Factorization of $x^5+x$

Question

I need to make the decomposition in $\mathbb{R}$ of:

$$x^5+x$$

Here my steps:

$$x(x^4+1)$$ $$x\big[ (x^2)^2+1 \big]$$ $$x\big[(x^2+1)^2-2x^2\big]$$

how should I proceed?

score 4 · Accepted Answer · answered May 26 '16 at 08:56

4

Just use $$b^2-a^2=(b-a)(b+a).$$

answered May 26 '16 at 08:56

Paul

20,553

thanks, I get: $x\big[(x^2+1+\sqrt{2}x)(x^2+1-\sqrt{2}x)\big]$ but didn't match with Wolfram result – Christian Giupponi May 26 '16 at 09:04
@ChristianGiupponi Next use Sridhar Acharya's formula...Search Google if you don't know it. – User Not Found May 26 '16 at 09:05
@ArghyaChakraborty I get $\Delta < 0$ – Christian Giupponi May 26 '16 at 09:10
1

@ChristianGiupponi That means that it is irreducible any further over $\mathbb{R}$...remember the point...over $\mathbb{R}$ – User Not Found May 26 '16 at 09:13
@ChristianGiupponi Hey one of them is reducible by Sridhar Acharya isn't it? – User Not Found May 26 '16 at 09:14
I get $\Delta < 0$ with both – Christian Giupponi May 26 '16 at 09:16

score 0 · Answer 2 · answered May 26 '16 at 09:35

To factor $x^4+1$ into linear factors find the roots of that polynomial! So we are looking for complex numbers satisfying $x^4=-1$. Squaring gives us $x^8=1$. So we are looking for 8th roots of unity. Since $x^4=-1\ne 1$ we are looking for primitive 8th roots of unity. There are exactly four of them: $$\frac{\pm 1 \pm i}{\sqrt{2}}.$$

score 0 · Answer 3 · answered May 26 '16 at 10:18

Another (less ingenious but more systematic?) way is to first work over complex numbers:
$$\begin{align}x^4+1&=(x^2)^2-i^2=(x^2-i)(x^2+i)\\ &=(x+\frac{1+i}{\sqrt2})(x-\frac{1+i}{\sqrt2})\cdot(x+\frac{-1+i}{\sqrt2})(x-\frac{-1+i}{\sqrt2})\\ &=(x-\frac{1+i}{\sqrt2})(x+\frac{-1+i}{\sqrt2})\cdot(x-\frac{-1+i}{\sqrt2})(x+\frac{1+i}{\sqrt2})\\ &=(x^2-\sqrt2x+1)(x^2+\sqrt2x+1)\end{align}$$ So this obtains the decomposition by permuting the two hidden factors over complex numbers.
Hope this helps.

Factorization of $x^5+x$

3 Answers3