p-adic number for polynomial I have three polynomials: 
$$f=x^5+x^4+4x^3+3x^2+3x$$
$$g=x^5+4x^3+2x^2+3x+6$$
$$p=x^2+3$$
Question: what is $v_p(fg)$?
First, I have multiplied f and g: 
$h :=f \cdot g = x^{10}+x^9+8x^8+9x^7+24x^6+29x^5+36x^4+39x^3+27x^2+18x $
Then I have divided that $h$ with $x^2+3$ to get a factor to know how often $h$ contains $x^2+3 $:
$(x^2+3) \cdot (x^6+x^5+2x^4+3x^3+3x^2+2x)$
When I divide $x^6+x^5+2x^4+3x^3+3x^2+2x$ again by $x^2+3$ then I get a rest: 
$(x^4+x^3-x^2+6) \cdot (x^2+3) + 2x-18$
Does that mean that $h$ contains $x^2+3$ only once and 
$v_p(f \cdot g)=1$?
Is that the correct way to find the p-adic number? Or is there a better conventional one?
I appreciate every hint. 

Because $x^2+3$ is irreducible in $\Bbb{Z}[x]$ (Eisenstein) it is the minimal polynomial of $i\sqrt3$. You can then check that


*

*$f(i\sqrt3)=0$, so $f(x)$ must be divisible by $x^2+3$. A long division shows that $f(x)/(x^2+3)=x+x^2+x^3$, and you can check this is not a multiple of $x^2+3$.

*$g(i\sqrt3)=0$, so $g(x)$ is also divisible by $x^2+3$. A long division gives a quotient that I leave to you.


Anyway, looks like $fg$ should be divisible by $(x^2+3)^2$, so you have made a mistake somewhere. As Hagen pointed out, we have for all discrete valuations
$\nu_p$ the formula $\nu_p(fg)=\nu_p(f)+\nu_p(g)$ that comes in handy.
