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 April 19th, 2014, 09:48 PM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 factor factor x^12-x^4-x^3+1
 April 20th, 2014, 10:46 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Divide by x-1
 April 21st, 2014, 05:42 PM #3 Newbie   Joined: Apr 2014 From: South Florida Posts: 8 Thanks: 1 $\displaystyle x(x-1)(x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5-x^2-x-1)$
April 21st, 2014, 08:08 PM   #4
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Quote:
 Originally Posted by scomora $\displaystyle x(x-1)(x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5-x^2-x-1)$
Shouldn't that be:
$\displaystyle (x-1)(x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4-x^2-x-1)$
?

 April 22nd, 2014, 07:26 PM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 > factor x^12 - x^4 - x^3 + 1 Re-write: (x^12 - x^4) - (x^3 - 1) Using geometric series: u = (x^12 - x^4) / (x - 1) = x^4 + x^5 .... + x^11 v = (x^3 - x^0) / (x - 1) = x^0 + x^1 + x^2 So factoring results: (x - 1)(u - v) = (x - 1)(x^11 + x^10 ... + x^4 - x^2 - x^1 - x^0) Is that a valid "way"? Never seen it before...was just playing around... Thanks from agentredlum and Olinguito
 April 23rd, 2014, 04:11 AM #6 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 I like it and i've never seen it before either Denis.

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