What's the inverse of 1 in (mod x)? (Working in F3)

Question

I'm struggling with a few basic inverses because the Euclidean algorhithm is not working.

1) Inverse of 1 in mod (x)

2) Inverse of -2x in mod (x^2+2x+2)

3) Inverse of -4 in mod (x^2 +2)

The Euclidean algorthin to find inverses is not working because there is only one step. For part 2 I rewrote the equation as inverse of x in mod (x^2+2x+2) but then the euclidean algortihim just leaves me with :

(x^2+2x+2) = (x)(x+2) +2 x = (-1)(-x)+0 And when I do it backwards I get the inverse as (2-x) which doesn't seem to work.

Answer 1

The first question has an easy answer: the inverse of $1$ is $1$, as in all rings.

Question $2)$: as we're in $\mathbf F_3$, $-2x=x$, and the euclidean division yields: $$x^2+2x+2=x^2-x-1=x(x-1)-1,\enspace\text{whence}\enspace x(x-1)=1+x^2-x-1,$$ so $(2x)^{-1}=x-1$.

Question $3)$: $-4=-1$ in $\mathbf F_3$, hence its inverse is $-1$.