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
$$\deg(f) \leq \deg(g)+\deg(h),\quad \deg(I) &lt; \deg(g), \quad\deg(J) &lt; \deg(h).$$
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.
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 $\deg(U)+\deg(R)&lt; \deg(P), \deg(V)+\deg(R)&lt;\deg(Q)$

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.
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) $&lt;\deg$(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) $&lt;\deg($irreducible polynomial). In the case of an irreducible quadratic polynomial, this means the numerator will have degree at most $1$.