Solving a recurrence of polynomials

Notice that the recurrence equation is constant coefficients, with respect to $n$. It can be solved using the generating function approach. Multiply both sides of the equation with $t^n$ and sum over $n\geqslant 2$: $$ \sum_{n=2}^\infty t^n p_{n+2} = -x \sum_{n=2}^\infty t^n p_{n+1} + \sum_{n=2}^\infty t^n p_{n-2} $$ Defining $g(t,x) = \sum_{n=0}^\infty t^n p_n(x)$ this translates into: $$ \frac{1}{t^2} \left( g(t,x) - p_0(x) -t p_1(x) -t^2 p_2(x) - t^3 p_3(x) \right) = -\frac{x}{t} \left(g(t,x)-p_0(x) - t p_1(x) - t^2 p_2(x) \right) + t^2 g(t,x) $$ giving: $$ g(t,x) = \frac{1+t x}{1+t x-t^4} \left( p_0(x) + t p_1(x) + t^2 p_2(x) \right) + \frac{t^3}{1+t x-t^4} p_3(x) $$ The general solution $p_n(x)$ is obtained by extracting series coefficient: $$ p_n(x) = [t]^n g(t,x) $$

Here is verification in Mathematica:

related problem: (I), (II), (III). Here is the solution of the corrected recurrence relation computed by maple

$$ \left( -3\,x-\sqrt {{x}^{2}+4} \right)\left( -2\, \left( -x-\sqrt{{x}^{2}+4} \right)^{-1} \right) ^{n}{\frac {1}{\sqrt {{x}^{2}+4}}}\left( -x-\sqrt {{x}^{2}+4} \right)^{-1} $$ $$+ \left( 3\,x-\sqrt {{x}^{2}+4} \right)\left( -2\,\left( -x+\sqrt {{x}^{2}+4} \right)^{1} \right) ^{n}{\frac {1}{\sqrt {{x}^{2}+4}}} \left( -x+\sqrt {{x}^{2}+4}\right)^{-1}$$

If you are interested in how you can solve it. Assume your solution $p_n = r^n\,, $ for some $r$ to be determined and substitute in your recurrence relation which gives

$$ r^{n+2} + x r^{n+1} - r^n = 0 \Rightarrow r^n(r^2+xr-1)=0 \Rightarrow (r^2+xr-1)=0$$ $$ r_1 = -\frac{1}{2}\,x+\frac{1}{2}\sqrt {{x}^{2}+4}\,, \,\,\, \, r_2 = -\frac{1}{2}\,x - \frac{1}{2}\,\sqrt {{x}^{2}+4} $$ Now, you construct the general solution as $$ p_n(x) = c_1 {r_{1}}^n + c_2 {r_{2}}^n = c_1 {\left( -\frac{1}{2}\,x+\frac{1}{2}\sqrt {{x}^{2}+4} \right)}^n + {\left( -\frac{1}{2}\,x - \frac{1}{2}\,\sqrt {{x}^{2}+4} \right)}^n $$ To determine the constants $c_1$ and $c_2$, you just need to exploit the initial conditions $p_0(x)$ and $p_1(x)$ and solve for $c_1$ and $c_2$.

Alternatively, compare your recurrence to known recurrences.
In this case, the Chebyshev polynomials ... result:

$$ p_n(x) = i^{n} \Biggl(4 T_n \biggl(\frac{ix}{2}\biggr) - 3 U_n \biggl(\frac{ix}{2}\biggr)\Biggr) $$