# Factor $z^5 + z +1 =0$

The problem stated in the title is:

Factor $z^5 + z + 1 = 0$ (naturally without the use of computers or calculators)

How do I go about solving this? Is there a more systematic approach than simply guessing a root and then applying polynomial division etc. ? Thank you for any help!

## 3 Answers

If $z^5 + z + 1 = f(z)g(z)$, then:

• $\deg f =1, \deg g=4$, or

• $\deg f =2, \deg g=3$

In both cases, you can try to determine the coefficients. Perhaps simplify things by first trying monic polynomials with independent term $1$.

The second case gives $$(z^2+a z+1)(z^3+bz^2+c z+ 1) = z^5 + (a + b)z^4 + (a b + c + 1)z^3 + (a c + b + 1)z^2 + (a + c)z + 1$$ Forcing the coefficients of $z^4,z^3,z^2$ to be zero and $a+c=1$, we get $$z^5 + z + 1 = (z^2 + z + 1) (z^3 - z^2 + 1)$$

• Ah OK, I've never seen this approach before! This looks quite good, thanks. – wittbluenote Dec 8 '17 at 10:55
• Thank you this very helpful! I suppose this technique breaks down though when the degree of the polynomial becomes large. But to be fair I guess very few techniques work for large degree polynomials. Thanks – wittbluenote Dec 8 '17 at 11:13

Hint: \begin{eqnarray} z^5+z+1 &=& z^5-z^2+z^2+z+1\\ &=& z^2(z^3-1)+z ^2+z+1 \\ &=&... \end{eqnarray}

• – Maria Mazur Dec 8 '17 at 11:00
• Thank you! I am not sure how to go on right now, but the first thing I am thinking of is expressing the terms $z^2 + z + 1$ as a geometric series $(z^3-1)/(z-1)$ and then factoring... Is this a good idea? I guess it at least established that we have a cube root of unity?! – wittbluenote Dec 8 '17 at 11:01
• Is this idea of "creatively adding zero" as you did a general approach. What told you that adding $z^2$ would be helpful? Thank you – wittbluenote Dec 8 '17 at 11:03
• Yes, but not everything will work. You have to have some personal experience to see what to add. – Maria Mazur Dec 8 '17 at 11:43
• I see. Thanks ! – wittbluenote Dec 8 '17 at 11:52

Hint:

If $z=w$ where $w$ is a complex cube root of unity

$z^5+z+1=0$

• How did you find that result? Did you guess? Thanks for your reply. – wittbluenote Dec 8 '17 at 10:46
• Why the downvotes? This is a perfectly valid hint (though it could have been a comment...) – Sebastian Schoennenbeck Dec 8 '17 at 10:51
• @wittbluenote, This works for $$z^{3m+2}+z^{3n+1}+z^{3r}$$ – lab bhattacharjee Dec 8 '17 at 11:04
• @SebastianSchoennenbeck, This is a very common phenomenon. I don't mind until I find something really wrong in my post – lab bhattacharjee Dec 8 '17 at 11:06
• Thanks! I suppose you argued like @John Watson... – wittbluenote Dec 8 '17 at 11:06