Consider the ring $R=\mathbb{Z}_5[x]/\langle x^3+x^2+x+3\rangle.$
$1.$ Find a $f(x)\in\mathbb{Z}_5[x]$ with $\deg(f(x)) <3$ such that $[f(x)] = [x^5 + 3x^3+2x+4]$ in $R.$
$2.$ Find the multiplicative inverse of $[x^4+2x^3+x+4]$ in $R.$
$3.$ Find the multiplicative inverse of $[x^2+1]$ in $R.$
$1.$
In $R,$ $[x^3+x^2+x+3]=[0].$ So $[x^5+x^4+x^3+3x^2]=[0].$ We now proceed to simplify $[x^5 + 3x^3+2x+4]$ in $R.$ $$[x^5 + 3x^3+2x+4]-[x^5+x^4+x^3+3x^2]=[4x^4+2x^3+2x^2+2x+4]-[4x^4+4x^3+4x^2+2x]=[3x^3+3x^2+4]-[3x^3+3x^2+3x+4]=[2x].$$
Since $\deg (2x)<3$ and $[2x]=[x^5+3x^3+2x+4],$ $\boxed{f(x)=2x}.$
$2.$
We simplify $[x^4+2x^3+x+4]$ in $R.$ We have that $$[x^4+2x^3+x+4]=[x^4+2x^3+x+4]-[x^4+x^3+x^2+3x]\\ =[x^3+4x^2+3x+4]-[x^3+x^2+x+3]\\ =[3x^2+2x+1].$$
By long division, in $\mathbb{Z}_5[x],(3x^2+2x+1)(2x+4)+4+x=x^3+x^2+x+3\; (1).$
By long division, in $\mathbb{Z}_5[x],$ $3x^2+2x+1=(4+x)(3x)+1\Rightarrow 4(3x^2+2x+1)=(4+x)(4(3x))+4.$
So in $\mathbb{Z}_5[x]/\langle3x^2+2x+1\rangle, [4][0]=[0]=[4+x][4(3x)]+4\Rightarrow [4+x][2x]=[1]\Rightarrow [4+x]^{-1}=[2x].$
So, multiplying both sides of $(1)$ by $2x,$ we have that in $\mathbb{Z}_5[x],$ $$(2x)(x^3+x^2+x+3)=(3x^2+2x+1)(2x+4)(2x)+(4+x)(2x)\\ =(3x^2+2x+1)(4x^2+3x)+4(3x^2+2x+1)+1\\ \Rightarrow 4(2x)(x^3+x^2+x+3) = (3x^2+2x+1)(4(4x^2+3x+4))+4\\ =(3x^2+2x+1)(x^2+2x+1)+4.$$ Thus, in $R,$ $$[4][2x][0]=[0]=[3x^2+2x+1][x^2+2x+1]+[4]\Rightarrow [3x^2+2x+1][x^2+2x+1]=[1]\Rightarrow \boxed{[x^4+2x^3+x+4]^{-1}=[3x^2+2x+1]^{-1}=[x^2+2x+1]}.$$
$3.$
Since $\deg(x^2+1)<3,$ we just need to divide $x^3+x^2+x+3$ by $x^2+1$ in $\mathbb{Z}_5[x]$ to determine the inverse. Doing so yields $x^3+x^2+x+3=(x^2+1)(x+1)+2\Rightarrow (x^2+1)\cdot2(x+1)+4=2(x^3+x^2+x+3).$ Thus, in $R,$ $[x^2+1][2(x+1)]+[4]=[0]\Rightarrow [x^2+1][2x+2]=[1].$ Thus, $\boxed{[x^2+1]^{-1} = [2x+2]}.$
Edit: I think I finally found the inverses! yay. However, are these the only inverses? I’m pretty sure multiplicative inverses are unique in a ring.