Setting $x=1$ we get that $P(0) = c/2$ so write $P(x) = xQ(x) + \frac{c}{2}$ and putting into the given equation and expanding we arrive at the equation $$(x+1)Q(x-1) = xQ(x)$$Setting $x = -1$ we get that $Q(-1) = 0$ so again write $Q(x) = (x+1)R(x)$ and put in the second equation to get $$R(x) = R(x-1)$$hence $R(x) = a$ for some constant $a$ and therefore $P(x) = ax(x+1) + \frac{c}{2}$ and easy to check that this works. $\blacksquare$