Help with factor groups and domains I can't see it. 
$\Large{\frac{\mathbb{Z}[x]}{(x^2-29)}}$
I know you take general polynoimals with coefficients in $\mathbb{Z}$ and add $(x^2-29)$. However, I'm a bit confused on what it is. Like I image it as a big coset. 
Also, how do you show that it is isomorphic to a subring of $\mathbb{R}$? wouldn't you need to do the subring test. 
 A: Well, do you have any ideas as to what subring of $\mathbb{R}$ it is isomorphic to? For example, what real number(s) does the element $x$ "act like"? Specifically, what is $x^2$ in the ring $\mathbb{Z}[x]/(x^2-29)$?
That should then suggest a homomorphism $f:\mathbb{Z}[x]/(x^2-29)\to \mathbb{R}$ that is an isomorphism onto its image (which will be a subring of $\mathbb{R}$).
The ring $\mathbb{Z}[x]/(x^2-29)$ consists of equivalence classes of polynomials in $\mathbb{Z}[x]$, modulo $x^2-29$. What does that mean? Well, given any two polynomials
$$f=a_0+a_1x+a_2x^2+\cdots+a_nx^n\in\mathbb{Z}[x]$$
$$g=b_0+b_1x+b_2x^2+\cdots+b_mx^m\in\mathbb{Z}[x]$$
we say that $f\equiv g\bmod (x^2-29)$ when
$$f-g=(x^2-29)h,\;\;\text{ for some }h\in\mathbb{Z}[x].$$
Consider, by analogy, the ring $\mathbb{Z}/(n)$, which consists of equivalence classes $[a]$  (where $a\in\mathbb{Z}$) such that $[a]=[b]$ iff $a\equiv b\bmod n$, and satisfying $[a]+[b]=[a+b]$ and $[a][b]=[ab]$. When does $[a]=[0]$?
Similarly, the ring $\mathbb{Z}[x]/(x^2-29)$ consists of equivalence classes $[f]$ such that $[f]=[g]$ iff $f\equiv g\bmod (x^2-29)$, and $[f]+[g]=[f+g]$ and $[f][g]=[fg]$. When does $[f]=[0]$ in $\mathbb{Z}[x]/(x^2-29)$? Can you characterize that via the roots of $f$? Consider the Factor Theorem.
