Elementary questions about polynomials and field extensions Let $$f(x)=x^2+x+1.$$ This is irreducible in $\mathbb{Z_2}[x]$, and thus $\mathbb{Z_2}[x]/(f(x))$ is a field $K$ where $(f(x))$ is a principle ideal. I don't quite understand how I find that $\overline{0}$,$\overline{1}$, $\overline{X}$, and $\overline{X+1}$ are the elements of this new field.
Also, would I be able to rewrite $f(x)$ as a product of factors of degree $1$ in $K[x]$? If so how would I go about this? 
This is from Dan Saracino's Abstract Algebra: A First Course.

In fact, $(f(x))$ is a maximal ideal, and this is equivalent to $\mathbb{Z}_2[x] / (f(x))$ being a field. Anyway, in this quotient $\bar{X}^2 + \bar{X} + \bar{1} = \bar{0}$, so, $\bar{X}^2 = \bar{X} + \bar{1}$, and by induction any power of $\bar{X}$ can be written as a $\mathbb{Z}_2$-linear combination of $\bar{1}$ and $\bar{X}$. (Neither of these is a multiple of the other, otherwise, by the below comment, $\bar{0}$ or $\bar{1}$ would be a root of $f$ in $\mathbb{Z}_2$, and hence $f$ would not be irreducible, a contradiction.) Thus, the field $\mathbb{Z}_2[x] / (f(x))$ is a vector space over $\mathbb{Z}_2$ with basis $\bar{1}, \bar{X}$, and its elements are precisely $\bar{0}, \bar{1}, \bar{X}, \bar{X} + \bar{1}$.
You can always write $f(x)$ as a product of linear factors over the splitting field $K$ of $f$ (in fact, this is why it's called a splitting field). By construction $f(\bar{X}) = \bar{X}^2 + \bar{X} + \bar{1} = \bar{0}$, so $\bar{X}$ is a root of $f$, and hence $x - \bar{X}$ is a factor of $f$ in $K[x]$. Since $\deg f = 2$, it hence has two linear factors. (We don't need to know the other factor to answer the question as written, but we can show that $f(x) = (x - \bar{X})(x - (\bar{X} + \bar{1}))$ in $K[x]$, which we can find readily with polynomial long division.)
