Quick FE. $\color{black}\rule{25cm}{1pt}$ The only answer is $f(x)=x(x^2+c)$, where $c$ is any real number. We just set $x-y=1$ to get that: $$f(2y+1)=(2y+1)(f(1)+4y(y+1))$$ Since $2y+1$ is a surjection in $\mathbb{R}$, we can set $t=2y+1$, and then we have that $y=\frac{t-1}{2}$, which implies that $f(x)=x(x^2+c)$, where $c$ is any real number.

We can re-arrange this to get, $$\frac{f(x+y)}{x+y}-(x+y)^2=\frac{f(x-y)}{x-y}-(x-y)^2$$for all reals $x,y$ such that $x\neq \pm y$. Since, $x+y$ is independent of $x-y$. $\frac{f(x+y)}{x+y}-(x+y)^2$ is a constant $c$. Hence $f(x)=x(x^2+c)$. Substituting back, it works.

$P\left(\frac{x+1}2,\frac{x-1}2\right)$