Question on the complex plane I am recently reviewing some basic abstract algebra, and I got a new understanding of what the complex plane, $\mathbb{C}$, is.
In $\mathbb{R}$, there are many polynomials that are irreducible, i.e., they don't have any root. So in order to make every polynomial reducible, we need to apply the theorem that every irreducible polynomial, $p(x)$, has a root in a field extension of $\mathbb{R}[x]$:
$$\mathbb{R}[x]/\left<p(x)\right>$$
I understand that $x$ in this quotient group is defined by $x^2=-1$ (which is actually the definition of $i$) and it can be regarded as a basis in the vector space $\mathbb{R}[x]/\left<x^2+1\right>$.
Now, my question is, how to prove that every polynomial is reducible in $\mathbb{R}[x]/\left<x^2+1\right>\cong \mathbb{C}$?
My intuitive guess is that all polynomials of degrees higher than $2$ can be written as a linear form $kx+b$ where $k,b\in\mathbb{R}$, which therefore implies it has a solution in $\mathbb{R}[x]/\left<x^2+1\right>$. Is this an available idea? Can we do this rigorously?
 A: Moved here from comments (with slightly more detail since I have the space):
In $\mathbb{R}$, $x^2+1$ is the only obstruction to every polynomial having a root. It is somewhat magical (and very particular to $\mathbb{R}$!) that allowing this one polynomial to split actually lets every polynomial split. 
As a counterexample, consider $\mathbb{Z}/2$. Here $x^2+1 = (x+1)(x+1)$ splits, and yet there are polynomials which still don't factor. Indeed for every degree $d$ there is an irreducible polynomial of degree $d$.
For another, perhaps more relatable example, we can make $x^2 + 1$ split in $\mathbb{Q}$ by passing to the field $\mathbb{Q}(i)$. However in this new field 
$x^2 - 2$ still doesn't factor, since $\sqrt{2}$ is not in it. 
In both of these examples, we can still pass to an algebraic closure in which every polynomial factors into linear terms. Unfortunately, it is much more complicated to build one of these algebraic closures than to build $\mathbb{C}$ from $\mathbb{R}$ by adding $i$ (cf. this article by Keith Conrad on the topic).

I hope this helps ^_^
