How does the quotient ring $Bbb Z[x]/(x^2-x,4x+2)$ look like?

We have $8(x^2-x)-2x(4x+2)=-12x=-3(4x+2)+6$, so $6in (x^2-x,4x+2)$ ,and $(x^2-x,4x+2)=(6,x^2-x,4x+2)$.

Hence, your ring $R$ is isomorphic to $mathbb{Z}/6 [x]/(x^2-x,-bar{2}x+bar{2})$.

The map given by the Chinese remainder theorem yields an isomorphism $Rsimeq mathbb{F}_2[x]/(x^2-x)timesmathbb{F}_3[x]/(x^2-x,x-bar{1})=mathbb{F}_2[x]/(x^2-x)timesmathbb{F}_3[x]/(x-bar{1})$, which finally yields $$Rsimeqmathbb{F}_2timesmathbb{F}_2timesmathbb{F}_3.$$

If we follow carefully the proof, we get an explicit isomorphism.

Let $f$ be the ring morphism $f:Rto Pinmathbb{Z}[x]mapsto ([P(0)]_2,[P(1)]_2, [P(1)]_3)inmathbb{F}_2timesmathbb{F}_2times mathbb{F}_3$.

It is easy to see $(x^2-x,4x+2)$ lies in the kernel of $f$. The induced map is then the desired isomorphism.

We can also prove this last fact directly, just to double check that everything is correct.

Claim. $f$ is surjective.

Given $a,b,cinmathbb{Z}$, we need to find $Pinmathbb{Z}[x]$ such that $P(0)equiv a [2], P(1)equiv b [2]$ and $P(1) equiv c [3]$.

We can try $P=ux+v$. We can choose $v=a$, $u=(b-a)+2k$, so the two first equations are satisfied. Now we want $(b-a)+2k+aequiv c [3]$, and we take $k=b-c$. To sum up $P=(3b-a-2c)x+a$ does the job.

Claim. $ker(f)=(x^2-x,4x-2)$.

As we said before, one inclusion is clear, so let $Pinmathbb{Z}[x]$ such that $f(P)$ is trivial. We want to prove that $Pin (x^2-x,4x+2)$. Dividing by $x^2-x$, and replacing $P$ by the corresponding remainder,one may assume that $P=ux+v$. By assumption, $v$ is even, and $u+v$ is a multiple of $2$ and $3$, so $v=2m$ and $u+v=6n$, that is $u=6n-v=6n-2m$. Hence $P=-2mx+ 6nx+2m=m(2-2x)-n 6x$. Now $6x$ lies in our ideal (since $6$ does), and $2-2x=4x+2-6x$ also lies in our ideal, so we are done.

Now apply the first isomorphism theorem.