What rule is used in the factorization $z^5 - 1 = (z - 1)(z^4 + z^3 + z^2 + z + 1)$?

Question

What rule used in the following factorization? Could anyone tell me please? $$z^5 - 1 = (z - 1)(z^4 + z^3 + z^2 + z + 1)$$

Answer 1

$z^n-1 = (z-1)(z^{n-1}+z^{n-2}+z^{n-3}...+z^1+1)$

$z^n+1 = (z+1)(z^{n-1}-z^{n-2}+z^{n-3}...-z^1+1)$ (For n is odd.)

Answer 2

Just multiply it out and check it.

If you mean how did someone guess that it could be factored like that? Well, $z^5 -1$ has the obvious root of $1$ so $(z - 1)$ must be a factor. You can divide polynomials in a process similar to long division of integers. Since, we know in advance that $(z - 1)$ is a factor, we should get no remainder.

Answer 3

You may also use the Ruffini's rule: 1 is a zero of $z^5-1$, then

$1 0 0 0 0 -1  (=1z^5+0z^4+0z^3+0z^2+0z-1)$

$1 1 1 1 1 1$ $1 1 1 1 1 0$

result of the division by $(z-1)$: $1z^4+1z^3+1z^2+1z+1$

Answer 4

It's just a sum of a geometric progression:

For $z\neq1$ we obtain: $$1+z+z^2+z^3+z^4=\frac{z^5-1}{z-1},$$ which gives $$z^5-1=(z-1)(z^4+z^3+z^2+z+1),$$ which is true also for $z=1$ and we are done!