9. Let $P(x)=x^{2020}+x+2$, which has 2020 distinct roots. Let $Q(x)$ be the monic polynomial of degree $\binom{2020}{2}$ whose roots are the pairwise products of the roots of $P(x)$. Let $\alpha$ satisfy $P(\alpha)=4$. Compute the sum of all possible values of $Q\left(\alpha^{2}\right)^{2}$.
Answer: $2020 \cdot 2^{2019}$
Solution: Let $P(x)$ have degree $n=2020$ with roots $r_{1}, \ldots, r_{n}$. Let $R(x)=\prod_{i}\left(x-r_{i}^{2}\right)$. Then
$$
\prod_{i} r_{i}^{n} P\left(\frac{x}{r_{i}}\right)=\prod_{i} \prod_{j}\left(x-r_{i} r_{j}\right)=Q(x)^{2} R(x) .
$$

Using $R\left(x^{2}\right)=(-1)^{n} P(x) P(-x)$ and Vieta, we obtain
$$
P(x) P(-x) Q\left(x^{2}\right)^{2}=P(0)^{n} \prod_{i} P\left(\frac{x^{2}}{r_{i}}\right) .
$$

Plugging in $x=\alpha$, we use the facts that $P(\alpha)=4, P(-\alpha)=4-2 \alpha$, and also
$$
P\left(\frac{\alpha^{2}}{r_{i}}\right)=\frac{\alpha^{4040}}{r_{i}^{2020}}+\frac{\alpha^{2}}{r_{i}}+2=-\frac{(\alpha-2)^{2}}{r_{i}+2}+\frac{\alpha^{2}}{r_{i}}+2=\frac{2\left(-\alpha-r_{i}\right)^{2}}{r_{i}\left(r_{i}+2\right)} \text {. }
$$

This will give us
$$
P(\alpha) P(-\alpha) Q\left(\alpha^{2}\right)^{2}=2^{n} \prod_{i} \frac{2\left(-\alpha-r_{i}\right)^{2}}{r_{i}\left(r_{i}+2\right)}=2^{n} \cdot \frac{2^{n} P(-\alpha)^{2}}{P(0) P(-2)} .
$$

Therefore,
$$
\begin{aligned}
Q\left(\alpha^{2}\right)^{2} & =\frac{4^{n} P(-\alpha)}{P(0) P(-2) P(\alpha)} \\
& =\frac{4^{n}(4-2 \alpha)}{2 \cdot 2^{2020} \cdot 4} \\
& =\frac{2^{4041}(2-\alpha)}{2^{2023}} \\
& =2^{2018}(2-\alpha)
\end{aligned}
$$

We can check that $P(x)-4$ has no double roots (e.g. by checking that it shares no roots with its derivative), which means that all possible $\alpha$ are distinct. Therefore, adding over all $\alpha$ gives $2020 \cdot 2^{2019}$, because the sum of the roots of $P(x)-4$ is 0 .