The inverse of $2x^2+2$ in $\mathbb{Z}_3[x]/( x^3+2x^2+2)$ What is the independent coefficient in the inverse of $2x^2+2$ in $\mathbb{Z}_3[x]/(x^3+2x^2+2)$ ?
I have been calculating some combinations, but I don't know how I can calculate the inverse.
 A: It must take the form $ax^2+bx+c$.
Multiply this by $2x^2+2$, and you get a degree-four polynomial.
Since $x^3+2x+2=0$, it follows that $x^3=x+1$ and so $x^4=x(x^3)=x^2+x$.
So your degree-four polynomial becomes a quadratic.
This quadratic equals 1, or $0x^2+0x+1$.
Equate the three coefficients, and get three linear equations in $a,b,c$.
A: Set $f(x) = 2x^2+2$, $g(x) = x^3+2x^2+2$. Since $f(x),g(x)$ are coprime, find
$$p(x)f(x)+q(x)g(x) = 1$$
using Euclidean division. 
Then $p(x)$ is the inverse of $f(x)$ in $F[x]/\langle g(x)\rangle$.
A: This is in no way a standard solution, but I just want to include it for fun. 
Note that $x^3=-(2x^2+2)$, we will find $x^{-3}$. Now
$$x^3-x^2=-2=1$$
implies 
$$x(x^2-x)=x^2(x-1)=1.$$
Thus 
$$x^{-1}=x(x-1)\text{ and }x^{-2}=x-1.$$
Multiplying the two we get
$$x^{-3} = x(x-1)^2=x(x^2+x+1)=x^3+x^2+x.$$
Using $x^3=x^2+1$ on the RHS, we see that
$$x^{-3}=2x^2+x+1.$$
Thus
$$(2x^2+2)^{-1} = -x^{-3} = -2x^2-x-1 = x^2+2x+2.$$
A: $$
(ax^2 + bx + c)(2x^2 + 2) + (dx + e)(x^3 + 2x + 2) = 1\\
(2a + d)x^4 + (2b + e)x^3 + (2a + 2c + 2d)x^2 + (2b + 2d + 2e)x + (2c + 2e) = 1\\
$$ 
$$\left[ \begin{array}{ccccc} 
2 & 0 & 0 & 1 & 0 \\ 
0 & 2 & 0 & 0 & 1 \\
2 & 0 & 2 & 2 & 0 \\
0 & 2 & 0 & 2 & 2 \\
0 & 0 & 2 & 0 & 2 \\
\end{array} \right] 
\left[ \begin{array}{c}
a \\ b \\ c \\ d \\ e \\ \end{array} \right]
=
\left[ \begin{array}{c}
0 \\ 0 \\ 0 \\ 0 \\ 1 \\ \end{array} \right]
$$
Solve for $a$, $b$, and $c$ and you find the inverse.
A: Hint $\, {\rm mod}\ 3\!:\ g = x^3\!+\!2x\!+\!2 \equiv  (x\!+\!1)(x^2\!+\!x\!-\!1).\, $ Compute $\, f = \dfrac{1}{2x^2\!+\!2}\,$ by CRT:
${\rm mod}\,\ \color{#0a0}{x^2\!+\!x\!-\!1}\!:\ \color{brown}{x^2}\equiv 1\!-\!x\,\Rightarrow\,f = \dfrac{1}{2\color{brown}{x^2}\!+\!2} \equiv \dfrac{1}{x+1} \equiv \color{#c00}x,\ $ by $\ x(x\!+\!1) \equiv 1$
${\rm mod}\,\ x\!+\!1\!:\ x\equiv -1\,\Rightarrow\,1\equiv f\equiv \color{#c00}x+(\color{#0a0}{x^2\!+\!x\!-\!1})c \equiv -1 -c\,\,\Rightarrow\, \color{#c0f}{c \equiv  1}$
Therefore, we conclude that  $\ f \equiv x+(x^2\!+\!x\!-\!1)\color{#c0f}{(1)}\, \pmod{g} $
