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

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### 1 Answer

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.**