# Factor $x^5 + x + 1$

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

• 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 $(x^5-x^2)+(x^2+x+1)$. – Gerry Myerson Aug 27 '13 at 13:10
• The Maple code $$infolevel[factor] := 5; factor(x^5+x+1)$$ produces $factor/polynom: polynomial factorization: number of terms 3 factor/unifactor: entering factor/unifactor: polynomial has degree 5 with 1 digit coefficients factor/linfacts: computing the linear factors factor/linfacts: there are 0 roots mod 2 factor/fac1mod: entering factor/fac1mod: found prime 2 factor/fac1mod: distinct degree factorization ... factor/lift: the product of the true factors is (x^2+x+1)*(x^3-x^2+1)factor/fac1mod: factorization is (x^2+x+1)*(x^3-x^2+1) factor/unifactor: exiting(x^2 + x + 1)( x^3 - x + 1)$ – user64494 Aug 27 '13 at 13:31
• @user64494, just exactly how is this supposed to help OP (or anyone else, for that matter)? – Gerry Myerson Aug 27 '13 at 13:40

## 2 Answers

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 $$\Big(x^2(x-1) + 1\Big)(x^2 + x + 1) = (x^3 - x^2 + 1)(x^2 + x + 1)$$

• 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 – salman Aug 27 '13 at 13:30
• 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! – Namaste Aug 27 '13 at 13:37
• 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. – Gerry Myerson Aug 27 '13 at 13:38
• @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 $r=x^{-1}$? – salman Aug 27 '13 at 14:17
• @BabakS. any help please? :) – salman Aug 27 '13 at 14:49

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 $(x^2+x+1)(x^3-x^2+1)$.

• This is fine --- if OP knows what a "primitive 3rd root of unity" is. – Gerry Myerson Aug 27 '13 at 13:36
• @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. – Potato Aug 27 '13 at 13:39
• Well, that's what happens when Physics professors try to teach math. He'd've been on safe ground with $x^5-x-1$. – Gerry Myerson Aug 27 '13 at 13:43