June 14th, 2012, 05:29 PM  #1 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  factor x^5+x+1
Please try this one: Factor : x^2+x+1)(x^3x^2+1)" /> 
June 14th, 2012, 07:32 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond  Re: factor x^5+x+1 Substitution confirms edit: polynomial long division and the remainder theorem show that x^5 + x + 1 does not have any linear factors, so initial terms of x^ 4 and x may be ruled out. A method similar to the one above rules out the signs of the 1's being both negative (there are no real roots for a). 
June 14th, 2012, 08:30 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: factor x^5+x+1 Hello, Albert.Teng! Quote:
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June 14th, 2012, 09:27 PM  #4 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  Re: factor x^5+x+1
Hi soroban : a very skillful solution ! 
June 15th, 2012, 03:48 AM  #5 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: factor x^5+x+1 
June 15th, 2012, 05:58 AM  #6 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  Re: factor x^5+x+1 
June 15th, 2012, 06:04 AM  #7  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: factor x^5+x+1 Quote:
Well done! Albert.Teng!  
June 16th, 2012, 04:30 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,813 Thanks: 2155 
Any polynomial of the form x^(3m1) + x^(3n2) + 1, where m and n are natural numbers, has x² + x + 1 as a factor.


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