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
 A: [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
A: 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
$$\begin{align*}\frac{1}{2}\left(f(x)+f(-x)\right)&=\frac{1}{2}\left(\frac{Ax+B}{x^2+4}+\frac{Cx+D}{x^2+9}+\frac{-Ax+B}{x^2+4}+\frac{-Cx+D}{x^2+9}\right)\\
&=\frac{B}{x^2+4}+\frac{D}{x^2+9}\end{align*}.$$
A: $$
\frac{3}{(x^2+4)(x^2+9)} = \frac{3}{(u+4)(u+9)} = \frac{B}{u+4} + \frac{D}{u+9}
$$
A: Opening up the right hand side, we get:
\begin{align*}
\frac{Ax+B}{(x^2+4)}+\frac{Cx+D}{(x^{2}+9)} &=\frac{(Ax+B)(x^{2}+9)+(Cx+D)(x^2+4)}{(x^2+4)(x^{2}+9)} \\
&=\frac{(A+C)x^3+(B+D)x^2+(9A+4C)x+(9B+4D)}{(x^2+4)(x^{2}+9)}
\end{align*}
If this would have to equal $\frac{3}{(x^2+4)(x^{2}+9)}$ for all $x\in\mathbb{R}$, then obviously $A+C=0$, $B+D=0$, $9A+4C=0$ and $9B+4D=3$. From the two equations concerning the relationship of $A$ and $C$ it follows that $A=C=0$.
A: Actually you don't have to expand the whole equation to see why A and C must be 0. Just imagine that you have multiplied both side already by the two denominators. Then the left-hand side would just be 3.
Now looking at the right-hand side, if A and C were non-zero, then the right-hand side would contain terms of the third degree (because a first-degree term of the numerator multiplied by the second-degree term of the denominator equals a third-degree term). Namely the coefficients of the third degree would be A+C. Since the left-hand side does not have any third degree terms, A+C = 0.
Therefore, we know that A and C are of equal magnitude but with opposite signs. Now look at the first-degree term on the right-hand side and we know that 9A+4C=0, which means A and C can't actually be of equal magnitude if they were non-zero. Therefore they are both zero.
So your textbook could simply have said examine the odd powers of x on the right-hand side of the equation to anticipate that A=C=0.
