algebra precalculus - 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

Answer

[Migrated from comment] It's simpler than that really. If you put $y=x^2$ you can get a partial fraction decomposition in terms of $y$. You wouldn't think of trying to include a square root term (corresponding to $x$) in the numerator. You know that the $y$ version can be done in terms of y and you get the $x^2$ version by substituting back