Factor $1+x+x^2+x^3+…+x^{14}$

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?

• $-1$ 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

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 $(1+x+x^2)$

• 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:

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

• 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. – Namaste 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.

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 $1+x+x^2+x^3+...+x^{n}=\prod_{d \mid n+1, d\neq1}\Phi_d(x).$

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

• You mean $x^{15}$ right? – Nameless Dec 3 '12 at 15:30
• @Nameless: Yes. – Mhenni Benghorbal Dec 3 '12 at 15:31