Algebra Pre-Algebra and Basic Algebra Math Forum

 April 19th, 2014, 10:48 PM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 factor factor x^12-x^4-x^3+1 April 20th, 2014, 11:46 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Divide by x-1 April 21st, 2014, 06:42 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, 09:08 PM   #4
Math Team

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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, 08:26 PM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 > 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, 05:11 AM #6 Math Team   Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 234 I like it and i've never seen it before either Denis.  Tags factor Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mared Algebra 1 April 5th, 2014 12:18 PM mared Algebra 4 March 25th, 2014 06:09 PM HellBunny Algebra 3 February 18th, 2012 11:31 AM haebin Calculus 2 September 14th, 2009 10:25 PM profetas Algebra 3 August 25th, 2009 10:20 AM

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