What are polynomials that have real coefficients such as x*f(x)=(x-3)(f(x+1)) ?

Let x = 0. The given relation will
become:

0*f(0) =
(0-3)*f(0+1)

0 = -3*f(1)

The
product of two factors is cancelling if one of them is zero. It is obvious that f(1) =
0.

If f(1) = 0 => x = 1 is one of the roots of the
requested polynomial.

Let x =
1.

1*f(1) = (1-3)*f(1+1)

But
f(1) = 0 => 1*0 = -2*f(2) => -2*f(2) = 0

If
f(2) = 0, then x = 2 is another root of the requested
polynomial.

Let x = 2

2*f(2) =
-1*f(3)

But f(2) = 0 => -1*f(3) = 0 => f(3) =
0 => x = 3 is the next root of polynomial.

Let x =
3.

3*f(3) = 0*f(4) => 0 =
0

We notice that from this point further we cannot find any
infos about the other roots of the required polynomial.

All
we know, so far, is the following:

f(x) =
x*(x-1)*(x-2)*g(x)

(x-3)*x*(x-1)*(x-2)*g(x) =
(x-3)*(x+1)*x*(x-1)*g(x+1)

We'll simplify and we'll
get:

g(x) = g(x+1)

Let's
consider the function h(x) = g(x) - g(x+1) = 0

If the order
of the polynomial function h(x) is n, then if h(x)=0 => g(x) is a constant
function.

The requested polynomials that
respect all imposed conditions are: f(x) = k*X*(X-1)*(X-2), where K is a constant
function.