<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] Assign different values to $x$, like $-1, 0, +1, +2$ and solve the linear system of 4 equations in 4 unknowns.
$$\left[\begin{array}{cccc}
1/1& 1/1& 1/3& 1/9\\
1/2& 1/4& 1/4& 1/16\\
1/3& 1/9& 1/5& 1/25\\
1/4& 1/16& 1/6& 1/36
\end{array}\right]
\left[\begin{array}{c}
A\\
B\\
C\\
D\end{array}\right]=
\left[\begin{array}{c}
1/9\\
1/64\\
1/225\\
1/576\end{array}\right]$$
$$\left[\begin{array}{c}
A\\
B\\
C\\
D\end{array}\right]=
\left[\begin{array}{c}
-1/4\\
1/4\\
1/4\\
1/4\end{array}\right]$$
</s>