Express in partial fraction form: $\displaystyle{\frac{2x}{(x+2)^2}}$
I think is $\displaystyle{\frac{2x}{(x+2)^{2}} = \frac{A}{x+2}+\frac{B}{(x+2)^2}}$
However when identifying $A$ and $B$, I'm not sure how to calculate A.
E.g. $$2x = A\cdot (x+2) + B$$
Substitute $x=-2$
$2\cdot(-2)$ = $A\cdot (2-2) +B$
$-4 = B$
In other questions there is always another factor to multiply by at this stage.
If you go on to rewrite $2x=A(x+2)+B$ as $2x=Ax+(2A+B)$, then you can match up the coefficients of the polynomials (of $x$) on each side. You must have
$$2=A\qquad\text{and}\qquad 0=2A+B$$
Now solve for $A$ and $B$. The first equation (obviously!) says $A=2$. When you plug this into the second, obtaining $0=2\cdot2+B$, you get $B=-4$.
Incidentally, where you write "$2x={A\over x+2}+{B\over(x+2)^2}$," the left hand side of the equation should be $2x\over(x+2)^2$, not just $2x$.
Starting from (as you almost wrote) $$\frac{2x}{(x+2)^2} = \frac{A}{x+2}+\frac{B}{(x+2)^2},$$ rewrite as $$\frac{2x}{(x+2)^2} = \frac{A(x+2)+B}{(x+2)^2} = \frac{Ax+(2A+B)}{(x+2)^2}.$$ From this, one sees that $2x = Ax + (2A+B)$, so that $A=2$ and $2A+B=0$ and thus $B=-4$ (as you correctly derived).
An alternative method, using substitution, starts again from $$\frac{2x}{(x+2)^2} = \frac{A(x+2)+B}{(x+2)^2}$$ so that $2x = A(x+2)+B$. Substituting $x=-2$ gives $B=-4$ and thus $2x = A(x+2)-4$. Now substitute any value other than $-2$ (say $0$) for $x$, giving $0 = 2A-4$, so that $A=2$.
May be somehow tricky but I love this kind ... $$\quad{\frac{2x}{(x+2)^2} = \\\frac{2(x)}{(x+2)^2} = \\ \frac{2(x+2)-4}{(x+2)^2} = \\ \frac{2(x+2)}{(x+2)^2} +\frac{-4}{(x+2)^2} = \\ \frac{2}{(x+2)} +\frac{-4}{(x+2)^2} \\}$$
Arriving at the formula
$$2x = A\cdot (x+2) + B$$
to reduce the problem to that of solving this polynomial equation for $A$ and $B$ is good start to the problem. So is solving for the value of $B$.
The method you're using to solve this equation — substituting $x = -2$ into your polynomial equation — is a shortcut.
You don't have to use the shortcut. (especially in a situation where it doesn't help!)
The two most common options to continue are:
Another simple option is to just solve the equation for $A$ in the normal fashion, after substituting in the known value for $B$. It turns out things cancel out and you get a number at the end.
For instance, you can multiply both sides by $(x+2)^2$ in
$$\frac{2x}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}$$
then
$$(x+2)^2\frac{2x}{(x+2)^2}=(x+2)^2\frac{A}{x+2}+(x+2)^2\frac{B}{(x+2)^2}$$
now simplify
$$(x+2)^2\frac{2x}{(x+2)^2}=(x+2)\cdot A+(x+2)^2\frac{B}{(x+2)^2}$$
and you can take the limit as $x\to 2$ both sides to show that:
$$ B = \underset{x\to -2}{\lim}{\ (x+2)^{2}\frac{2x}{(x+2)^2}}=-4. $$
for $A$ take before the limit the derivative w.r.t $x$ in both sides to get
$$ A = \underset{x\to -2}{\lim}{\ \frac{d}{dx}{2x}=2. }$$
Recall that, if $\ \displaystyle{f(x)=\frac{p(x)}{(x-r)^{k}}}$ and $(x-r)$ doesn't divide $p(x)$ you have: $$f(x)=\frac{p(x)}{(x-r)^{k}}=\sum_{s=1}^{k}\frac{A_{s}}{(x-r)^{s}}$$ if you multiply by $(x-r)^{k}$ in both sides you will have: $$ (x-r)^{k}f(x)=p(x)=\sum_{s=1}^{k}A_{s}\cdot(x-r)^{k-s}$$ and from that you have: $$ A_{k}=\underset{x\to r}{\lim}{\ (x-r)^{k}f(x)} $$ and the others $$A_{i}=\frac{1}{(k-i)!}\underset{x\to r}{\lim}{\ \frac{d^{k-i}}{dx^{k-i}}\left((x-r)^{k}f(x)\right)}$$
You're actually almost there. Just substitute your inferred value for $B$ into your equation, and you get $$2x=A(x+2)-4$$ ...now rearrange... $$\require{cancel} A=\frac{2x+4}{x+2}=\frac{2\cancel{(x+2)}}{\cancel{x+2}}=2$$ ..done!
There is a nice method for rational functions that only have one repeated factor at the denominator: let the fraction be $f(x)=\frac{N(x)}{D(x)}$ where $D(x)=d(x)^k$.
The general form of the partial fraction decomposition is
$$\displaystyle f(x)=\frac{a_1(x)}{d(x)}+\frac{a_2(x)}{d(x)^2}+\dots+\frac{a_k(x)}{d(x)^k}$$
Then, if you perform the division of $N(x)$ by $d(x)$, you can write $f(x)$ as
$$\displaystyle f(x)=q(x)d(x)+r_1(x)$$
where $\deg(r_1)<\deg(d)$. We then got
$$\displaystyle f(x)=\frac{q(x)d(x)+r_1(x)}{d(x)}=\frac{r_1(x)}{d(x)}+\frac{q(x)}{d(x)^{k-1}}$$
so $r_1$ is actually $a_1$. By repeating the process with $q$, we can get all the other functions $a_2$,...,$a_k$.
Example with $\frac{2x}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}$: $2x=2(x+2)-4$, so $A=-4$ and $q(x)=2$. As $q(x)=0(x+2)+2$, we obviously have $B=2$. We obtain
$$f(x)=\frac{2(x+2)-4}{(x+2)^2}=\frac{2}{(x+2)}+\frac{-4}{(x+2)^2}$$
When $d(x)$ has degree $1$, we can use Synthetic Division to perform the division, so it is very fast.
$$2x = A(x+2) + B$$
You just found that $B=-4$ and now you want to find the value of $A$. Well, let $B=-4$ and see what happens...
\begin{align} 2x &= A(x+2)-4 \\ 2x+4 &= A(x+2) \\ 2(x+2) &= A(x+2) \end{align}
Since this has to be true for all $x$, then $A=2$.
This is not a lucky coincidence. Substituting back into the original equation will always result in something solvable.
