Show that in ℝ[x], no polynomial of odd degree > 1 is irreducible. I think that logically, I understand the concept because no matter what polynomial you have you can always factor it into something with a x to a power plus or minus some real number, and that real number can be a fraction, or an irrational number, or a whole number. However, I am not sure how to write this in a reasonable mathematical way.
 A: A polynomial of odd degree always has a root. Indeed by looking at the highest degree term, $\lim_{x \to \infty} p(x) = \infty$ and $\lim_{x \to -\infty} p(x) = -\infty$ (or the reverse if the leading coefficient is negative), so you can apply the intermediate value theorem. And of course, if a polynomial has a root it's reducible.
A: Hint: You only need $\deg p\ge1$ odd. Can you compute $\lim\limits_{x\to-\infty}p(x)$ and $\lim\limits_{x\to+\infty}p(x)?$ What does this tell you?
A: A polynomial $p(x) \in \Bbb R[x]$ with $\deg p$ odd always has a real zero $\alpha$; thus we have $p(x) = (x - \alpha)q(x)$ with $q(x) \in \Bbb R[x]$, showing that $p(x)$ is reducible in $\Bbb R[x]$.
A complete proof such $p(x)$ must have a real root may be found in my answer to this question, only a mouse click away!
Hope this helps.  Cheerio,
and of course, 
Fiat Lux!!!
A: Hint:  Complex roots of polynomials in $\mathbb{R}[x]$ always come in pairs.  Further, the total number of roots (counting multiplicity) is always $deg(f)$.  So what can you deduce if $deg(f)$ is odd?
