2+x+1$. It satisfies the relation $\omega^2=\omega+1$, and indeed, $\mathbf F_4$ is a field with $4$ elements: $\;0,1,\omega,\omega+1$.
The polynomial $x^2+x+1$ has two roots in $\mathbf F_4$: $\;\omega$ and a root $\omega'$, such tthat, by Vieta's relations, $\;\omega+\omega'=1$, hence $\omega'=\omega+1$, so that $$x^2+x+1=(x+\omega)(x+\omega+1).$$