Find all polynomials $Q(x)(x^2-6x+8) =Q(x-2)(x^2-6x)$ for $x\in \Bbb R$
I've tried different substitutions like put $Q(x)$ $=$ $k$ for some $k$ and getting the equation $k(x^2-6x+8)$ $=$ $(k-2)(x^2-6x)$ $<=>$ $k$=$(6x-x^2)/4$, but that doesn't give equlity. Thank you for your help
maybe Hint:Some information come out from $Q(x)(x^2-6x+8) =Q(x-2)(x^2-6x)$ $$x=0 \to Q(0)(8)=Q(-2)0 \to Q(0)=0\\ x=6 \to Q(6)(8)=Q(6-2)0 \to Q(6)=0\\ x=4 \to Q(4)(0)=Q(2)(16-24) \to Q(4)=0$$ so $Q(x)$ at least contain $$Q(x)=(x-0)(x-4)(x-6)$$ or $$Q(x)=(x-0)(x-6)(x-4)q(x)$$
$$\begin{align*} (x^2-6x+8) Q(x) &\equiv (x^2 -6x)Q(x-2)\\ (x^2 - 6x)[Q(x)-Q(x-2)] &\equiv -8Q(x) \end{align*}$$
If $Q(x)$ has degree $n > 0$, $Q(x) - Q(x-2)$ has degree $n-1$, and so the left hand side has degree $n+1$. However, the right hand side has degree $n$.
So $n\not>0$, and $Q(x)$ can only be a constant function with degree $0$. Let $Q(x) \equiv q$,
$$\begin{align*} (x^2-6x+8)q &\equiv (x^2 -6x)q\\ 8q &\equiv 0\\ q &\equiv 0\\ Q(x) &\equiv 0 \end{align*}$$
Notes regarding that $Q(x) - Q(x-2)$ has degree $n-1$: Let $$Q(x) \equiv ax^n + bx^{n-1} + \cdots$$ where $a \ne 0$ and all lower degree terms are omitted. Then $Q(x-2)$ is
$$\begin{align*} Q(x-2) &\equiv a(x^n - 2nx^{n-1} + \cdots) + b(x^{n-1} - \cdots) + \cdots\\ &\equiv ax^n -2anx^{n-1} + bx^{n-1} + \cdots\\ Q(x)-Q(x-2)&\equiv 2anx^{n-1} + \cdots \end{align*}$$
Since $a\ne 0$ and $n> 0$, the coefficient of $x^{n-1}$ is non-zero, and so the right hand side has degree $n-1$.
