Find an inverse in a quotient of a Polynomial Ring I know this question is so easy that should be a first year algebra, but It has been a long time since I last studied this argument. 

I have to find the inverse of $x+2$ in the ring $\mathbb{Q}[x]/\left<x^6 + x^4 + x^2 + 1\right>$.

I tried to divide $(x^6 + x^4 + x^2 + 1)+1$ for $x + 2$ but I obtained a reminder of $86$. Then I tried to solve 
$$(x + 2)(ax^5+bx^4+cx^3+dx^2+ex+f)=q(x^6 + x^4 + x^2 + 1)+1$$
and obtained that q must be a complex number. So now I'm wondering if there's a better way to proceed. Can anybody give me an hint?
Thanks 

You can write $x^6+x^4+x^2+1=(x^5-2x^4+5x^3-10x^2+21x-42)(x+2)+85$
So, from here, $85=(x^6+x^4+x^2+1)-(x^5-2x^4+5x^3-10x^2+21x-42)(x+2)$, i.e.,
$85\equiv -(x^5-2x^4+5x^3-10x^2+21x-42)(x+2) \pmod{x^6+x^4+x^2+1}$
So, $(x+2)^{-1}=-\frac1{85}(x^5-2x^4+5x^3-10x^2+21x-42)$ in the quotient ring.
