[Algebra]
Problem: Let $\mathcal{S}$ denote the set of all real polynomials $A(x)$ with leading coefficient 1 such that there exists a real polynomial $B(x)$ that satisfies

$$
\frac{1}{A(x)}+\frac{1}{B(x)}+\frac{1}{x+10}=\frac{1}{x}
$$

for all real numbers $x$ for which $A(x) \neq 0, B(x) \neq 0$, and $x \neq-10,0$. Compute $\sum_{A \in \mathcal{S}} A(10)$.

Solution: For brevity, $P$ will be used to represent the polynomial $P(x)$, and let $\operatorname{deg}(P)$ represent the degree of $P$. Rewrite the given condition as follows:

$$
\begin{aligned}
\frac{1}{A(x)}+\frac{1}{B(x)}+\frac{1}{x+10}=\frac{1}{x} & \Longrightarrow \frac{A+B}{A B}=\frac{10}{x(x+10)} \\
& \Longrightarrow A B-\frac{x(x+10)}{10} A-\frac{x(x+10)}{10} B=0 \\
& \Longrightarrow\left(A-\frac{x(x+10)}{10}\right)\left(B-\frac{x(x+10)}{10}\right)=\frac{x^{2}(x+10)^{2}}{100} .
\end{aligned}
$$

Because $A$ and $B$ are both polynomials, $A-\frac{x(x+10)}{10}$ must be some factor $F$ of $\frac{x^{2}(x+10)^{2}}{100}$. Furthermore, if $\operatorname{deg}(F) \leq 1$, then $A$ has leading coefficient $\frac{1}{10}$, which violates the condition that $A$ has leading coefficient 1 . So



$\operatorname{deg}(F) \geq 2$. Thus $F$ must be a nonzero constant times one of

$$
\left\{x^{2}, x(x+10),(x+10)^{2}, x^{2}(x+10), x(x+10)^{2}, x^{2}(x+10)^{2}\right\} .
$$

The degree of $A$ determines what this constant must be.

- If $\operatorname{deg}(A) \geq 3$, then $\operatorname{deg}(F)=\operatorname{deg}(A) \geq 3$ and $F$ has leading coefficient 1 . Any such $F$ is valid.
- If $\operatorname{deg}(A)=2$, then $\operatorname{deg}(F)=2$ and $F$ has leading coefficient $\frac{9}{10}$. Again, any such $F$ is valid.
- If $\operatorname{deg}(A) \leq 1$, then $\operatorname{deg}(F)=2$ and $F$ has leading coefficient $-\frac{1}{10}$. But not all $F$ are valid, because this does not guarantee that the leading coefficient of $A$ is 1 . Among $-\frac{1}{10} x^{2},-\frac{1}{10} x(x+10)$, and $-\frac{1}{10}(x+10)^{2}$ as possible values of $F$, only $-\frac{1}{10} x^{2}$ gives a valid $A$ with leading coefficient 1 .

Thus $F$ is one of

$$
\left\{-\frac{1}{10} x^{2}, \frac{9}{10} x^{2}, \frac{9}{10} x(x+10), \frac{9}{10}(x+10)^{2}, x^{2}(x+10), x(x+10)^{2}, x^{2}(x+10)^{2}\right\} .
$$

Then

$$
\sum\left(A(10)-\frac{10 \cdot 20}{10}\right)=-\frac{1}{10} \cdot 10^{2}+\frac{9}{10} \cdot\left(10^{2}+10 \cdot 20+20^{2}\right)+\left(10^{2} \cdot 20+10 \cdot 20^{2}+10^{2} \cdot 20^{2}\right)=46620
$$

so $\sum A(10)=7 \cdot \frac{10 \cdot 20}{10}+46620=\mathbf{4 6 7 6 0}$, as desired.

Final answer: 46760