# Show that in ℝ[x], no polynomial of odd degree > 1 is irreducible. [duplicate]

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.

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## marked as duplicate by user2345215, Sami Ben Romdhane, T. Bongers, user86418, MagdiragdagApr 17 '14 at 20:06

How can you «factor into something with a x to a power plus or minus some real number» the polynomial $x^2+x+1$? –  Mariano Suárez-Alvarez Apr 17 '14 at 19:16
@MarianoSuárez-Alvarez odd degree. –  user141763 Apr 17 '14 at 19:17
This is usually proved using the intermediate value theorem. Do you have access to this theorem? –  Ayman Hourieh Apr 17 '14 at 19:19
Right. This question just screams "I'm a duplicate". –  user2345215 Apr 17 '14 at 19:29

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?

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limx→−∞ for a polynomial of odd degree would be -∞ and limx→+∞ would be ∞. –  mmm Apr 17 '14 at 19:26
@Megan: Yes. And polynomials are continuous. So the graph must cross the $y=0$ line and has a root. –  user2345215 Apr 17 '14 at 19:26
Ah yes I do see how the intermediate value theorem works here now. Thank you for the assistance –  mmm Apr 17 '14 at 19:30

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.

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Is this because, for example if you have a factor of x^5+1 then x=-1 is a root, but for a polynomial of even degree, say x^4+1 does not have a root because there is no number x such that x^4=-1 in the real numbers? –  mmm Apr 17 '14 at 19:20
I think you're a bit confused. A polynomial of even degree can of course has a root, for example $x^4-1$ has roots $1, -1$. But some don't. However, a polynomial of odd degree always has a root because of the intermediate value theorem. –  Najib Idrissi Apr 17 '14 at 19:22
Oh yes, sorry I should have been clearer. I do understand how polynomials of even degrees work. Can you explain how the IVT is applied here though, I don't quite understand how that works with this. –  mmm Apr 17 '14 at 19:24
$x^4+1$ does not have a real root, but is still reducible over $\mathbb{R}$, as it factors as $$x^4+1 = (x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$$ –  Nicholas Stull Apr 17 '14 at 19:24
@FlybyNight Sorry if my comment seemed irrelevant. I was simply trying to clarify that not having a real root is not enough to conclude a polynomial is irreducible. It is true that every odd degree polynomial must have a real root, hence cannot be irreducible, but no polynomial of degree $\geq 3$ is irreducible over $\mathbb{R}$ either, which is what I was hinting at. –  Nicholas Stull Apr 17 '14 at 19:42

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!!!

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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?

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