<s>[INST] Find all polynomials with real coefficients that satisfy $(x^2-6x+8)P(x)=(x^2+2x)P(x-2)$ Find all polynomials with real coefficients that satisfy $$(x^2-6x+8)P(x)=(x^2+2x)P(x-2)\forall x\in\Bbb R$$
My work;
$$\frac{P(x)}{P(x-2)}=-\frac{4}{x-2}+\frac{12}{x-4}+1\tag{1}$$
$$\frac{P(x-2)}{P(x)}=-\frac{12}{x+2}+\frac{4}{x}+1\tag{2}$$
I also factorised the two known polynomial which didn't give anything useful. What should I extract from these two ratios?
 [/INST] $$(x^2-6x+8)P(x)=(x^2+2x)P(x-2)\iff (x-2)(x-4)P(x)=(x+2)xP(x-2)$$
When $x=2,4$ LHS=0, so $RHS=0$, hence $x=0,2$ is a root for $P(x)$.
When $x=0,-2$ RHS=0, so $LHS=0$, hence $x=0,-2$ is a root for $P(x)$.
Write $P(x)=a(x)\cdot(x-2)x(x+2)$, plug in the assumption, we have
$$a(x)\cdot(x-4)(x-2)^2x(x+2)=a(x-2)\cdot(x-4)(x-2)x^2(x+2)$$
$$\implies a(x)\cdot(x-2)=a(x-2)\cdot x$$
$$\implies a(x)=ax$$
For the last step, we first argue $a(x)$ can't have constant term, since $x=0$ is a root. Hence $a(x)=xb(x)$, which implies $b(x)=b(x−2)\forall x\in \mathbb{R}$.
Then we use the fact if polynomial $p(x)$ takes value $p_0$ infinitely many times (which means $p(x)-p_0=0$ has infinite zeros), then $p(x)$ is constant.
Hence $P(x)=ax^2(x-2)(x+2)$
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