Partial fraction (doubt) I have this partial fraction
$$\displaystyle\frac{1}{(2+x)^2(4+x)^2}$$
I tried to resolve using this method:
$$\displaystyle\frac{A}{2+x}+\displaystyle\frac{B}{(2+x)^2}+\displaystyle\frac{C}{4+x}+\displaystyle\frac{D}{(4+x)^2}$$
$$1=A(2+x)(4+x)^2+B(4+x)^2+C(4+x)(2+x)^2+D(2+x)^2$$
When x=-2
$$1=B(4-2)^2$$
$$B=\displaystyle\frac{1}{4}$$
When x=-4
$$1=D(2-4)^2$$ 
$$D=\displaystyle\frac{1}{4}$$ 
When x=0
$$1=A(2)(16)+B(16)+C(4)(4)+D(4)$$
$$1=A(32)+B(16)+C(16)+D(4)$$
Replacing the values for B y D
$$1=A(32)+4+C(16)+1$$
$$1-4-1=A(32)+c(16)$$
$$-4=A(32)+C(16)$$
How I can get the values ​​of $A$ and $D$?

Heaviside cover-up is an alternative technique worth exploring.
It directly gives $\begin{align}B = \frac{1}{(4+x)^2}\Bigg|_{x=-2} =  \frac{1}{4} \text{ and } D = \frac{1}{(2+x)^2}\Bigg|_{x=-4} =  \frac{1}{4}\end{align}\tag{1}$
Next, it requires that the $B$ and $D$ terms be moved to the other side.
The partial fraction identity then becomes 
$\begin{eqnarray}\frac{A}{2+x} + \frac{C}{4+x} &= &\frac{1}{(2+x)^2(4+x)^2} - \frac{1/4}{(2+x)^2} - \frac{1/4}{(4+x)^2}\\&=&\frac{4-(4+x)^2-(2+x)^2}{4(2+x)^2(4+x)^2}\\&=&\frac{-2(8 + 6x + x^2)}{4(2+x)^2(4+x)^2}\\&=&\frac{-1/2}{(2+x)(4+x)}\end{eqnarray}$
which gives $\begin{align}A = \frac{-1/2}{(4+x)}\Bigg|_{x=-2} =  -\frac{1}{4} \text{ and } C = \frac{-1/2}{(2+x)}\Bigg|_{x=-4} =  \frac{1}{4}\end{align}\tag{2}$
Remark: Note how this technique eliminates the need to solve a system of linear equations for finding out the partial fraction decomposition coefficients (which can often be tedious).
See here for some more examples.
