TITLE: Philosophy of simple field extensions
QUESTION [12 upvotes]: In B. L. van der Waerden's Algebra stuck on the problem 6.9:

The polynomial $f(x) = x^4 + 1$ is irreducible in the field of rationals.
  Adjoin a root $\theta$ and resolve the polynomial in the extended
  field $\mathbb Q (\theta)$ into prime factors.

Seems like I haven't nailed the idea of extending fields, so I'm asking to control my thoughts and help with troubled places.
We can obtain the desired extension by two ways: either we use the fact, which states that we know, in which field would that polynomial have a root (nonsymbolic adjunction), or we can build residue class field modulo that polynomial (symbolic adjunction).
I don't realize what I'm supposed to do in both cases.

In nonsymbolic way, should I just completely factorize $f(x)$ over $\mathbb C$, or am I to find only one (any shall do) root?
I can write
$f(x) = (x^2-i)(x^2+i) = (x-\sqrt i)(x + \sqrt i)(x - \sqrt{-i})(x + \sqrt{-i})$
but what would that give me?

Or I could simply say that $\theta = \sqrt{\sqrt{-1}}$ is an obvious root, divide $f(x)$ by $(x - \theta)$ and think hard what to do next with the quotient $x^3 + \theta x^2 + {\theta}^2 x + {\theta}^3$?
In symbolic way, I have to find residue class field modulo $f(x)$. Is this possible at all? I haven't seen yet a single example of making that with polynomials.

I'm really sorry for the size of the question. The systematic ignorance reminds of itself. However, I feel, that a proper answer will amend many other problems in my knowledge.
I am really grateful at least for the time you spent on reading it.

REPLY [2 votes]: Here we use nonsymbolic adjunction and state some facts that can be verified.
We set
$\tag 1 \displaystyle \theta = e^{\frac{\pi i}{4}}$
since $\theta^4 + 1 = 0$. 
The set
$\tag 2 \Bbb Q [\theta] = \{ f(\theta) \mid f(x) \text{ is a polynomial over } \Bbb Q \}$
is closed under the operations of addition and multiplication defined on $\Bbb C$.
The algebraic structure $(\Bbb Q [\theta], + , \times)$ is a field.
There is a natural field inclusion mapping, 
${\displaystyle \iota :\Bbb Q \hookrightarrow \Bbb Q [\theta]}$.
The polynomial $f(x) = x^4 + 1$ can be completely factored into linear factors in the field $\Bbb Q [\theta]$,
$\tag 3 f(x) = (x - \theta) (x - \theta^3)(x - \theta^5)(x - \theta^7)$