<s>[INST] Explain this logic to me please $$
\  \frac{3}{(x^2+4)(x^2+9)} = \frac{Ax + B}{(x^2+4)} + \frac{Cx+D}{(x^2+9)}
$$
Instructions say that "we can anticipate that $$ A = C = 0,$$ because neither the numerator nor the denominator involves odd powers of x, whereas nonzero values of A or C would lead to odd degree terms on the right"
I understand what they're saying,  but I don't follow the logic. Can someone please explain in layman's terms? Thanks
 [/INST] The following seems to be in the spirit of the instructions, but precise. (However, the substitution $y=x^2$ is better.)
Let $f(x)$ be the function on the left-hand side. Then $f(-x)=f(x)$ for all $x$, that is, $f(x)$ is an even function. So 
$$f(x)=\frac{1}{2}\left(f(x)+f(-x)\right).$$
Now look at the right-hand side. We have
$$\begin{align*}\frac{1}{2}\left(f(x)+f(-x)\right)&=\frac{1}{2}\left(\frac{Ax+B}{x^2+4}+\frac{Cx+D}{x^2+9}+\frac{-Ax+B}{x^2+4}+\frac{-Cx+D}{x^2+9}\right)\\
&=\frac{B}{x^2+4}+\frac{D}{x^2+9}\end{align*}.$$
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