Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In a previous task, I was asked to factor $1+x+x^2+x^3$ for $x \in \mathbb{R}$, which I accomplished by solving

$1+x+x^2+x^3 = 0 \to $

$1+x(1+x+x^2) = 0 \to $

$x(1+x+x^2) = -1$

which has a solution $x = -1$, and thus I knew $(x+1)$ was a factor. A bit of guesswork gave me $(x+1)(x^2+1)$.

Now I'm asked to factor $1+x+x^2+x^3+...+x^{14}$ for $x \in \mathbb{R}$ and I'm a bit stuck. Again, we have the implication $x(1+x+x^2+...+x^{13}) = -1$, for which $x=-1$ is a solution, so again we have a factor $x+1$. But now I cannot apply guesswork to determining the rest of the factors, so I feel there is some kind of conclusion I can draw about the powers (perhaps their parity) to solve this problem?

share|cite|improve this question
$-1$ is not a root of your equation. Plugging it into the equation gives $1$, not $0$ as desired. – Greg Muller Dec 3 '12 at 14:47
Ross's answer below raises an important question: over which coefficients do you want to factorise it? If you want to factorise it over the complex numbers, it is fairly trivial (Hint: multiply it by 1-x), but over the rational numbers, or a finite field, less so. – Rhys Dec 3 '12 at 15:18
up vote 3 down vote accepted

Note that the polynomial has 15 terms, so try grouping it in 5 groups of 3:

$$ 1 + x + x^2 + \cdots + x^{14} =\\ =(1+x+x^2) + x^3(1+x+x^2)+ x^6(1+x+x^2) + x^9(1+x+x^2)+ x^{12}(1+x+x^2)$$

and then factor out $(1+x+x^2)$

share|cite|improve this answer
That's only the easy part. You get $\rm\:x^{14}+\cdots+x+1 = (x^2+x+1)(x^{12} + x^9 + x^6+x^3+1).\:$ How do you propose to factor the degree $12$ cofactor? – Bill Dubuque Dec 3 '12 at 17:06
@BillDubuque : I'd do $y=x^3$ and the go to the complex (pick conjugate roots of unity), but I guess this technique exceeds the OP current level (precalculus) – leonbloy Dec 3 '12 at 18:39

Careful about concluding from $x(1+x+x^2+...+x^{13}) = -1$ that $x=-1$ must be a solution:

$$\textrm{When}\;\; x = -1,\;\; x(1+x+x^2+...+x^{13}) = -1 [7(1) + 7(-1)] = -1(0) = 0\neq -1$$

share|cite|improve this answer
Whoops! Well, that leaves me even more stuck since I assumed that if $1+x+x^2+...+x^{14}=0$, then $x(1+x+x^2+...+x^{13}=-1$, but it's harder to find a solution for $x$, let alone an answer to the actual problem. – njp Dec 3 '12 at 15:02
@DesmondWolf:that step is fine. The error comes from concluding $x-1$ is a root. In your previous example, there were an even number of terms, so $x-1$ was a factor (see my answer). In this case the number of terms is odd. – Ross Millikan Dec 3 '12 at 15:16
@Desmond you're correct that $x(1+x+x^2+...+x^{13}=-1$; it is your assumption that $x = -1$ must be a root that is problematic. – amWhy Dec 3 '12 at 15:20

When you have a geometric series with a composite number of terms, you can factor it into two series with a number of terms matching the factors. In your case $1+x+x^2+x^3+…+x^{14}=(1+x+x^2)(1+x^3+x^6+x^9+x^{12})$. There is another factorization along this direction-can you find it? Now you have two different factorizations-they must be composed of the same irreducible polynomials-try taking greatest common divisors.

share|cite|improve this answer

The best way is to use cyclotomic polynomials. $$1+x+x^2+x^3+...+x^{14}=\Phi_3(x)\cdot\Phi_5(x)\cdot\Phi_{15}(x)\ .$$ From $x^{n+1}-1=\prod_{d \mid n+1}\Phi_d(x)$ (if we factor out $\Phi_1(x)=x-1$) we get $1+x+x^2+x^3+...+x^{n}=\prod_{d \mid n+1, d\neq1}\Phi_d(x).$

share|cite|improve this answer
Unless I'm missing something, this product cannot be equal to the desired polynomial, because its constant term has the wrong sign. – Eric Stucky Dec 3 '12 at 14:44
@EricStucky: Thanks! – P.. Dec 3 '12 at 14:47
That still doesn't quite work out, since $\Phi_1$ has constant -1 and the others have constant +1. It would probably help the OP if you explained a bit of the theory that leads you to your answers. – Eric Stucky Dec 3 '12 at 14:49

Using the fact

$$ x^n-1 = (x-1)(1+x+x^2+\dots+x^{n-1}),$$

our polynomial can be written in the form

$$ 1+x+x^2+x^3+…+x^{14} = \frac{x^{15}-1}{x-1}. $$

Now, we can find the roots of $ x^{15} - 1 $ using the complex variable tecniques

$$ x^{15}=1=e^{i2k\pi} \implies x = e^{\frac{i2k\pi}{15}},\quad k=0,1,2,\dots,14. $$

So, our polynomial can be written as

$$ 1+x+x^2+\dots+x^{14} =(x-e^{\frac{i2\pi}{15}})(x-e^{\frac{i4\pi}{15}})\dots (x-e^{\frac{i28\pi}{15}})$$

$$ = \Pi_{m=1}^{14}(x-e^{\frac{i2m\pi}{15}}). $$

Note that, $$ e^{i\theta}= cos(\theta)+i\sin(\theta) $$ $$ e^{2k\pi i} = 1,\quad k\in \mathbb{Z}. $$

share|cite|improve this answer
You mean $x^{15}$ right? – Nameless Dec 3 '12 at 15:30
@Nameless: Yes. – Mhenni Benghorbal Dec 3 '12 at 15:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.