Findindg the properties of a given Quotient Ring As part of an exercise given in an undergrad Algebra course, we were tasked by obtaining as much info as possible about the quotient ring: $ℝ[x]/<h(x)>$
, where: $$h(x)=11x^4+8x^3+12x^2+7x+7$$
The task is a bit vague (or rather precise, depending on how you look at it I guess), but my approach so far has been the following:

The most obvious fact is that every element of $ℝ[x]/<h(x)>$ will be of the form:
$ax^3+bx^2+cx+d$
Now, we can easily verify using available freeware, or by employing more traditional but extremely time-consuming methods, that: 
$$h(x)=11(x-(0.217647+0.886533i)(x-(0.217647-0.886533i))(x+(0.581283-0.652508i)(x-(0.581283+0.652508i))$$
and thus it's complex roots are: $0.217647+0.886533i, 0.217647-0.886533i,  0.581283-0.652508i, 0.581283+0.652508i$ 
So, $h(x)$ is irreducible over $ℝ[x]$ and it follows that since $ℝ$ is a field (thus $ℝ[x]$ is a principal ideal domain), the quotient ring $$ℝ[x]/<h(x)>$$ is a field as well.
Moreover, this quotient ring will contain (an isomorphic copy of) $ℝ$ and a root of $h(x)$.
Indeed, since h(x) is irreducible, the main ideal $I=<h(x)>$ is a non-zero prime ideal. Since $ℝ[x]$ is a principal ideal domain, the ideal $I$ is maximal and thus, the ring $ℝ[x]/<h(x)>$ is a field. Now, if we take the mapping: $a\rightarrow a+I$, this is an isomorphism from $ℝ$ to $a+I$: $a\inℝ⊂ℝ[x]/<h(x)>$. Let $θ=x+I\inℝ[x]/<h(x)>$
 We claim that $θ$ is a root of $h(x)$. In $ℝ[x]/<h(x)>$ we have:
$$h(θ)=(7+Ι)+(7+Ι)θ+...+(11+I)θ^4=$$
$$(7+I)+(7+I)(x+I+...+(11+I)(x+I)^4=$$
$$(7+I)+(7x+I)+...+(11x^4+I)=$$
$$11x^4+8x^3+12x^2+7x+7+I=$$
$$h(x)+I=I$$
Is the approach so far correct and if so, does it seem exhaustive? Any ideas that enrich it will be more than welcome.
 A: Hint 1
Study the graph of $h$ to see it has no real roots.
Hint 2
Use the derivative test to show that $h$ has no multiple roots.
Hint 3
Deduce that $h$ is the product of two distinct irreducible polynomials in $\mathbb{R}[x]$ of degree $2$.
Hint 4
Deduce that your quotient ring is isomorphic to $\mathbb{C} \oplus \mathbb{C}$.
A: One problem is that $h$ is actually not irreducible over $\mathbb{R}$; indeed your four roots can be written as $z_1,\overline{z_1},z_2,\overline{z_2}$. Now you can take the product $q_1=(x-z_1)(x-\overline{z_1})$ and find that it is equal to $$x^2-(z_1+\overline{z_1})x+z_1\overline{z_1}=x^2-2\Re(z_1)x+|z|^2,$$which is a polynomial with real coefficients. The same holds with $z_2$, so $h$ is a product of two polynomials of lesser degree and thus not irreducible.
As a result it is not true that the quotient ring is a domain (or a field). However, $I=\langle h\rangle $ is the intersection of $\langle q_1\rangle$ and $\langle q_2\rangle$, and $q_1$ and $q_2$ are coprime, so by the Chinese Remainder Theorem you get that
$$\mathbb{R}[x]/\langle h\rangle\cong \mathbb{R}[x]/\langle q_1\rangle \times \mathbb{R}[x]/\langle q_2\rangle,$$and moreover each of these factors is isomorphic to $\mathbb{C}$.
