My Math Forum factor x^5+x+1

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 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 :$x^5+x+1$ $Ansx^2+x+1)(x^3-x^2+1)" />
 June 14th, 2012, 07:32 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: factor x^5+x+1 $(x^2\,+\,ax\,+\,1)(x^3\,+\,bx^2\,+\,cx\,+\,1)$ $=\,x^5\,+\,bx^4\,+\,cx^3\,+\,x^2\,+\,ax^4\,+\,abx^ 3\,+\,acx^2\,+\,ax\,+\,x^3\,+\,bx^2\,+\,cx\,+\,1.$ $a\,+\,b\,=\,0 \\ ab\,+\,c\,+\,1\,=\,0 \\ ac\,+\,b\,+\,1\,=\,0 (*)\\ a\,+\,c\,=\,1$ $b\,=\,-a\,(**) \\ c\,=\,1\,-\,a\,(***)$ $\text{From (*),\,(**) and (***)},\,a\,-\,a^2\,-\,a\,+\,1\,=\,0\,\Rightarrow\,a\,=\,\pm1$ Substitution confirms $a\,=\,1 \\ b\,=\,-1 \\ c\,=\,0$ $x^5\,+\,x\,+\,1\,=\,(x^2\,+\,ax\,+\,1)(x^3\,+\,bx^ 2\,+\,cx\,+\,1)=\,(x^2\,+\,x\,+\,1)(x^3\,-\,x^2\,+\,1)$ 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
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Re: factor x^5+x+1

Hello, Albert.Teng!

Quote:
 $\text{Factor: }\:x^5\,+\,x\,+\,1$ $\text{Answer: }\:(x^2\,+\,x\,+\,1)(x^3\,-\,x^2\,+\,1)$

$\text{W\!e have: }\:x^5\,+\,x\,+\,1$

$\text{Subtract and add }x^3,\,\text{ subtract and add }x^2$

[color=beige]. . [/color]$x^5\,-\,x^3\,+\,x^3\,-\,x^2\,+\,x^2\,+\,x\,+\,1$

[color=beige]. . [/color]$-\; \underbrace{x^5\,-\,x^3}\,+\,\underbrace{x^3\,-\,x^2}\,+\,\underbrace{x^2\,+\,x}\,+\,1$

[color=beige]. . [/color]$=\;x^3\underbrace{(x^2\,-\,1)}\,+\,x^2(x\,-\,1)\,+\,x(x\,+\,1)\,+\,1$

[color=beige]. . [/color]$=\;\underbrace{x^3(x\,-\,1)(x\,+\,1)\,+\,x^2(x\,-\,1)}\,+\,x(x\,+\,1)\,+\,1$

[color=beige]. . [/color]$=\;x^2(x\,-\,1)\,\underbrace{\big[x(x\,+\,1)\.+\,1\big]}\,+\,\underbrace{\big[x(x\,+\,1)\,+\,1\big]}$

[color=beige]. . [/color]$=\;\big[x(x\,+\,1)\,+\,1\big]\,\big[x^2(x\,-\,1)\,+\,1\big]$

[color=beige]. . [/color]$=\;\big[x^2\,+\,x\,+\,1\big]\,\big[x^3\,-\,x^2\,+\,1\big]$

 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 $x^5+x+1=x^5+x^4+x^3+x^2+x+1 - x^4 - x^3 - x^2 = x^5+x^4+x^3+x^2+x+1 - x^2(x^2+x+1)$ $\text{we know the first part is related with sixth root of unity So, we can easily say that x=-1 satisfies x^5+x^4+x^3+x^2+x+1=0 }$ $\text{now divide x^5+x^4+x^3+x^2+x+1 by x+1 , use synthetic division to get x^4+x^2+1, add and substract x^2}$ $x^4+x^2+1= (x^2)^2+2x^2+1-x^2 = (x^2+1)^2-x^2 = (x^2+x+1)(x^2-x+1)$ $\text{then } x^5+x^4+x^3+x^2+x+1= (x+1)(x^4+x^2+1) = (x+1)(x^2-x+1)(x^2+x+1)$ $x^5+x+1=(x+1)(x^2-x+1)(x^2+x+1)-x^2(x^2+x+1) = (x^2+x+1)((x+1)(x^2-x+1)-x^2) = (x^2+x+1)(x^3-x^2+1)$
 June 15th, 2012, 05:58 AM #6 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Re: factor x^5+x+1 $x^5+x+1=x^5+x^4+x^3+x^2+x+1-x^2(x^2+x+1)$ $=x^3(x^2+x+1)+(x^2+x+1)-x^2(x^2+x+1)$ $=(x^2+x+1)(x^3-x^2+1)$
June 15th, 2012, 06:04 AM   #7
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Re: factor x^5+x+1

Quote:
 Originally Posted by Albert.Teng $x^5+x+1=x^5+x^4+x^3+x^2+x+1-x^2(x^2+x+1)=x^3(x^2+x+1)+(x^2+x+1)-x^2(x^2+x+1)=(x^2+x+1)(x^3-x^2+1)$
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!

 June 16th, 2012, 04:30 AM #8 Global Moderator   Joined: Dec 2006 Posts: 21,015 Thanks: 2250 Any polynomial of the form x^(3m-1) + x^(3n-2) + 1, where m and n are natural numbers, has x² + x + 1 as a factor.

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