I have this:
$ (x^5+1) (x^5-1) $
And I don't know how to continue factor.
Geogebra's Factor says:
$(x+1)(x-1)(x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$
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Are you factoring over $\mathbb{Q}, \mathbb{R}$ or $\mathbb{C}$? – James Harrison Sep 21 '14 at 00:36
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@JamesHarrison $\mathbb{Q}$, probably. – Akiva Weinberger Sep 21 '14 at 00:38
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@JamesHarrison I don't know what you say, or yes, but I don't have a selection. I'd like it to be like Geogebra's result so I think I need it in N – JorgeeFG Sep 21 '14 at 00:38
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The factors of $(x^5+1)$ are covered in Question 138382. – Jam Nov 21 '19 at 23:05
3 Answers
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Since $-1$ is a root of $x^5+1$, we know that it is divisible by $x+1$. Likewise, since $1$ is a root of $x^5-1$, we know that it is divisible by $x-1$. You can use polynomial long division to obtain the other factors.
Josh B
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Hint:
$$x^n-1=(x - 1)[x^{n - 1} + x^{n - 2} + ... + x^2 + x + 1] $$
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1Thanks, I noted that $x^4-x^3+x^2-x+1$ cannot be factored, is it true? how do I know WHEN something cannot be expressed as factors? – JorgeeFG Sep 21 '14 at 01:24
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The divisors of the constant term, $1$, are $\pm 1$. So, you have to apply the division of $x^4-x^3+x^2-x+1$ and $x-1$ and the division of $x^4-x^3+x^2-x+1$ and $x+1$. When the remainder is not equal to $0$, then it cannot be expressed as factors. – Sep 21 '14 at 13:00
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I assume that you are trying to factor these polynomials over $\mathbb{Q}$.
What you need is to check that $f(x)=x^{p-1}+\ldots+1$ is irreducible for any prime $p$. To see this, consider $f(x+1)=\frac{(x+1)^p-1}{x}$ for which Eisenstein's criterion works.
user2097
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