Let F be a field and let f(x) exist in F[x] such that deg f(x) = 3. Prove that f(x) is reducible in F[x] if and only if f(x) has a root in F.
Well I can visualize this proof with graphs, but I am unable write it in a proof. It makes sense that something that is reducible with a deg of 3 would cross the x-axis 2 times, which would make it reducible. But all of this is not formal. How do I begin?
You can try one direction at a time. Suppose that $\displaystyle f$ is reducible. This means that you can write $\displaystyle f(x)=g(x)h(x)$ where the degree of $\displaystyle g$ and $\displaystyle h$ is at least 1 and less than 3. Can you say something about the degrees of $\displaystyle g$ and $\displaystyle h$ in particular?
well since deg f(x) is 3 the deg g(x) and deg h(x) can only be either 1 or 3, since exponents are added together when multiplying. So the forward method is making more sense but going the other direction with the roots throws me off. I know that f(x) will cross the axis 3 times but how does that declare that it is reducible?
The degrees actually have to be 1 and 2. Since you have a degree 1, it is of the form $\displaystyle x-a$ for some $\displaystyle a$. Therefore, a root is $\displaystyle x=a$.
For the other direction, you should recall the factor theorem. If $\displaystyle f(x)$ is a polynomial and $\displaystyle f(a)=0$ (i.e. $\displaystyle a$ is a root), then $\displaystyle f(x)=(x-a)g(x)$ for some polynomial $\displaystyle g$. The result is immediate.