Suppose $\;f(x)=a_0+a_1x+\cdots+a_{n-1}x^{n-1}\;,\;\;a_0\neq0\;$ , so $\gcd(x^n,\,f(x))=1$ , and thus there exist $\;p(x),\,q(x)\in\Bbb F[x]\;$ with $\;p(x) x^n+q(x) f(x)=1\;$ , and then
$$(f(x)+(x^n))(q(x)+(x^n))=1+(x^n)$$
I have to admit that I have hard time to really understand the philosophy behind your question. Let's try however to provide feedback.
For whatever field $K$ and an irreducible polynomial $f \in K(x)$, $(f) \subseteq K(x)$ is a prime ideal and $K(x)/(f)$ a field. Moreover if $\overline K$ is an algebraic closure of $K$ and $\theta \in \overline K$ a root of $f$, the rupture field $K(\theta)$ is isomorphic to the field $K(x)/(f)$ and if $\varphi$ is such isomorphism, $\varphi(\theta) = \overline x$ (the class of $x$).
Coming back to your example, we have $K = \mathbb Q$, $\overline K= \mathbb C$ and $f(x) = x^4+1$.
If I understand well, what you name the non symbolic adjunction is $\mathbb Q(\theta)$ and the symbolic one $\mathbb Q(x)/(f)$. Working in one or the other is equivalent. It just means working on one side or the other of the isomorphism $\varphi$. At the end the important topic is that $\theta^4 +1 = 0$ in $\mathbb Q(\theta)$ and ${\overline x}^4+1 = 0$ in $\mathbb Q(x)/(f)$... where in last equality $\overline x$ stands for the class of $x$ in $\mathbb Q(x)/(f)$!
Now, the question you're asked to solve is: factor the quotient $f/(x-\theta)$ in $\mathbb Q(\theta)$. To be coherent with the question and to avoid manipulating equivalence classes, I think it is easier to work in what you call the non symbolic adjunction $\mathbb Q(\theta)$.
Knowing that $f(x)=x^4+1 =(x-\theta)(x + \theta)p(x)$ where $p(x) = x^2+\theta^2$, the problem boils down to decide whether $p$ is irreducible over $\mathbb Q(\theta)$ or have another root in that field. In $\mathbb C$, the set of roots of $f$ is $S=\{\theta, -\theta, i\theta, -i \theta\}$ as $i$ is a primitive fourth root of unity.
You'll verify that both $i \theta$ and $-i\theta$ are roots of $p$ as $\left( \pm i \theta\right)^2=-1 \theta^2=i^2 \theta^2 = -\theta^2$. Hence $f$ factors as
$$f(x) = (x-\theta)(x+\theta)(x-i\theta)(x+i\theta)$$ in $\mathbb Q(\theta)$. As an additional bonus, you gain that $i^2 = \theta^4$ and therefore $i$ is either $\theta^2$ or $-\theta^2$. Which can also be retrieved as all roots of $f$ in $\mathbb C$ are eighth root of unity as mentioned in other responses. Therefore we have the factorization in prime factors
$$f(x) = (x-\theta)(x+\theta)(x-\theta^3)(x+\theta^3)$$ of $f$ in $\mathbb Q(\theta)$.
Notes: an important thing to notice, and that I used above, is that
$-\theta$ is also a root of $f$ and therefore of $f/(x-\theta)$.
For the fun, let's work in the so called symbolic adjunction $L = \mathbb Q(x)/(f)$.
In $L$, you have $f(\overline x) = {\overline x}^4+1 = 0$. I omit the bars over the elements of $\mathbb Q$, which I shouldn't...
Also in $L(y)$:
$$y^4+1 = (y-{\overline x})(y+{\overline x})(y^2 + {\overline x}^2)$$ and in $L$
$$\left({\overline x}^3\right)^2 + {\overline x}^2= {\overline x}^6 +{\overline x}^2= {\overline x}^4 {\overline x}^2 +{\overline x}^2= -{\overline x}^2+{\overline x}^2=0$$
You retrieve the fact that ${\overline x}^3$ is a root of $p(y) = y^2 +{\overline x}^2$. Finally, the factorization of $f(y)$ in $L(y)$ in prime factors is:
$$f(y) = (y-{\overline x})(y+{\overline x})(y-{\overline x}^3)(y+{\overline x}^3)$$
Best Answer
Not sure what is the easiest solution, but here's one I like.
Since $g$ is of positive degree, $R=T[x]/(g)$ is a finite-dimensional vector space over $T$. Let $a_0\in R$ be an element which is not a zero divisor. Then the map $R\to R$ given by the formula $a\mapsto a_0a$ is injective and $T$-linear, and thus it is surjective (since $R$ is a finite-dimensional vector space over $T$). In particular, there is some $a$ such that $a_0a=1$.