Find the polynomials $P \in \Bbb{R}[X]$ so that $(X+4) P(X)=X P(X+1)$. 
Find the polynomials $P \in \Bbb{R}[X]$ so that $(X+4) P(X)=X P(X+1)$.

I haven't gotten very far:
$$(0+4)P(0)=0P(0+1) \iff P(0)=0$$
Through $P(0)=0$, we can derive $P(-1)=0$:
$$(-1+4)P(-1)=-1P(-1+1)=-P(0)=0 \iff P(-1)=0,$$
also $P(-2)=0$ and $P(-3)=0$.
What do I do now?
 A: I think you're most of the way there. Factor theorem tells us that $X(X + 1)(X + 2)(X + 3)$ is a factor of your polynomial, that is,
$$P(X) = X(X + 1)(X + 2)(X + 3)Q(X) \tag{$\star$}$$
for some polynomial $Q \in \Bbb{R}[X]$. We need to know which of these polynomials are solutions. Maybe we will find that all such polynomials are solutions, in which case, $(\star)$ is our final answer!
So, let's suppose that $P(X)$ takes the form $(\star)$. Then
$$(X + 4)P(X) = X(X + 1)(X + 2)(X + 3)(X + 4)Q(X)$$
and
$$XP(X + 1) = X(X + 1)(X + 2)(X + 3)(X + 4)Q(X + 1).$$
The only way these could be equal is if $Q(X) = Q(X + 1)$. But, the only way this could possibly happen is if $Q$ is constant (this would imply $Q$ is periodic, hence bounded, hence constant). So, our refined form of $(\star)$ is:
$$P(X) = kX(X + 1)(X + 2)(X + 3),$$
for some $k \in \Bbb{R}$, and the above working (substituting $Q(X) = k$) shows that all such $P$ work. Thus, this is the general solution.
A: $\require{cancel}$
An adaptation of an excellent solution to the more general problem of finding polynomial solutions to
$$xP(x-a)=(x-b)P(x) \ \ \text{for} \ \ a,b \in \mathbb Z, \ a \ne b \tag{1}$$
(here with $a=-1, b=-4$).
Indeed such issues appear rather often on Math Stack Exchange (Calculation of polynomial $g(x)$ satisfies $x\cdot g(x+1)=(x-3)\cdot g(x)$),(Find all polynomials $p(x)$ such that: $xp(x-1) = (x-30)p(x)$), (Determine the polynomials $p(x)$ satisfying $x\cdot p(x-1) = (x-26)\cdot p(x)$), etc. and artof problemsolving for particular cases of $a$ and $b$.
I found question (1) and its solution by a certain bobthesmarty here.
Here is an adapted form of this solution where the central issue is the relationship $b=ka$ with $k>0$ for having nonzero solutions.
Let us consider first the special case $a=0$ for which clearly the only solution is $P(x)=0$.
Let us assume for the following that $a \ne 0$.
Plug in $x=0$ in (1) to get $P(0)=0$, so $x|P(x)$. Therefore, there exists $P_1(x)$ such that $P(x):=xP_1(x)$.
(1) becomes:
$$\cancel{x}(x-a)P_1(x-a)=(x-b)\cancel{x}P_1(x)\tag{2}$$
Plugging $x=a$ in (2) gives $P_1(a)=0$ ; so $(x-a) | P_1(x)$
Therefore: $P_1(x)=(x-a)P_2(x)$ for a certain $P_2(x)$.
In this way, (2) becomes
$$\cancel{(x-a)}(x-2a)P_2(x-a)=(x-b)\cancel{(x-a)}P_2(x)\tag{3}$$
Continuing this process, we will accumulate roots $0,a,2a,\cdots$ with (generalizing (2) and (3)) a polynomial $P_k$ such that:
$$(x-ka)P_k(x-a)=(x-b)P_k(x)\tag{4}$$
Does this process stops ?

*

*Either $ka$ (for $k$ positive) is never equal to $b$, meaning that $b$ is not a multiple of $a$ ; if such is the case, we will get an infinity of roots, which is impossible for a polynomial.


*Or $b=ka$ for a certain value of $k$. In this case, (3) becomes
$$P_k(x-a)=P_k(x) \tag{5}$$
As a polynomial cannot be periodic, the consequence of (5) is that $P_k(x)=c$ for a certain constant $c$.
Substituting in a reverse order, we get the general solution of (1):
$$\text{If} \ b=ka, \ \text{then} \ \ \boxed{P(x)=cx(x-a)(x-2a)\cdots(x-(k-1)a)}$$
