Finding partial fractions expansions mentally

$$ \frac{6x^4 - 12 x^3 + 5 x^3 - 10 x^2 + 8 x^2 - 16 x + 2 x^2 + 3x-6 }{x^2(x-2)} $$

$$ \frac{(6x^4 - 12 x^3) + (5 x^3 - 10 x^2) + (8 x^2 - 16 x) + 2 x^2 + (3x-6) }{x^2(x-2)} $$

My favorite high school math teacher called the rule a "propitious zero," adding and subtracting the same thing to get a more attractive grouping.

In the order I wrote things, $$ 6x + 5 + \frac{8}{x} + \frac{2}{x-2} + \frac{3}{x^2} $$

The residue theorem, for instance. If $f(x)$ is a polynomial with degree $4$, $$ g(x)=\frac{f(x)}{x^2(x-2)} = \frac{A}{x^2}+\frac{B}{x}+\frac{C}{x-2}+D+Ex \tag{1}$$ is granted, and $$ A=\text{Res}\left(x\cdot g(x),x=0\right),\quad B = \text{Res}(g(x),x=0),\quad C=\text{Res}(g(x),x=2)\tag{2} $$ are straightforward to compute through simple limits. Then, by computing $$ g(x)-\frac{A}{x^2}-\frac{B}{x}-\frac{C}{x-2}, \tag{3} $$ that we know in advance to be a polynomial with degree $\leq 1$, we recover $D$ and $E$, too.

As an alternative, it is enough to notice that: $$\begin{eqnarray*}6x^4-7x^3-13x-6 &=& 6x\cdot x^2(x-2)+5x^3-13x-6\\&=&6x\cdot x^2(x-2)+5\cdot x^2(x-2)+10x^2-13x-6 \end{eqnarray*}$$ and: $$ 10x^2-13x-6 = 10\cdot x^2 + 3(x-2)-10 x,\quad \frac{1}{x(x-2)}=\frac{1}{2}\left(\frac{1}{x-2}-\frac{1}{x}\right).$$

At a schematic level, what makes this problem "ugly" is the $x-2$ in the denominator. So the problem solving strategy is to isolate the $x-2$; the rest should then be easy to do mentally. To this end, you know from the principle of partial fraction decomposition that $\frac{6x^4-7x^3-13x-6}{x^2(x-2)}$ can be written in the form $\frac{p(x)}{x^2} + \frac{c}{x-2}$ for some constant $c$. Thus, $$6x^4-7x^3-13x-6 = p(x)(x-2) + cx^2,$$ so in particular $6x^4-7x^3 - cx^2 -13x-6$ is divisible by $x-2$.

There is a standard algorithm for determining when a polynomial $a x^4 + bx^3 + cx^2 + dx+e$ is divisible by $x-2$: It is divisible if and only if $e+2(d+2(c+2(b+2(a)))) = 0$. In the case at hand, this translates to $0=-6+2(-13+2(-c+2(-7+2(6)))) = 8-4c$, which means $c=2$.

Now we can determine $p(x) = \frac{6x^4-7x^3-2x^2-13x-6}{x-2} = 6x^3 + 5x^2 + 8x + 3$. Dividing this by $x^2$ gives all the other terms in the decomposition of the original fraction.

There are really only two computations to do here: figuring out $c$, and dividing $6x^4-7x^3-2x^2-13x-6$ by $x-2$, both of which can reasonably be done in one's head.