Consider $\mathbb R(\theta)$ with $\theta \notin \mathbb R$ and $\theta^3 -4\theta^2 +6\theta +6$. What is the degree of [$\mathbb R (\theta):\mathbb R$]?
I figure, if $\theta \notin \mathbb R$, then $\theta \in \mathbb C$. Thus if i can reduce $\theta^3 -4\theta^2 +6\theta +6$ in $\mathbb C$, then is its degree the value of [$\mathbb R (\theta):\mathbb R$]? Or does it simply mean $\theta$ is not reducible in $\mathbb R$ so only the polynomial needs reducing?
However, I'm not sure how to reduce it. I have tried reducing it in $\mathbb R $ but there isn't an easy factorisation. Where am I going wrong?
Irreducibility is irrelevant. There are no fields intermediate between $\mathbb R$ and $\mathbb C$, so if $\theta \notin \mathbb R$ then $\mathbb R(\theta) = \mathbb C$.
Edit: And just to answer the irreducibility part: You know that polynomials with real coefficients can have complex roots but they come in conjugate pairs. This polynomial has odd degree so it can have at most $1$ complex pair of roots, the other root must be real. So it is irreducible, it can be written as a product of a degree one polynomial and an irreducible degree two polynomial.
Of course, that doesn't mean finding that real root will be easy. There is a formula for degree $3$ polynomials but it's not a simple one liner like the quadratic equation is.