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. 

Since $p$ is square-free, you can use $v_p(fg)=v_p(f)+v_p(g)$, which might be less work ... 
