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How to find roots of $X^5 - 1$? (Or any polynomial of that form where $X$ has an odd power.)

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You mean complex roots? Just write an arbitrary complex number as $re^{i\theta}$, substitute into the polynomial and solve first for $r$, then for $\theta$. – Alex B. May 15 '11 at 12:29
Note that the comment of Alex B applies to any polynomial of the form $x^n-c$ where $n$ is a positive integer and $c$ is a complex number. – Gerry Myerson May 15 '11 at 23:35
Over which domain? In other words, what is the type of $X$? Does $X \in \mathbb{Z}$ or $\in \mathbb{C}$ etc? – user2468 May 16 '11 at 16:28
@M.S., I think we've all been assuming what's wanted is solutions in the complex numbers, and Paul hasn't raised any objections, so it's safe to conclude that that's the intention. I can't imagine anyone needing help to find the roots in the integers. – Gerry Myerson May 16 '11 at 23:19
up vote 15 down vote accepted

Edit 2: Since the equation $x^n-1=0$ is equivalent to $x=\sqrt[n]{1}$, it has the following $n$ solutions:

$$e^{i\dfrac{2\pi (1+k)}{n}}=\cos\left( \dfrac{2\pi (1+k)}{n}\right)+i\sin\left( \dfrac{2\pi (1+k)}{n}\right),\qquad k=0,1,\ldots ,n-1,$$

where $n$ is a positive integer, odd or even.

See comment by Gerry Myerson indicating how the Galois group for an irreducible quartic determines whether the method exposed below works.

Although late I present another algebraic solution, less elegant but more "automatic", in the sense that no clever trick is necessary. [Edit in response to Gerry Myerson's comment] We are going to factor it into two factors, one being linear and the other quadratic. This method works for quintic polynomials, as long as we are able to find one solution and solve the quadratic polynomial factor. If it fails, in principle we can solve the quadratic polynomial by means of a resolvent cubic equation (also in this post in Portuguese) [end of edit]. For the general polynomial of 7th degree or higher it is not applicable, because it depends on finding one root by inspection (or numerically) and the remaining ones algebraically.

By inspection we see that $x=1$ is a zero of $x^{5}-1$. By long division or Ruffini's rule, we find

$$x^{5}-1=\left( x-1\right) \left( x^{4}+x^{3}+x^{2}+x+1\right) .$$

Now we factor $x^{4}+x^{3}+x^{2}+x+1$ into two quadratic polynomials

$$x^{4}+x^{3}+x^{2}+x+1=\left( x^{2}+bx+c\right) \left( x^{2}+Bx+C\right) ,$$

whose (real) coefficients have to be found by comparing LHS with the expanded RHS. Since

$$\left( x^{2}+bx+c\right) \left( x^{2}+Bx+C\right)$$

$$ =x^{4}+\left( B+b\right) x^{3}+\left( C+bB+c\right) x^{2}+\left(bC+cB\right) x+cC,$$

the coefficients must satisfy

$$\left\{ \begin{array}{c} B+b=1 \\ C+bB+c=1 \\ bC+cB=1 \\ cC=1% \end{array}% \right. \Leftrightarrow \left\{ \begin{array}{c} B=1-b \\ 1/c+b\left( 1-b\right) +c=1 \\ b/c+c\left( 1-b\right) =1 \\ C=1/c% \end{array}% \right. $$

One of the solutions of $b/c+c\left( 1-b\right) =1$ is $c=1$. By substitution we find the remaining coefficients:

$$\left\{ \begin{array}{c} B=\frac{1}{2}\mp \frac{1}{2}\sqrt{5} \\ b=\frac{1}{2}\pm \frac{1}{2}\sqrt{5} \\ c=1 \\ C=1% \end{array}% \right. $$


$$x^{4}+x^{3}+x^{2}+x+1=\left( x^{2}+\left( \frac{1}{2}+\frac{1}{2}\sqrt{5}% \right) x+1\right) \left( x^{2}+\left( \frac{1}{2}-\frac{1}{2}\sqrt{5}% \right) x+1\right) $$

and finally

$$x^{5}-1=\left( x-1\right) \left( x^{2}+\left( \frac{1}{2}+\frac{1}{2}\sqrt{5% }\right) x+1\right) \left( x^{2}+\left( \frac{1}{2}-\frac{1}{2}\sqrt{5}% \right) x+1\right) .$$

Hence the roots of $x^{5}-1$ are the roots of these three factors. They are, respectively, $x_1=1$ and

$$x_{2,3}=-\frac{1}{4}-\frac{1}{4}\sqrt{5}\pm \frac{1}{4}\sqrt{-10+2\sqrt{5}},\quad x_{4,5}=-\frac{1}{4}+\frac{1}{4}\sqrt{5}\pm \frac{1}{4}\sqrt{-10-2\sqrt{5}}.$$

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How general is this "general method"? It's not going to work for $x^{11}-1$, is it, and that would seem to be the kind of polynomial Paul has in mind. – Gerry Myerson May 15 '11 at 23:30
@Gerry Myerson: See the revised answer. – Américo Tavares May 15 '11 at 23:36
@Americo, thanks. I suspect that as a "general method for quintic polynomials, as long as we are able to find one solution" it's going to run into some difficulties with, say, $(x-1)(x^4-x-1)$. – Gerry Myerson May 16 '11 at 0:03
@Gerry: That's right. Applying this method to $(x^{4}-x-1)=\left( x^{2}+bx+c\right) \left( x^{2}+Bx+C\right) $ I was left with a system of equations, one of which is $-1/c-c^{2}/\left( 1+c^{2}\right) ^{2}+c=0$. – Américo Tavares May 16 '11 at 10:15
... Does Abstract Algebra provides some criteria to know in advance if a quartic polynomial can be factored in this way? I know nothing of Abstract Algebra. Anyway, one can solve quartic equations by a resolvent cubic equation. I didn't check if we run into troubles when applied to $x^{4}-x-1=0$. – Américo Tavares May 16 '11 at 10:15

It is easy to write down the solution in trigonometric functions, namely, $x=e^{2\pi i n/5}$ for $n=0,\ldots, 4$. Here is an algebraic solution. First factor the polynomial as $$x^5-1=(x-1)(x^4+x^3+x^2+x+1).$$ The first factor gives you the obvious solution $x=1$. To find other roots, we need to solve the equation $$ x^4+x^3+x^2+x+1=0.$$ Divides both sides by $x^2$, and use the fact that $(x+\frac{1}{x})^2=x^2+2+\frac{1}{x^2}$, we get $$ (x+\frac{1}{x})^2 +(x+\frac{1}{x})-1=0$$ Set $y=x+\frac{1}{x}$, and we get a quadratic equation $$ y^2+y-1=0,$$ which one may easily solve using quadratic formulas. Once we have solved for $y=y_0, y_1$, the equation reduces into two other quadratic equations $$ x+ \frac{1}{x}= y_i, \qquad i=0, 1 $$ or equivalently, $$ x^2-y_i x+1=0.$$ We just apply the quadratic formula again.

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@Jiangwei: I like your solution :) – user9413 May 15 '11 at 12:53
Very nice. But Paul asked about "any polynomial of that form where $x$ has an odd power." If you took $x^{11}-1$, you'd run into something a lot worse than quadratics. The trig functions are, of course, still available. – Gerry Myerson May 15 '11 at 23:32
Very elegant and nice. +1 – Américo Tavares May 16 '11 at 12:12
@Gerry: Trig functions equivalent to a complex variable as in Chandru's answer or of other kind? – Américo Tavares May 16 '11 at 12:17
@Americo, Trig functions as in the first line of Jiangwei's answer. – Gerry Myerson May 16 '11 at 13:03

Hint: Use De Moivre's theorem and note that $$1^{1/5} = \Bigl( \cos 2m\pi + i \sin{2m\pi} \Bigr)^{1/5} \ ;\qquad m \in \mathbb{Z}$$

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