How do we find the roots of ${x^3 + x + 1}$ in $ {Z_2[x]} $
The elements of the congruence class are: $$0, 1, x, x + 1, x^2, x^2 + 1, x^2 + x, x^2 + x + 1$$ as they have to be of the form $ax^2 + bx + c \in Z_2$.
However, I am unable to find the roots.
Specifically, I am unsure why $x^2$ and $x^2 + x + 1$ are roots in $x^3 + x + 1$
Any help would be appreciated.
On
Do you mean finding the other roots of $x^3+x+1$ in $\mathbf F_8=\mathbf Z_2[x]/(x^3+x+1)$, in addition to the class $\omega$ of $x$?
$\omega^2$ is another root, because of the Frobenius morphism: $$(\omega^2)^3+\omega^2+1=(\omega^3+\omega+1)^2=0.$$ The sum of the roots is $0$,so the third root is the sum of the first two: $\,\omega^2+\omega$.
If you mean finding the roots of $x^3+x+1$ in $\mathbf Z_2$, there are none, as is easily checked, hence the polynomial is irreducible in $\mathbf Z_2[x]$ (a reducible polynomial of degree $3$ in a field has a root in that field).
Older Answer left as extra
If I take you correctly, what you are after are the roots of the polynomial $t^3+t+1$ viewed as an element of $\frac{\mathbb{Z}_2[x]}{(x^3+x+1)}[t]$. Let me name that ideal $I$. You are asking why $x^2+I$ and $x^2+x+1+I$ are roots. Let us just substitute those roots into the polynomial $f(t)=t^3+t+1$, shall we? \begin{align*} f(x^2)={}&x^6+x^2+1=x^2(-x^2-x)-x^2+1=-x^4-x^3-x^2+1={} \\ {}={}&x^2+x+x+1-x^2+1=0, \end{align*} definitely a root then, and: \begin{align*} f(x^2+x+1)={}&(x^2+x+1)^3+x^2+x+1+1=(x^2+x+1)(x^4+x^2+1)+x^2+x={} \\ {}={}&x^6+x^4+x^2+x^5+x^3+x+x^4+x^2+1+x^2+x={} \\ {}={}&x^2(-x^2-x)-x^2-x+x^2+x(-x^2-x)-x-1+x-x^2-x+x^2+1+x^2+x={} \\ {}={}&x^2+x+x+1-x^2-x+x^2+x(-x^2-x)-x-1+x-x^2-x+x^2+1+x^2+x={} \\ {}={}&1-x+x^2-x^2+x+1-x-1+x-x^2-x+x^2+1+x^2+x={} \\ {}={}&-x-1+x-x^2-x+x^2+1+x^2+x=-1-x+1+x^2=-1, \end{align*} definitely not a root then. $x$ is the another root, as always in this kind of quotients. As noted by Bernard, the last root is close to the one tried above: it's $x^2+x$: \begin{align*} f(x^2+x)={}&(x^2+x)^3+x^2+x+1=(x^2+x)(-x^2-x+x^2)+x^2+x+1={} \\ {}={}&-x^3-x^2+x^2+x+1=-x^3+x+1=x+1+x+1=0. \end{align*} Let us now factor the polynomial since we have its three roots: $$f(t)=(t-x)(t-x^2-x)(t-x^2).$$ In carrying out my calculations, I used the fact that: \begin{align*} x^4=x\cdot x^3=x(-x-1)=-x^2-x, && x^3=-x-1, \\ \begin{aligned} x^6=x^2\cdot x^4={}&x^2(-x^2-x)=-x^4-x^3={} \\ {}={}&x^2+x+x+1=x^2+1 \end{aligned} && \begin{aligned} x^5={}&x\cdot x^4=x(-x^2-x)={} \\ {}={}&x+1-x^2=x^2+x+1 \end{aligned} \end{align*} which is because $x^3+x+1=0$ in your quotient field.
End of old answer which answers actual question, with a little addition
Yes indeed, this is a field, since the polynomial generating $I$ is clearly irreducible over $\mathbb{Z}_2$, as it has no roots. We could have seen that from the start, but now that we have the 3 roots in the quotient we see none of them is in $\mathbb{Z}_2$. Also, this quotient is the splitting field of that polynomial.
$f(x)$ has no roots in $\mathbb{Z}_2$ since $\mathbb{Z}_2=\{0,1\}$, and $f(0)=0^2+0+1=1$ and $f(1)=1^2+1+1=1+1+1=0+1=1$. It is a third-degree polynomial, so having no roots implies being irreducible, as if it were reducible it would have to factor as a first-degree polynomial times a second-degree one, and hence have a root. Being $f$ irreducible, the quotient of $\mathbb{Z}_2[x]$ with respect to $I=(f(x))$ is a field.