Factor $1+x+x^2+x^3+...+x^{14}$

Question

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?

Answer 1

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 $(1+x+x^2)$

Answer 2

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

Answer 3

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

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

Answer 4

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.

Answer 5

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