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 $D$?

If you clear fractions you get $$A(2+x)(4+x)^2+B(4+x)^2+C(4+x)(2+x)^2+D(2+x)^2=1$$
The easy ones are $x=-2$ which gives $B=\frac 14$ and $x=-4$ which gives $D=\frac 14$
Now equate the coefficients of $x^3$ on each side to give $$A+C=0$$ and set $x=0$ (or equivalently equate constant coefficients), to give $$32A+16B+16C+4D=1$$ which becomes $$16A+4+0+1=1$$
Where the $0$ comes from $16(A+C)=0$
I think it is the equating coefficients of $x^3$ - or using a value like $x=1$ which you've missed - you need four data points to identify the four unknown values.
