Let $k$ be a field, and $K=k(t)$ be the field of rational functions (where $t$ is indeterminate). Let $F=k(t^2)$. A typical element of $F$ will look like: $$ \frac{\displaystyle \sum_{i=0}^{n} a_{2i} t^{2i}}{\displaystyle \sum_{j=0}^{m} b_{2i} t^{2i}} \ \ \ \textrm{where } a_{2i}, b_{2i}\in k $$ It is a fact that $F\subseteq K$ is a field extension of degree $2$, where $\{1, t\}$ is a basis of $K$ over $F$.
I 'understand' the proof of this fact (which seems to rely on the fact that $g(x)=x^2-t^2 \in F[x]$ is the minimal polynomial of $t\in K$ over $F$). But I am interested in how to explicitly write elements of $K$ in the basis $\{1, t\}$. For example, how would you write $$ \frac{1}{t^5+t^4+4t^3+7t^2+3} $$ as $b_{1} t + b_{0}$ where $b_{0}, b_{1}\in F$? I have reduced this problem to the following:
Let $f(x)$ be a polynomial in $x$ over some field. Then there exists another polynomial $g(x)$ (over the same field) such that $f(x) g(x)$ has no terms $x^{m}$ with $m$ odd.
How would I go about proving this (if it is true..)?
If $f\in K$ then $$f(t)=\frac{f(t)+f(-t)}{2}\cdot1+\frac{f(t)-f(-t)}{2t}\cdot t,$$ where $\frac{f(t)+f(-t)}{2}$ and $\frac{f(t)-f(-t)}{2t}$ are elements of $F$ because they are both even polynomials.
If $f(x)$ is given then we can take $g(x):=(-1)^{\text{deg}(f)}f(-x)$. We will have $f(x)g(x)=(-1)^{\text{deg}(f)}f(x)f(-x)$, which is even, and therefore has only even powers.