Describe the multiplication in the ring $\Bbb Q[x]/(x^2+x+1)$. Describe the multiplication in the ring $\Bbb Q[x]/(x^2+x+1)$. What is the multiplicative inverse of $[x]$?
To figure this out, I found that $x = -\frac {1}{2} + i \sqrt {\frac{3}{4}}$. So there clearly aren't any real solutions to this. But I am trying to figure out what elements of this ring look like in order to find an inverse and to see what multiplication looks like. I am new to field extensions and am a little confused as to how to work with them. Any help is appreciated.

There are a few ways to think about it. 
The first is quite literal. The elements of $\frac{\mathbb{Q}[x]}{(x^2+x+1)}$ are residues of polynomials, $\overline{f(x)}$. In this ring $\overline{f(x)} = \overline{g(x)}$ if and only if $f(x)-g(x)$ is divisible by $x^2+x+1$.
To multiply $\overline{f(x)}$ and $\overline{g(x)}$, multiply them as polynomials in $\mathbb{Q}[x]$, then divide by $x^2+x+1$ and take the remainder.

Another way: $$\frac{\mathbb{Q}[x]}{(x^2+x+1)} \cong \{a+b\alpha |\ a,b \in \mathbb{Q}, \text{ and } \alpha^2 + \alpha + 1 = 0 \}$$
$\alpha$ is some "thing" that satisfies the polynomial $x^2 + x +1$. Then multiplication becomes:
\begin{align*} (a + b\alpha)(c + d\alpha) &= ac + (ad + bc)\alpha + bd\alpha^2\\
&= ac + (ad + bc)\alpha + bd(-\alpha - 1)\\
&= (ac-bd) + (ad + bc - bd)\alpha.
\end{align*}

Per user1952009's suggestion:
How could we express $\alpha^3$ and $\alpha^4$ as something of the form $a+b\alpha$?
\begin{align*}
\alpha^3 &= \alpha \alpha^2\\
&= \alpha(-\alpha-1)\\
&= -\alpha^2 - \alpha\\
&= 1.
\end{align*}
In this cool turn of events, we see that $\alpha^3 = 1$. Another way to see that this would happen is that $x^3-1 = (x-1)(x^2+x+1)$ and since $\alpha$ satisfies $x^2+x+1$, it will then satisfy $x^3-1$.
From this we see that $\alpha^4 = \alpha$ and in fact for $n \in \mathbb{N}$:
$$
\alpha^n = \begin{cases}
1 \ &\text{ if } n \equiv 0 \pmod 3 \\
\alpha \ &\text{ if } n \equiv 1 \pmod 3 \\
-1 - \alpha \ &\text{ if } n \equiv 2 \pmod 3
\end{cases}
$$
