Partial Fraction Decomposition When Denominator Has an Irreducible Factor of Degree 2 When doing partial fraction decomposition, there are three cases:
Case 1: Denominator has distinct linear factors.
$f(x) = \dfrac{P(x)}{(x - a_1)...(x - a_k)} = \dfrac{A}{x - a_2} + \dfrac{B}{x - a_k}$ where $a_1, ..., a_k$ are pairwise distinct. 
Case 2: Denominator has repeated linear factors.
$f(x) = \dfrac{P(x)}{(x - a)^c} = \dfrac{B_1}{x - a} + ... + \dfrac{B_c}{(c - a)^c}$
Case 3: Denominator has an irreducible factor of degree 2:
$f(x) = \dfrac{P(x)}{(x - a)(x^2 + bx + c)} = \dfrac{A_1}{x - a} + \dfrac{C_1x + C_2}{x^2 + bx + c}$
In cases 1 and 2, the numerators are all given constant placeholders ($A, B$), but in case 3, the numerator is given a polynomial of degree one as the placeholder ($C_1x + C_2$). I'm curious as to why this is the case. What is the difference between case 1 and 2 and 3 that necessitates this difference in the numerator? What is the reasoning behind this?
I would greatly appreciate it if people could please take the time to elaborate on this.
 A: I will offer a few explanations (some of which are obviously related, but may be more intuitive from one point of view than another.
"Because you need it"
One obvious thing to do is to try without and see what happens. You have, multiplying up and equating coefficients,
$$ 1 = (x^2+2bx+c)A_1 + (x-a)P(x), $$
where $P(x)$ is the numerator of the quadratic fraction. If it's constant, there's no way to eliminate the $A_1 x^2$ without (absurdly) taking $A_1=0$, so the left-hand side would not be constant. If you take $P(x)=C_1x+C_2$, you can get rid of the $x^2$ term.
Asymptotic behaviour
For very large $x$, $x-a \approx x $, $x^2 + 2bx+c \approx x^2$: only the highest-degree terms are important. But this means that
$$ \frac{1}{(x-a)(x^2+2bx+c)} \approx \frac{1}{x^3}, $$
while
$$ \frac{A_1}{x-a}+\frac{C_2}{x^2+2bx+c} \approx \frac{A_1}{x}. $$
Since both sides need to have the same asymptotics, there needs to be an extra term to cancel the $A_1/x$.
Pretend the denominator is factorisable
Finally, we can pretend the denominator is actually factorisable, and see what happens when we put the fractions back together.
$$ \frac{A}{x-a_1} + \frac{B}{x-a_2} = \frac{(A+B)x-a_1 B + a_2 A}{(x-a_1)(x-a_2)}, $$
so if $A+B \neq 0$, there is an $x$ term anyway. (This is actually legitimate, since the Fundamental Theorem of Algebra says any quadratic has two complex roots. Note the $x$ term does not involve the roots at all anyway.) It would be weird if this property didn't carry over to unfactorisable denominators, since we can always just not factor a factorisable quadratic when doing partial fractions.
