If $\beta$ is a zero of $f(x)=x^2+x+2$ over $\mathbb{Z}_3$, find the other zero If $\beta$ is a zero of $f(x)=x^2+x+2$ over $\mathbb{Z}_3$, find the other zero.
What I tried:
Suppose $B$ is a zero of $f(x)$, so $f(\beta)=\beta^2+\beta+2=0$. I know that $f(x)$ is irreducible in $\mathbb{Z}_3$. Any idea how to proceed?
 A: Call the other root $\gamma$. We have:
$x^2 + x + 2 = (x - \beta)(x - \gamma) = x^2 - (\beta + \gamma)x + \beta\gamma$.
Thus we have:
$\gamma = \dfrac{2}{\beta} = 2 - \beta$
Note that $\dfrac{2}{\beta} = 2 - \beta$ tells us that:
$\beta^2 + 2 = 2\beta$, adding $\beta$ to both sides gives us:
$\beta^2 + \beta + 2 = 0$, as we already knew.
We can check that both forms of $\gamma$ give an actual root:
$\left(\dfrac{2}{\beta}\right)^2 + \dfrac{2}{\beta} + 2 = \dfrac{1}{\beta^2} + \dfrac{2\beta}{\beta^2} + \dfrac{2\beta^2}{\beta^2} = \dfrac{2(1 + 2\beta + 2\beta^2)}{2\beta^2}$
$= \dfrac{2 + \beta + \beta^2}{2\beta^2} = \dfrac{0}{2\beta^2} = 0$, and:
$(2 - \beta)^2 + (2 - \beta) + 2 = 1 - \beta + \beta^2 + 2 - \beta + 2 = \beta^2 - 2\beta + 2 = \beta^2 + 3\beta - 2\beta + 2$
$ = \beta^2 + \beta + 2 = 0$.
Finally, it should be clear that:
$(x - \beta)(x - (2 - \beta)) = x^2 - (2 - \beta + \beta)x + (\beta)(2 - \beta)$
$= x^2 - 2x + 2\beta - \beta^2 = x^2 + x  -(\beta + \beta^2) = x^2 + x + 2$.
