# Thread: [SOLVED] Factor x^n - 1

1. ## [SOLVED] Factor x^n - 1

Factor x^n - 1 for n = 6, 7, 8,..., 22.

Do I need to show the answer for each n? What is the proper way to show this?

2. Originally Posted by john_n82
Factor x^n - 1 for n = 6, 7, 8,..., 22.

Do I need to show the answer for each n? What is the proper way to show this?
$\displaystyle x^n-1 = (x-1)(1+x+x^2+x^3+...+x^{n-1})$.

So $\displaystyle x^6-1 = (x-1)(1+x+x^2+x^3+x^4+x^5)$.

I don't know if that's what you're looking for.

3. Originally Posted by redsoxfan325
$\displaystyle x^n-1 = (x-1)(1+x+x^2+x^3+...+x^{n-1})$.

So $\displaystyle x^6-1 = (x-1)(1+x+x^2+x^3+x^4+x^5)$.

I don't know if that's what you're looking for.
because the question askes me to find x^n - 1 for n = 6, 7, 8, ..., 22 so I think I have to do for each n.

4. Originally Posted by john_n82
because the question askes me to find x^n - 1 for n = 6, 7, 8, ..., 22 so I think I have to do for each n.
Well you've been shown the general way to do it, so write them all I guess.

5. alright, thank you for helping.

6. Originally Posted by redsoxfan325
$\displaystyle x^n-1 = (x-1)(1+x+x^2+x^3+...+x^{n-1})$.

So $\displaystyle x^6-1 = (x-1)(1+x+x^2+x^3+x^4+x^5)$.

I don't know if that's what you're looking for.
But notice that if the exponent is not prime then some further factorisation may be possible. For example, $\displaystyle x^6-1 = (x^3)^2-1 = (x^3+1)(x^3-1) = (x+1)(x^2-x+1)(x-1)(x^2+x+1)$.

[I'm assuming that you are looking for factorisations where the coefficients are rational numbers. If you're working over the complex numbers then you can factorise $\displaystyle x^n-1$ into n linear factors. If you are doing it over the real numbers then you can get a factorisation into linear and quadratic factors (but that is just as hard as doing it over the complex numbers).]

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# factorization of ^n-1

Click on a term to search for related topics.