Find polynomials : $ xP(x-1)=(x-11)P(x)$ Find all polynomials $P(x) \in\mathbb{R}[x]$ satisfying $$ xP(x-1)=(x-11)P(x)$$
(Attempted work has been moved to answer)

Substitute $x=0$, we have $P(0)=0$ and substitute $x=11$, we have $P(10)=0$,
so $P(x)$ has $0$ and $10$ as its roots, i.e., $\exists Q(x)$, $P(x)=x(x-10)Q(x)$
Since $ xP(x-1)=(x-11)P(x)$, so $x(x-1)(x-11)Q(x-1)=(x-11)x(x-10)Q(x)$
thus, $(x-1)Q(x-1) = (x-10)Q(x)$ ---[2]
Similarly, substitute $x=1$ and $x=10$ in [2], $P(x)$ has $1$ and $9$ as its roots.
$\exists R(x)$, $Q(x)=(x-1)(x-9)R(x)$ substitute in [2], we have
$R(x-1)(x-1)(x-2)(x-10)=(x-10)(x-1)(x-9)R(x)$
so $R(x-1)(x-2)=(x-9)R(x)$
Finally,we have $A(x)=A(x+1)$, $\forall x \in \mathbb{R}$
Since $A$ is continuous function so $A(x) = c$, $\forall x \in \mathbb{R}$
Therefore, $P(x)=c(x)(x-1)(x-2)\ldots(x-10)$, where $c$ is constant.
