<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] 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 $A=C=0$.
</s>