Is $f(x)=x^3-9x^2+6x-12$ irreducible over $\mathbb{R}$? I want to show that $f(x)=x^3-9x^2+6x-12$ is reducible over $\mathbb{R}$. Since $\mathbb{R}$ is a field I know that if $f$ has a root in $\mathbb{R}$ then $f$ has a linear factor and hence it is reducible. My way of checking this seems like "cheating" because we don't talk about the notion of continuity in ring theory. One can easily show that $f(0)<0$ and $f(3)>0$, and since polynomials are continuous, the intermediate value theorem guarantees that $f$ has a root in $\mathbb{R}$. How could I show this in another way? Thanks!
 A: Irreducible polynomials over the real numbers have degree $1$ or $2$, see here. So any real cubic polynomial is reducible.
A: If you don't talk about the notion of continuity, you aren't talking about $\mathbb{R}$. A defining feature of $\mathbb{R}$ is its completeness, and without it I think there's no way to show that $f(x)$ is reducible over $\mathbb{R}$.
Suppose you managed to show $f(x)$ is reducible over $\mathbb{R}$, without using completeness of $\mathbb{R}$, then the same argument would work if you replace each with $\mathbb{R}$ with $\mathbb{Q}$.
The reason is that the axioms for the real numbers are the same as those for the rational numbers, other than the axiom of completeness. So any argument about $\mathbb{R}$ that doesn't use completeness must work for $\mathbb{Q}$ as well.
Since $f(x)$ is irreducible over $\mathbb{Q}$, there's no way to prove $f(x)$ is reducible over $\mathbb{R}$ without using its completeness.
A: Cardano's method gives nice numbers for this one. the real root is
$$ 3 + \sqrt[3]{24 + \sqrt{233}} +  \sqrt[3]{24 - \sqrt{233}} \; \; \approx \; \;  8.458372474   $$
A: The continuity argument is the easiest. However here is another one:
Your polynomial $P$ has three roots in $\mathbb{C}$, if $\lambda$ is a root which is not real, then $\bar{lambda}$ too, because $P(\bar{\lambda}) = \bar{P(\lambda)}$. So $P$ = $(X - \lambda)(X- \bar{\lambda})(X-\mu)$. By identifying the degree 2 coefficient, you have  $-\mu - lambda - \bar{\lambda} \in \mathbb{R}$ and so $\mu \in \mathbb{R}$.
