For $a \in \mathbb R$, find polynomials $P$ such that $(x^2-ax+18)P(x)-(x^2+3x)P(x-3)=0$ Find all polynomials $P(x)$ with real coefficients such that:-    $(x^2-ax+18)P(x)-(x^2+3x)P(x-3)=0$
This is $a$ creating some big problems for me. I don't know what to do.
I am not able to figure out anything because of that $a$. The best I can figure out is that I will find the roots of $P(x)$ because $a$ is not computable as there is no way of finding out the zeros of $P(x-3)$. If I would have been able to find the roots of $(x^2-ax+18)$ then I would have been able to figure out what to do.
If there would have been no $a$ I would have found of the roots of $P(x)$ like for example $\alpha, \beta$ then I would have written out $P(x)$ in the form of $(x-\alpha)(x-\beta)Q(x)$ for $Q(x)$ being any polynomial. Then I would have tried to calculate the answer.
Any help would be appreciated

At first we have $(x^2-ax+18)P(x)=(x^2+3x)P(x-3)=x(x+3)P(x-3)$. Now since $x \nmid (x^2-ax+18)$, we have $2$ cases:
$(x^2-ax+18)$ divisible by $(x+3)$:
$$(x+3) | (x^2-ax+18) \Rightarrow (x^2-ax+18)=(x+3)(x+6) \Rightarrow$$
$$\Rightarrow (x+6)P(x) = xP(x-3) \Rightarrow$$
$$\Rightarrow \left\{\begin{array}{c} x | P(x) \\ (x+6) | P(x-3) \end{array}\right\}  \Rightarrow \left\{\begin{array}{c} (x-3) | P(x-3) \\ (x+9) | P(x) \end{array}\right\}  \Rightarrow$$
$$\Rightarrow (x+9)Q(x) = (x-3)Q(x-3) \Rightarrow \dots$$
That means $P(x)$ must have infinitely many factor that's not the case!
$(x^2-ax+18)$ not divisible by $(x+3)$:
$$(x+3) \nmid (x^2-ax+18) \Rightarrow \left\{\begin{array}{c} x(x+3) | P(x) \\ (x^2-ax+18) | P(x-3) \end{array}\right\} \Rightarrow$$
$$\Rightarrow \left\{\begin{array}{c} (x-3)x | P(x-3) \\ (x^2+(6-a)x+(27-3a)) | P(x) \end{array}\right\}  \Rightarrow x^2 | P(x) \Rightarrow \dots$$
That means $P(x)$ must have infinitely many $x$ that's not the case!
So there is no $P(x)$ except $P(x)=0$.
