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Hi im trying my hand on these types of expressions. In the book I'm using it mentions that for these types you should rewrite the x term such as $2x-x$. But I can't seem to make any use of this technique? What is this type of factorize called by the way?

thanks

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    $\begingroup$ I have no idea what the $2x-x$ is supposed to do for you. But you can get some mileage out of adding and subtracting $x^2$, that is, writing $(x^5-x^2)+(x^2+x+1)$. $\endgroup$ – Gerry Myerson Aug 27 '13 at 13:10
  • $\begingroup$ The Maple code $$infolevel[factor] := 5; factor(x^5+x+1) $$ produces $factor/polynom: polynomial factorization: number of terms 3 factor/unifactor: entering factor/unifactor: polynomial has degree 5 with 1 digit coefficients factor/linfacts: computing the linear factors factor/linfacts: there are 0 roots mod 2 factor/fac1mod: entering factor/fac1mod: found prime 2 factor/fac1mod: distinct degree factorization ... factor/lift: the product of the true factors is (x^2+x+1)*(x^3-x^2+1)factor/fac1mod: factorization is (x^2+x+1)*(x^3-x^2+1) factor/unifactor: exiting(x^2 + x + 1)( x^3 - x + 1) $ $\endgroup$ – user64494 Aug 27 '13 at 13:31
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    $\begingroup$ @user64494, just exactly how is this supposed to help OP (or anyone else, for that matter)? $\endgroup$ – Gerry Myerson Aug 27 '13 at 13:40
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As Gerry suggests, we can add and subtract $\;+ x^2 - x^2 = 0 $ without changing the expression:

$$\begin{align} x^5 + x + 1 & = (x^5-x^2)+(x^2+x+1)\\ \\ & = x^2(x^3 - 1) + (x^2 + x + 1) \\ \\ & = x^2(x-1)\color{blue}{\bf (x^2 + x + 1)} + \color{blue}{\bf (x^2 + x + 1)}\end{align}$$

Now factor out the common factor...which gives us $$\Big(x^2(x-1) + 1\Big)(x^2 + x + 1) = (x^3 - x^2 + 1)(x^2 + x + 1)$$

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  • $\begingroup$ I got $(x^2+x+1)(x^2(x-1)+1) = (x^2+x+1)(x^3-x^2+1)$ How do I spot that fact that I should add and subtract $x^2$? Also I didn't know that $(x^3-1)=(x-1)(x^2+x+1)$, are there any main identities like these that I should be aware of? thanks $\endgroup$ – salman Aug 27 '13 at 13:30
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    $\begingroup$ How to spot? It usually is a matter of practice! Also, whenever you encounter $x^n - 1$, you should check $x - 1$ as a factor! $\endgroup$ – Namaste Aug 27 '13 at 13:37
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    $\begingroup$ You should definitely be aware that $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1$$ There's a similar identity for $x^n+1$, though it only works when $n$ is odd --- you might try to work it out. $\endgroup$ – Gerry Myerson Aug 27 '13 at 13:38
  • $\begingroup$ @GerryMyerson How do I use this when $n=2$? I tried: $x^2 -1 = (x-1)(x^1+x^0+x^{-1}+x^{-2}+...+x+1) = (x-1)(2x+2+x^{-1}+x^{-2}+x^{-3}+...)$ Is the $x^{-1}+x^{-2}+...$ a sum to infinity with $a = x^{-1}$ and $r=x^{-1}$? $\endgroup$ – salman Aug 27 '13 at 14:17
  • $\begingroup$ @BabakS. any help please? :) $\endgroup$ – salman Aug 27 '13 at 14:49
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You can verify that both primitive $3$rd roots of unity are roots of this equation, since $5$ is $2$ mod $3$. This means the corresponding cyclotomic polynomial, $x^2+x+1$, divides $x^5+x+1$, and you can do polynomial division to find the other factor.

The answer is $(x^2+x+1)(x^3-x^2+1)$.

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    $\begingroup$ This is fine --- if OP knows what a "primitive 3rd root of unity" is. $\endgroup$ – Gerry Myerson Aug 27 '13 at 13:36
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    $\begingroup$ @GerryMyerson I admit this is probably a useless answer. I'm only posting it because one of my physics professors once used this polynomial as a example of a 5th degree equation that could not be solved in radicals. (It was in a lecture on perturbation theory, and the roots of the polynomial were then found numerically using a perturbation method.) I didn't catch his mistake until after the lecture, but I've been on the lookout for this polynomial ever since. $\endgroup$ – Potato Aug 27 '13 at 13:39
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    $\begingroup$ Well, that's what happens when Physics professors try to teach math. He'd've been on safe ground with $x^5-x-1$. $\endgroup$ – Gerry Myerson Aug 27 '13 at 13:43

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