$P(0)=r_1...r_{2023} \geq 0$ and $P(1)=(1-r_1)...(1-r_{2023}) \geq 0$ so $P(0)+P(1)=0 \to P(0)=0$

RagvaloD wrote: $P(0)=r_1...r_{2023} \geq 0$ and $P(1)=(1-r_1)...(1-r_{2023}) \geq 0$ so $P(0)+P(1)=0 \to P(0)=0$ $P(0) = - r_1...r_{2023} $

I replace $2023$ by any odd $N$ and prove that the answer is $2^{-N}$. Achievability. Just take $r_i = 1/2$, $1\le i\le N$. Upper bound. Let $R(x) =\textstyle \prod_{1\le i\le N}(x-r_i)$. Then, $(1-r_1)\cdots(1-r_N) = r_1\cdots r_N$. We can safely assume $r_i\ne 0$ or $r_i\ne 1$ as otherwise the product is zero. Then we get $q_1\cdots q_N = 1$, where $q_i = 1/r_i -1$. So, \[ r_1\cdots r_N = \Bigl((1+q_1)\cdots (1+q_N)\Bigl)^{-1}\le \Bigl(\bigl(1+(q_1\cdots q_N)^{1/N}\bigr)^N\Bigr)^{-1} = 2^{-N}, \]using Holder's inequality.

Notice that $$P(0)+P(1)=0\implies (1-r_1)(1-r_2)\cdots (1-r_{2023})=r_1 r_2\cdots r_{2023}.$$Now using AM-GM (as $0\leq r_i\leq 1$): $$\sqrt[2023]{r_1r_2\cdots r_{2023}}=\sqrt[2023]{(1-r_1)(1-r_2)\cdots (1-r_{2023})}\leq \frac{(1-r_1)+(1-r_2)+\dots+(1-r_{2023})}{2023}=1-\frac{r_1+\dots+r_{2023}}{2023}\leq 1-\sqrt[2023]{r_1r_2\cdots r_{2023}},$$thus $r_1r_2\cdots r_{2023}\leq \frac{1}{2^{2023}}$. Equality is attained for the polynomial $\left (x-\frac 12\right)^{2023}$.