Multiplicative inverse of $2x + 3 + I$ in $\mathbb{Z}_5[x]/I $? In $\mathbb{Z}_5[x]$, let $ I = \langle x^2 + x + 2\rangle$. Find the multiplicative inverse of $2x + 3 + I$ in $\mathbb{Z}_5[x]/I$. 
The answer is supposed to be $3x + 1 + I$. But I don't understand how you find this. I divided $x^2 + x + 2$ by $2x+3$ to get $$ x^2 + x + 2 = (2x + 3)(3x + 1) + 4.$$ But I'm not sure how this will help me. I know that the elements of $\mathbb{Z}_5[x]/I$ are of the form $ax + b$, with $a, b \in \mathbb{Z}_5$. So I want $$ 1 + I = (ax + b)(2x+3) + I = (2ax^2 + (3a+2b)x + 3b) + I. $$ I want to find the coefficients $a, b$ from this. We must have $3b = 1$, and the coefficients by $x$ and $x^2$ must vanish. Since $\mathbb{Z}_5$ is a field, the relation $3b = 1$ means $b$ is the multiplicative inverse of $3$. So $b = 2$. From the coefficients of $x$, I get also $3a = -2b$. I multiply this with $2$ to get $a = 6a = -4b = -8 = 2$. 
So I would say the inverse is $2x + 2$, in contradiction with the answer at the back of my book. So where is the mistake in my reasoning, and how to find correct answer? 

Your original approach was on the right track.  By dividing, you found that
$$
x^2 + x + 2 = (2x + 3)(3x + 1) + 4 \, .
$$
Rearranging, we have
\begin{align*}
(2x + 3)(3x + 1) = -4 + x^2 + x + 2 = 1 + x^2 + x + 2 \equiv 1 \pmod{x^2 + x + 2}
\end{align*}
so $3x + 1$ is the inverse of $2x+3$.
In general, if you have the equality $a f + b g= u$ where $u$ is a unit, you can multiply through by $u^{-1}$ to get a linear combination that yields $1$: $u^{-1} a f + u^{-1} b g = 1$.
