Complex zeros of polynomial with no real zeros

  • Thread starter bobus
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  • #1
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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!!!
 

Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi bobus! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)
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! :biggrin:

ok, you have x4 + x3 + x2 + x + 1 …

hmm :confused: … let's do a bit of lateral thinking :rolleyes:

what is it a factor of ? :wink:
 
  • #3
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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 dont work...
Any idea??
 
  • #4
tiny-tim
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no, you're looking down, look up! :rolleyes:

what fifth-order expression is it a factor of? :smile:
 
  • #5
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x^5+x^4+x^3+x^2+x
How does this help me?
 
  • #6
tiny-tim
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now subtract something :wink:
 
  • #7
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hmmm??? I guess I'm a little confused :(
 
  • #8
mathman
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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.
 

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