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 $D$?

take LCM of second expression 
you get and sum it
{A(2+x)(4+x)^2 +B(4+x)^2+C(2+x)^2(4+x)+D(2+x)^2}/{(x+2)^2(x+4)^2}= your 1st expression 
so you have {A(2+x)(4+x)^2 +B(4+x)^2+C(2+x)^2(4+x)+D(2+x)^2}=1 
compare coefficients of x^3 ,x^2,x^1 and x^0 which on right side of  aboveequation are equal to 0,0,0 and 1 respectively.
you have 
A+C=0
8A+B+4C+D=0
32A+8B + 20C+ 4D=0
32A + 16B + 16C +4D=1
soLVING YOU GET 
B=1/12, D=7/12 ,A =-1/6, C=1/6
