April 19th, 2014, 09:48 PM  #1 
Senior Member Joined: Nov 2013 Posts: 434 Thanks: 8  factor
factor x^12x^4x^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 x1

April 21st, 2014, 05:42 PM  #3 
Newbie Joined: Apr 2014 From: South Florida Posts: 8 Thanks: 1 
$\displaystyle x(x1)(x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5x^2x1)$

April 21st, 2014, 08:08 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  
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 Rewrite: (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... 
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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