Let's say that polynomial $P(x)$ sifts another polynomial $Q(x)$ if three conditions are satisfied:
- $\deg P(x) + 1 = \deg Q(x) = n$
- $Q(x)$ has $n$ distinct real roots, $P(x)$ has $n - 1$ distinct real roots
- Roots of $Q(x)$ and $P(x)$ alternate, so if $\alpha_1, \alpha_2, \dots, \alpha_n$ are roots of $Q(x)$, $\beta_1, \beta_2, \dots, \beta_{n - 1}$ are roots of $P(x)$, then $\alpha_1 < \beta_1 < \alpha_2 < \beta_2 < \dots < \beta_{n - 1} < \alpha_n$.
There's a statement that if $P(x)$ sifts $Q(x)$ and $R(x)$ is the remainder of $Q(x)$ divided by $P(x)$, then $R(x)$ sifts $P(x)$, but I failed to come up with an idea to prove it. I tried to build something used factorised forms of polynomials, but it seems to lead nowhere.