How Can You Factor \( z^7 + 1 \) into Four Non-Trivial Complex Factors?

[samer88]

Homework Statement

express z^7 + 1 as a product of four non-trivial factors and given that z is a complex number

Homework Equations

The Attempt at a Solution

 

[Char. Limit]

Well, you know that z^7+1=0 has the root z=-1. So first thing you can do is express z^7+1 as a product of (z+1) and a sixth-order polynomial.

 

[lurflurf]

consider
e^(j pi/7)
where j^2=-1

 

[samer88]

thnx all but i didnt get the good answer yet ! i need four non-trivial factors

 

[Mark44]

z⁷ + 1 can be factored into seven linear factors. Each of them is one of the seven complex roots of -1.

 

[HallsofIvy]

thnx all but i didnt get the good answer yet ! i need four non-trivial factors

You mean you are waiting for someone to tell you the answer? That isn't going to happen!

You have been given several very good suggestions. Have you factored out x+ 1? What sixth degree expression is the other factor?

Here's another way of doing it. For any n,
$$x^n- y^n= (x- y)(x^{n-1}+ x^{n-2}y+ x^{n-3}y^2+ \cdot\cdot\cdot+ xy^{n-2}+ y^{n-1}$$

Here, you have $x^7+ 1= x^7- (-1)^7$
Again, that is the same as factoring out x+1.

Now, what do you mean by "four non-trivial factors"? You can, of course, factor a 7th degree polynomial into 7 linear factors. Why combining some of them into four factors?

Your roots are, of course, the "roots of unity". Those were what lurflurf gave you.