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June 14th, 2012, 05:29 PM   #1
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factor x^5+x+1

Please try this one:
Factor :

x^2+x+1)(x^3-x^2+1)" />
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June 14th, 2012, 07:32 PM   #2
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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).
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June 14th, 2012, 08:30 PM   #3
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Re: factor x^5+x+1

Hello, Albert.Teng!

Quote:







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June 14th, 2012, 09:27 PM   #4
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Re: factor x^5+x+1

Hi soroban :

a very skillful solution !
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June 15th, 2012, 03:48 AM   #5
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Re: factor x^5+x+1












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June 15th, 2012, 05:58 AM   #6
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Re: factor x^5+x+1



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June 15th, 2012, 06:04 AM   #7
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Re: factor x^5+x+1

Quote:
Originally Posted by Albert.Teng
Sometimes you come too close but then goes very very far and end it by a very very very complicated way!
Well done! Albert.Teng!
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June 16th, 2012, 04:30 AM   #8
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Any polynomial of the form x^(3m-1) + x^(3n-2) + 1, where m and n are natural numbers, has x + x + 1 as a factor.
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