Plugging in $x=0$ and $x=4$ we find $P(0)=P(2)=0$. If there were any more roots besides these two, we would have an infinite number of roots, which is impossible for a polynomial, so we know that our polynomial only has two roots. I don't know how to continue from there.

Quadratic64 wrote: Plugging in $x=0$ and $x=4$ we find $P(0)=P(2)=0$. If there were any more roots besides these two, we would have an infinite number of roots, which is impossible for a polynomial, so we know that our polynomial only has two roots. I don't know how to continue from there. This is incorrect actually.

I cheated and Wolframmed the problem earlier today. Trying to reverse engineer a solution, I took the same tactic that Quadratic made, with the exception that I found ALL the roots of the polynomial $P(x)$. (Note that quadratics are polynomials, but not all polynomials are quadratic!) However, even though I was able to find all the roots, I can't find any good motivation for the next natural step: finding the multiplicity of the roots. Just because a polynomial has only $n$ distinct roots doesn't make it an $n$th degree polynomial - $f(x) = x^{9999}$ has only one distinct root, but is definitely not linear. Once you find the roots of $P(x)$, and you find the multiplicity of all of those roots, you need to do one of two things: A) Prove that if $P(x)$ is a solution, then $KP(x)$ is a solution (for a constant $K$), thus showing that ANY polynomial with those roots works OR B) Find the unique value(s) of $K$ such that $KP(x)$ is a solution to the functional equation.

talkinaway this is why I didn't get 5/5 from this problem.I didn't consider that some roots might be multiple.However if you know the roots it's easy to find the multiplicity of each one.Try it! Very helpful hint

The relation $ (x^2-6x+8)P(x)=(x^2+2x)P(x-2)$ writes $ (x-2)(x-4)P(x)=x(x+2)P(x-2)$. For $x=2$ we thus have $P(0) = P(2-2)=0$; for $x=4$ we have $P(2) = P(4-2)=0$; and for $x=-2$ we have $P(-2) =0$. Therefore $P(x) = x(x-2)(x+2)Q(x)$. Plugging in and cancelling equal factors, we are left with $ (x-2)Q(x)=xQ(x-2)$. For $x=0$ we thus have $Q(0) = 0$. Therefore $Q(x) = xR(x)$. Plugging in and cancelling equal factors, we are left with $ R(x)=R(x-2)$. This implies $R$ is a constant polynomial. Thus the solutins are $\boxed{P(X) = Rx^2(x^2-4)}$, for any $R\in \mathbb{R}$

mavropnevma wrote: The relation $ (x^2-6x+8)P(x)=(x^2+2x)P(x-2)$ writes $ (x-2)(x-4)P(x)=x(x+2)P(x-2)$. For $x=2$ we thus have $P(0) = P(2-2)=0$; for $x=4$ we have $P(2) = P(4-2)=0$; and for $x=-2$ we have $P(-2) =0$. Therefore $P(x) = x(x-2)(x+2)Q(x)$. Plugging in and cancelling equal factors, we are left with $ (x-2)Q(x)=xQ(x-2)$. For $x=0$ we thus have $Q(0) = 0$. Therefore $Q(x) = xR(x)$. Plugging in and cancelling equal factors, we are left with $ R(x)=R(x-2)$. This implies $R$ is a constant polynomial. Thus the solutins are $\boxed{P(X) = Rx^2(x^2-4)}$, for any $R\in \mathbb{R}$ This is what I did as well. Maybe people were just overthinking the problem and decided to take a different route.

one natural aproach is to see that P(0)=P(2)=P(-2)=0 and P(1)=P(-1).Next if P i polynomial of degree n we look at coefficients of x^(n+1) .From that we see that n=4.Next from P(0)=P(2)=P(-2)=0 and P(1)=P(-1) we can calculate that A3=A1=0 and A2=-4A4 where Ai is the coefficient with x^i.The last thing is to put this into given formula and see that this works for every A4.