Complex zeros of polynomial with no real zeros

[bobus]

I have to find the complex zeros of the following polynomial:
x^6+x^4+x^3+x^2+1
This evidently doesn't have any real solution so I tried to facto it with long division and by guessing I came up with:
(x^2-x+1)(x^4+x^3+x^2+x+1)

How can I factor the 4th degree polynomial now? Or how can I factor the original polynomial from beginning?

Thanks!

 

[tiny-tim]

welcome to pf!

hi bobus! welcome to pf!

(try using the X² icon just above the Reply box )

This evidently doesn't have any real solution so I tried to facto it with long division and by guessing I came up with:
(x^2-x+1)(x^4+x^3+x^2+x+1)

How can I factor the 4th degree polynomial now?

that was very good factoring!

ok, you have x⁴ + x³ + x² + x + 1 …

hmm … let's do a bit of lateral thinking …

what is it a factor of ? ​

 

[bobus]

Ok I'm pretty sure that not even X^4+x^3+x^2+x+1 has any real solution. So I guess there are 4 more complex solutions. So I was thinking about (ax^2+bx+c)(dx^2+ex+f) and trying to come up with some values for a, b, c, d, e and f but it seems to don't work...
Any idea??

 

[tiny-tim]

no, you're looking down, look up!

what fifth-order expression is it a factor of?

 

[bobus]

x^5+x^4+x^3+x^2+x
How does this help me?

 

[tiny-tim]

now subtract something

 

[bobus]

hmmm? I guess I'm a little confused :(

 

[mathman]

x^4 + x^3 +x^2 + x + 1 = (x^5 - 1)/(x - 1) (Geometric sum formula.)

You should be able to get the four roots you need.