<s>[INST] Calculation of polynomial $g(x)$ satisfies $x\cdot g(x+1)=(x-3)\cdot g(x)$ 
If a polynomial $g(x)$ satisfies $x\cdot g(x+1)=(x-3)\cdot g(x)$ for all $x$, and $g(3)=6$, then $g(25)=$?

My try: $x\cdot g(x+1)=(x-3)\cdot g(x)$,
Put $x=3$, we get $g(4)=0$, means $(x-4)$ is a factor of $g(x)$.
Similarly put $x=0$. We get $g(0)=0$, means $x$ is a factor of $g(x)$.
This means $g(x)=x\cdot (x-4)h(x)$, where $h(x)$ is a polynomial.
Then how can I calculate it?
Thanks.
 [/INST] A repeated use of $g(x+1)=\dfrac{x-3}x g(x)$ gives 
$$\begin{array} gg(25)&=&\dfrac{21}{24} g(24)\\ &=&\dfrac{22}{24}\dfrac{21}{23} g(23)\\&=&\cdots \\ &=&\dfrac{21}{24}\dfrac{20}{23}\cdots \frac{1}{4} g(3)\end{array}$$
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