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April 19th, 2014, 10:48 PM   #1
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factor

factor
x^12-x^4-x^3+1
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April 20th, 2014, 11:46 AM   #2
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Divide by x-1
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April 21st, 2014, 06:42 PM   #3
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$\displaystyle x(x-1)(x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5-x^2-x-1)$
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April 21st, 2014, 09:08 PM   #4
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Quote:
Originally Posted by scomora View Post
$\displaystyle x(x-1)(x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5-x^2-x-1)$
Shouldn't that be:
$\displaystyle (x-1)(x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4-x^2-x-1)$
?
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April 22nd, 2014, 08:26 PM   #5
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> 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...
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April 23rd, 2014, 05:11 AM   #6
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I like it and i've never seen it before either Denis.

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