Factor $x^4+1$ over $\mathbb{R}$

Well, I read this question first wrongly, because the reader is about complex analysis, I did it for $\mathbb{C}$ first.

I got. $x^4+1=(x-e^{\pi i/4 })(x-e^{3 \pi i/4})(x-e^{5\pi i/4})(x-e^{7\pi i/4})$.

My teacher told me that there is very smart way to do this for $\mathbb{R}$ that we already learned. But I only can think of trial and error kind of methods.


Group each complex root $\alpha$ with $\bar\alpha$: $$ (x-\alpha)(x-\bar\alpha)=x^2-(\alpha+\bar\alpha)x+\alpha\bar\alpha\in{\Bbb R}[x]. $$

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    $\begingroup$ @Kasper: This is the smart way that your teacher meant (most probably). $\endgroup$ – J.R. Mar 4 '14 at 11:46
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    $\begingroup$ Aah, that looks very smart indeed, thx ! $\endgroup$ – Kasper Mar 4 '14 at 11:49

It's like this: $x^4+1 = x^4 + 2x^2 + 1 - 2x^2 = (x^2+1)^2 - (\sqrt2x)^2 = (x^2+\sqrt2x+1)(x^2-\sqrt2x+1)$

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    $\begingroup$ My vote is that this was the smart way the teacher had in mind $\endgroup$ – Geoff Robinson Mar 4 '14 at 13:21

In complicated terms, the field extension $\mathbb C / \mathbb R$ has degree $2$, so you expect every quartic to be reducible over $\mathbb R$. In simple terms, since you have the four roots of this polynomial, there is a nice way to group them together : the roots $e^{i\pi/4}$ and $e^{7 i \pi/4}$ are conjugate, so they are the roots of the same quadratic ; group the two linear factors together and you will get a real quadratic. Similarly for the roots corresponding to $3$ and $5$.

Hope that helps,


In www.wolframalpha.com , I typed factor x^4 + 1 It gives 4 first order terms. 4 square roots of i

Hope this helps. Regards, Matt


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