Series of characteristic polynomials

From a very attenuated literature search, I see that the Lucas polynomials of the second kind obey the recursion $$ L_{k+1}(x) = x \ L_k(x) - L_{k-1}(x), \quad L_1(x) = 1 \ , L_2(x)=x$$ This matches the recursion for $Q_n,$ with an index shift of 1. The one reference I've seen (I don't have access to journal papers from home) states that this polynomial set obeys the relation

$$L_k(2 \cos(\theta) ) = 2 \cos( k \ \theta) $$

This functional relationship would explain the location of the zeros, and likely the envelope property as well. I suggest that the proposer start his research with 'Lucas polynomials of the second kind.'

The Chebyshev polynomials of the second kind satisfy the recurrence relation $$ \begin{cases} U_0(y) = 1 \\ U_1(y)=x\\ U_n(y) = 2y U_{n-1}(y)-U_{n-2}(y) \end{cases} $$ so that $Q_n(y) = U_n(y/2)$ and $P_n(x) = U_n(1-x/2)$.

The zeros of $U_n$ are $$ y_k = \cos\left( \pi \frac{k+1}{n+1}\right) \, , \, k = 0, \ldots, n $$ in the range $(-1, 1)$, so that $$ x_k = 2 - 2\cos\left( \pi \frac{k+1}{n+1}\right) \, , \, k = 0, \ldots, n $$ are the zeros of $P_n$ in the range $(0, 4)$.

Also for $|x| < 1$ $$ U_n(x) = \frac{\sin((n+1)\arccos(x))}{\sqrt{1-x^2}} $$ which implies $$ |U_n(x)| \le \frac{1}{\sqrt{1-x^2}} $$ and therefore $$ | P_n(x)| \le \frac{2}{\sqrt{x(4-x)}} \le \frac 1x + \frac{1}{4-x} $$ for $0 < x < 4$, the last estimate follows from the inequality between harmonic and geometric mean.