<s>[INST] 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 $D$?
 [/INST] Try another value of $x$ as well, say $x=2$, giving $1 = 144A + 9 + 64C + 4$, or $144A + 96C = -12$. Combined with the equation you already have, $32A+16C=-4$, solving gives $A=-\frac{1}{4}$, $C=\frac{1}{4}$.
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