Expression for the element $(a+u)^{-1}$ in the extension field $G = F(u)$, where $u$ is a root of $x^{2}+px +q$ over a field $F$ Let $f = x^{2}+px+q$ be an irreducible polynomial over a field $F$ and let $G = F(u)$ be the extension of $F$ by a root $u$ of $f$. Every element of $G$ can be written in the normal form $a+bu$ where $a, b \in F$. I need to find an expression (in the normal form) for thee element $(a+u)^{-1}$ where $a \in F$.
This problem has become the bane of my existence over the past few days, as I have tried to model my answer off of an example given to us in our lecture notes, which I suspect is incorrect itself. So, I suppose part of my question here is asking whether the example problem is actually incorrectly solved.
The example problem says that in the case of the extension of $\mathbb{Q}$ by a root $u$ of the irreducible polynomial $f = x^{2} + x + 1$, to find an expression for $(a+bu)^{-1}$, we have to divide $f$ by $a+bx$ with remainder by using the Extended Euclidean Algorithm. The result of this is the following sequence of steps, which I have tried myself and found to be correct:
$$x^{2} + x + 1 = \left( \frac{1}{b}x\right)(a+bx) + \left( 1 - \frac{a}{b}\right)x + 1 \\ =  \left( \frac{1}{b}x\right) (a+bx) + \frac{1}{b} \left( 1 - \frac{a}{b}\right)(a+bx) + \left( \frac{a^{2}}{b^{2}} - \frac{a}{b} + 1\right) \\ = \left(\frac{1}{b}x+\frac{1}{b}-\frac{a}{b^{2}} \right)(a+bx)+\left(  \frac{a^{2}}{b^{2}} - \frac{a}{b} + 1\right)$$
My professor then proceeds to say that $\frac{a^{2}}{b^{2}}- \frac{a}{b} + 1 \neq 0$ because then a rational number $-\frac{a}{b}$ would be a root of $f$. Also, even though technically, one additional step needs to be completed in the Euclidean Algorithm, we already have a polynomial $h \in \mathbb{Q}[x]$ such that $h(a+bx)-1$ is a multiple of $f$.
So, he multiplies the equality through by $\left(  \frac{a^{2}}{b^{2}} - \frac{a}{b} + 1\right)^{-1}$ to obtain 
$$ \left(  \frac{a^{2}}{b^{2}} - \frac{a}{b} + 1\right)^{-1}(x^{2}+x + 1) = \left(  \frac{a^{2}}{b^{2}} - \frac{a}{b} + 1\right)^{-1}\left(\frac{1}{b}x+\frac{1}{b}-\frac{a}{b^{2}} \right)(a+bx) + 1$$
and then takes $h = -\left(  \frac{a^{2}}{b^{2}} - \frac{a}{b} + 1\right)^{-1}\left(\frac{1}{b}x+\frac{1}{b}-\frac{a}{b^{2}} \right) = \frac{1}{a^{2}-ab+b^{2}}(a-b-bx) $, so that $h(u) = \frac{1}{a^{2}-ab+b^{2}}(a-b-bu)$.
However, $h(u)(a+bu) \neq 1$ like he says it is: I multiplied it out myself to get $$ h(u) (a+ bu) = \frac{a-b-bu}{a^{2}-ab+b^{2}}(a+bu) \\ = \frac{a^{2}-ab-abu + abu -b^{2}u - b^{2}u^{2}}{a^{2}-ab+b^{2}} \\ = \frac{a^{2}-ab-b^{2}u-b^{2}u^{2}}{a^{2}-ab+b^{2}}  \\ = \frac{a^{2}-ab-b^{2}(u+u^{2}}{a^{2}-ab+b^{2}} \neq 1 $$.
But, I can find absolutely no error in the process by which my professor found his function $h$! What am I missing here??
I need to know how to do that example problem in order to do the real problem I am trying to tackle, and that is finding $(a+u)^{-1}$ for $f = x^{2}+px+q$ over a general field $F$ (so, $p,q \in F$ in case you were wondering). I did a similar process to what my professor did, applying the extended Euclidean algorithm to divide $f$ by $a+x$:
$$x^{2}+px+q = x(a+x) + (p-a)(a+x) + q -(p-a)a \\ = (x+(p-a))(a+x)+(q-(p-a)a) $$
Like in the example problem, here the remainder $q - (p-a)a \neq 0$, because then otherwise $-a \in F$ would be a root of $f$ when we claimed it was irreducible over $F$ (a contradiction). 
So, because of this, I know that the remainder has an inverse, and I can multiply the equality through by said inverse, $(q - (p-a)a)^{-1} $ to obtain 
$$(q - (p-a)a)^{-1} (x^{2}+px+q) = (q - (p-a)a)^{-1}(x+(p-a))(a+x) + 1$$
Then, I took $$h = -  (q - (p-a)a)^{-1}(x+(p-a)) \\ = \frac{-x-(p-a)}{q-(p-a)a} = \frac{a - p - x}{a^{2}-pa+q} $$
Then, letting $h(u) = \frac{a-p-u}{a^{2}-pa+q}$, we should have $h(u)(a+u) = 1$, but instead I get $h(u)(a+u) = \left(\frac{a-p-u}{a^{2}-pa+q} \right)(a+u) = \frac{a^{2}-pa - up +u^{2}}{a^{2}-pa+q}$, and not $1$!
What am I doing wrong??! 

Looks like a sign error in the very last line.
The top multiplies out to be $a^2-pa-pu-u^2$, but then $-pu-u^2=q$, so that it is equal to the denominator.
In the other example you gave,
$$ \frac{a^{2}-ab-b^{2}u-b^{2}u^{2}}{a^{2}-ab+b^{2}}  \\ = \frac{a^{2}-ab-b^{2}(u+u^{2})}{a^{2}-ab+b^{2}}=\frac{a^2-ab-b^2(-1)}{a^2-ab+b^2} =1 $$.
You didn't make any mistakes. You simply failed to use the relation $u^2+u+1=0$.
