Quotient of two ideals equality

Question

I was following a sequence of ring isomorphisms, sorry about that but I cannot figure out why in the last step we have:

$ (x^2+ 1, 1+x) / (1+x) = ( 2) $ in the ring $ \mathbb{Z} [X] / (1+x) $

Answer 1

We have that:

$$x^2 + 1 = x(x+1) - x + 1 = x(x+1) - (x+1) + 2$$

Thus in $\mathbb{Z}[x]/(x+1)$ we have that $x^2 + 1 = 2$, so $(x^2 + 1) + I = 2 + I$, where $I = \langle 1+x \rangle$