1. Let $P(x)=x^{3}+x^{2}-r^{2} x-2020$ be a polynomial with roots $r, s, t$. What is $P(1)$ ?
Answer: $-4038$

Solution 1: Plugging in $x=r$ gives $r^{2}=2020$. This means $P(1)=2-r^{2}-2020=-4038$.
Solution 2: Vieta's formulas give the following equations:
$$
\begin{aligned}
r+s+t & =-1 \\
r s+s t+t r & =-r^{2} \\
r s t & =2020 .
\end{aligned}
$$

The second equation is $(r+t)(r+s)=0$. Without loss of generality, let $r+t=0$. Then $s=r+s+t=-1$. Finally $r^{2}=r s t=2020$, so $P(1)=-4038$