Forming Partial Fractions Suppose we have:
$ \dfrac{f(x)}{g(x)h(x)} $
and we want to break it down into:
$$ \frac{I(x)}{g(x)} +  \frac{J(x)}{h(x)}$$
and that
$$\deg(f) \leq \deg(g)+\deg(h),\quad \deg(I) < \deg(g), \quad\deg(J) < \deg(h).$$
What is the general way of doing this?
I don't understand the intuition behind having to express;
$\frac{ax^{2}+bx+c }{(dx+e)(f x^{2}+g) } = \frac{A}{dx+e} + \frac{Bx+C}{fx ^{2}+g }$.
Is there a general rule I'm missing?
Thanks.
Edit: Just to clarify, these are all polynomials.
 A: First what is the generalized theorem?

Theorem Given three polynomials $P,Q,R$such that $\gcd(P,Q)=1$ there exists $U,V$ such that:
$$\frac{R(x)}{P(x)Q(x)}=\frac{U(x)}{P(x)}+\frac{V(x)}{Q(x)} $$
with $\deg(U)+\deg(R)< \deg(P), \deg(V)+\deg(R)<\deg(Q)$

Proof It follows easily from Bézout's identity and extended GCD algorithm
As we observe the condition $P$ and $Q$ being relatively coprime are necessary for the validity of the theorem because we can find simple examples which does not satisfy the theorem.
Second how to compute $U$ and $V$?

*

*Algebraic method using the extended GCD algorithm, which is the general method and it always works


*Analytic method, for example, if we want to find $U$ and we know the roots of $P$ we know that $deg(U)\leq deg(P)$ so we only need to find the values of $U$ in at least $deg(P)$ points. This can be done using:
$$I(x)=\frac{R(x)}{Q(x)}=U(x)+\frac{V(x)}{Q(x)}P(x) \tag 1$$
It's easy to see that if we evaluate $I$ in the roots of $P$ we can find the values of $U$, if we have multiple roots we also evaluate the derivatives of $I$.


*There are other methods using for example the roots of $Q$ in formula $(1)$ in which case we use limits and evaluating in some remarkable points such as $a,1,-1$ the roots of $P,R,Q$.
Third why we need this decomposition, there are a lot of applications of the decomposition like this in computing integrals and solving differential equations.
A: As long as the denominator has only simple irreducible factors, it's rather straightforward. Here, for instance, I'll suppose $f,g$ have the same sign, so that $fx^2+g$  is irreducible and it has complex roots. The general strategy is to:

*

*Multiply both sides of equality by $(dx+e)(fx^2+g)$. You obtain a polynomial identity:
$$ax^2+bx+c=A(fx^2+g)+(Bx+C)(dx+e).$$

*Set $x=-\dfrac ed\,$ and get an equation for $A$.

*Set $x=\mathrm i\mkern 1.5mu\sqrt{-\dfrac gf}$. You get an equation with complex coefficients for $d$ and $e$, i.e. two equations with real coefficients.

Also remember you first have to divide (with remainder) the numerator by the denominator, so as to have a proper  rational function ($\deg$(numerator) $<\deg$(denominator)).
For a proper rational function, the decomposition theorem says they are are sums of proper rational functions with primary denominators (=powers of a single  irreducible polynomial) and $\deg$(numerator) $<\deg($irreducible polynomial). In the case of an irreducible quadratic polynomial, this means the numerator will have degree at most $1$.
