Great question! First, let's tackle just the case where the roots of $Q(x)$ are all distinct. One way to conceptualize what's going on is the following: if $r$ is a root of $Q(x)$, then as $x \to r$, the function $f(x) = \frac{P(x)}{Q(x)}$ (we always assume $P, Q$ have no common roots) goes to infinity. How quickly does it go to infinity? Well, write $Q(x) = (x - r) R(x)$. Then
$$\frac{P(x)}{Q(x)} = \frac{1}{x - r} \left( \frac{P(x)}{R(x)} \right)$$
and $R(r) \neq 0$. So we see that as $x \to r$, this expression goes to infinity like $\frac{1}{x - r}$; more precisely, it goes to infinity like $\frac{c_r}{x - r}$ where $c_r = \frac{P(r)}{R(r)}$. This number is referred to in complex analysis as the residue of the pole at $x = r$. So the upshot of all of this is that we can subtract this pole away and write
$$f(x) - \frac{c_r}{x - r} = \frac{1}{x - r} \left( \frac{P(x)}{R(x)} - \frac{P(r)}{R(r)} \right).$$
The expression in parentheses approaches $0$ as $x \to r$, and in fact it is a rational function whose numerator is divisible by $x - r$, so we can actually divide by $x - r$. The result is a new rational function which no longer has a pole at $r$.
We can repeat this process for every root of $Q$ until we get a rational function with no poles whatsoever. But this must be a polynomial. So now we've written $f$ as a sum of fractions of the form $\frac{c_r}{x - r}$ plus a polynomial. (Note that in general we need to consider the complex roots of $Q$.)
Okay, so what if $Q$ has repeated roots? Then $f$ might go to infinity more quickly as $x \to r$. If $r$ is a root with multiplicity $n$, then writing $Q(x) = (x - r)^n R(x)$ we now have
$$\frac{P(x)}{Q(x)} = \frac{1}{(x - r)^n} \left( \frac{P(x)}{R(x)} \right)$$
where $R(r) \neq 0$. So we see that as $x \to r$, this expression goes to infinity like $\frac{1}{(x - r)^n}$; more precisely, it goes to infinity like $\frac{c_{r,n}}{(x - r)^n}$ where $c_{r, n} = \frac{P(r)}{R(r)}$. So we can do the same thing as before and just subtract this off, getting
$$f(x) - \frac{c_{r,n}}{(x - r)^n} = \frac{1}{(x - r)^n} \left( \frac{P(x)}{R(x)} - \frac{P(r)}{R(r)} \right).$$
The expression in parentheses approaches $0$ as $x \to r$, so again it is divisible by $x - r$, but this time we're not done! We still have to subtract off terms that look like $\frac{1}{(x - r)^k}$ where $k < n$ until the resulting rational function no longer goes to infinity as $x \to r$.
The above is nice as far as it goes, but let me mention that algebraically partial fraction decomposition is more general than rational functions over $\mathbb{C}$. It also generalizes to, for example, rational numbers! Like rational functions, rational numbers also have partial fraction decompositions, like
$$\frac{5}{12} = \frac{2}{3} - \frac{1}{4}.$$
Explaining all of this in a unified framework requires the language of abstract algebra, in particular the notion of a group, of a field, and of a principal ideal domain. Partial fraction decomposition in this setting describes the additive group of the field of fractions of a principal ideal domain using essentially the Chinese remainder theorem, but that's a story for another day...
(It might not seem like uniqueness of partial fraction decomposition holds for rational numbers because we can also write $\frac{5}{12} = \frac{3}{4} - \frac{1}{3}$, but the correct notion of uniqueness here is subtle; it is uniqueness "mod $1$.")
Decomposing $1/(x+n)(x^2 + k)$ is only difficult (for me) when k is a positive quantity. Otherwise just use $a^2 - b^2$. For example:
$$\frac1{(x-3)(x^2-1)} = \frac1{(x-3)(x-1)(x+1)} = \frac{A}{x-3} + \frac{B}{x-1} + \frac{C}{x+1}$$
Here,
$$A = \frac1{(3-1)(3+1)} = \frac1{8}\\
B = \frac1{(1-3)(1+1)} =\frac{-1}{4} \\
C = \frac1{(-1-3)(-1-1)} = \frac{1}{8}$$
Now, for the guy I don't like:
$$\frac{1}{(x^2 +1)(x-3)} = \frac{Ax+B}{x^2 +1} + \frac{C}{x-3}$$
Try some easy values of $x$, I'll take $x=0$,
$$ B + \frac{C}{3} = \frac{-1}{3} \quad\dots(1)$$
Now, I'll take $x= 1$,
$$\frac{A + B}{2} - \frac{C}{2} = -\frac{1}{4} \quad\dots(2)$$
Let $x=2$,
$$\frac{2A + B}{5} - C= -\frac1{5}\quad\dots(3)$$
Solving $(1)$ and $(2)$ and $(3)$, we get
$$A = -\frac{1}{10},\, B = -\frac{3}{10},\, C = \frac{1}{10} $$
This is the easiest (and only) method I know to find partial fractions and it helps me immensely in indefinite integration.
Best Answer
The partial fraction expansion is
$$\frac{1}{(2+j\omega)^2(4+j\omega)}=\frac{A}{2+j\omega}+\frac{B}{(2+j\omega)^2}+\frac{C}{4+j\omega}\tag 1$$
To find $C$, we multiply both sides of $(1)$ by $4+j\omega$ and take the limit as $j\omega \to -4$. Proceeding, we find
$$\begin{align} \lim_{j\omega \to -4}\frac{(4+j\omega)}{(2+j\omega)^2(4+j\omega)}&=\frac14\\\\ &=\lim_{j\omega \to -4}\left(\frac{A(4+j\omega)}{2+j\omega}+\frac{B(4+j\omega)}{(2+j\omega)^2}+\frac{C(4+j\omega)}{4+j\omega}\right)\\\\ &=C \end{align}$$
To find $B$ , we multiply both sides of $(1)$ by $(2+j\omega)^2$ and take the limit as $j\omega \to -2$. Proceeding, we find
$$\begin{align} \lim_{j\omega \to -2}\frac{(2+j\omega)^2}{(2+j\omega)^2(4+j\omega)}&=\frac12\\\\ &=\lim_{j\omega \to -2}\left(\frac{A(2+j\omega)^2}{2+j\omega}+\frac{B(2+j\omega)^2}{(2+j\omega)^2}+\frac{C(2+j\omega)^2}{4+j\omega}\right)\\\\ &=B \end{align}$$
To find $A$ , we multiply both sides of $(1)$ by $(2+j\omega)^2$, take a derivative with respect to $j\omega$, and take the limit as $j\omega \to -2$. Proceeding, we find
$$\begin{align} \lim_{j\omega \to -2}\frac{d}{d(j\omega)}\left(\frac{(2+j\omega)^2}{(2+j\omega)^2(4+j\omega)}\right)&=-\frac14\\\\ &=\lim_{j\omega \to -2}\frac{d}{d(j\omega)}\left(\frac{A(2+j\omega)^2}{2+j\omega}+\frac{B(2+j\omega)^2}{(2+j\omega)^2}+\frac{C(2+j\omega)^2}{4+j\omega}\right)\\\\ &=A \end{align}$$