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 June 7th, 2011, 03:39 AM #1 Senior Member   Joined: Oct 2010 From: Vietnam Posts: 226 Thanks: 0 Factor I made this one myself. Factor: x^2.y^2(y-x)+y^2.z^2(z-y)-z^2.x^2(z-x)
June 7th, 2011, 08:00 AM   #2
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Re: Factor

Hello, Eminem_Recovery!

Quote:
 $\text{I made this one myself.}$ [color=beige] . [/color][color=blue]Good one![/color] $\text{Factor: }\:x^2y^2(y\,-\,x)\,+\,y^2z^2(z\,-\,y)\,-\,z^2x^2(z\,-\,x)$

$\text{W\!e have: }\;x^2y^3\,-\,x^3y^2\,+\,y^2z^3\,-\,y^3z^2\,-\,x^2z^3\,+\,x^3z^2$

$\text{Re-order: }\;-x^3y^2\,+\,x^3z^2\,+\,x^2y^3\,-\,x^2z^\,-\,y^3z^2\,+\,y^2z^3$

$\text{Factor: }\;-x^3(y^2\,-\,z^2)\,+\,x^2(y^3\,-\,z^3)\,-\,y^2z^2(y\,-\,z)$

$\text{Factor: }\;-x^3(y\,-\,x)(y\,+\,z)\,+\,x^2(y\,-\,z)(y^2\,+\,yz\,+\,z^2)\,-\,y^2z^2(y\,-\,z)$

$\text{Factor: }\;(y\,-\,z)\bigg[-x^3(y\,+\,z)\,+\,x^2(y^2\,+\,yz\,+\,z^2) \,-\,y^2z^2\bigg]$

$\text{Expand: }\;(y\,-\,z)\bigg[-x^3y\,-\,x^3z \,+\,x^2y^2\,+\,x^2yz \,+\,x^2z^2\,-\,y^2z^2\bigg]$

$\text{Re-order: }\;(y\,-\,z)\bigg[-y^2z^2\,+\,z^2y^2\,+\,x^2yz \,-\,x^3y \,+\,x^2z^2\,-\,x^3z\bigg]$

$\text{Factor: }\;(y\,-\,z)\bigg[-y^2(z^2\,-\,x^2)\,+\,x^2y(z\,-\,x)\,+\,x^2z(z\,-\,x)\bigg]$

$\text{Factor: }\;(y\,-\,z)\bigg[-y^2(z^2\,-\,x^2)\,+\,x^2y(z\,-\,x)\,+\,x^2z(z\,-\,x)\bigg]$

$\text{Factor: }\;(y\,-\,z)\bigg[-y^2(z\,-\,x)(z\,+\,x)\,+\,x^2y(z\,-\,x) \,+\,x^2z(z\,-\,x)\bigg]$

$\text{Factor: }\;(y\,-\,z)(z\,-\,x)\bigg[-y^2(z\,+\,x) \,+\,x^2y\,+\,x^2z\bigg]$

$\text{Expand: }\;(y\,-\,z)(z\,-\,x)\bigg[-y^2z\,-\,xy^2\,+\,x^2y \,+\,x^2z\bigg]$

$\text{Re-order: }\;(y\,-\,z)(z\,-\,z)\bigg[x^2z\,-\,y^2z\,+\,x^2y\,-\,xy^2\bigg]$

$\text{Factor: }\;(y\,-\,z)(z\,-\,x)\bigg[z(x^2\,-\,y^2)\,+\,xy(x\,-\,y)\bigg]$

$\text{Factor: }\;(y\,-\,z)(z\,-\,x)\bigg[z(x\,-\,y)(x\,+\,y) \,+\,xy(x\,-\,y)\bigg]$

$\text{Factor: }\;(y\,-\,z)(z\,-\,x)(x\,-\,y)\bigg[z(x\,+\,y)\,+\,xy\bigg]$

$\text{Expand: }\;(y\,-\,z)(z\,-\,x)(x\,-\,y)\bigg[xz\,+\,yz\,+\,xy\bigg]$

$\text{Re-order: }\;(x\,-\,y)(y\,-\,z)(z\,-\,x)(xy\,+\,yz\,+\,xz)$

 June 7th, 2011, 01:42 PM #3 Senior Member   Joined: Sep 2010 Posts: 101 Thanks: 0 Re: Factor Excellent!
 June 7th, 2011, 03:25 PM #4 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 soroban is the second smartest mathematician in this forum.
June 8th, 2011, 04:37 AM   #5
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From: Vietnam

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Re: Factor

Quote:
Originally Posted by soroban
Hello, Eminem_Recovery!

Quote:
 $\text{I made this one myself.}$ [color=beige] . [/color][color=blue]Good one![/color] $\text{Factor: }\^2y^2(y\,-\,x)\,+\,y^2z^2(z\,-\,y)\,-\,z^2x^2(z\,-\,x)" />

$\text{W\!e have: }\;x^2y^3\,-\,x^3y^2\,+\,y^2z^3\,-\,y^3z^2\,-\,x^2z^3\,+\,x^3z^2$

$\text{Re-order: }\;-x^3y^2\,+\,x^3z^2\,+\,x^2y^3\,-\,x^2z^\,-\,y^3z^2\,+\,y^2z^3$

$\text{Factor: }\;-x^3(y^2\,-\,z^2)\,+\,x^2(y^3\,-\,z^3)\,-\,y^2z^2(y\,-\,z)$

$\text{Factor: }\;-x^3(y\,-\,x)(y\,+\,z)\,+\,x^2(y\,-\,z)(y^2\,+\,yz\,+\,z^2)\,-\,y^2z^2(y\,-\,z)$

$\text{Factor: }\;(y\,-\,z)\bigg[-x^3(y\,+\,z)\,+\,x^2(y^2\,+\,yz\,+\,z^2) \,-\,y^2z^2\bigg]$

$\text{Expand: }\;(y\,-\,z)\bigg[-x^3y\,-\,x^3z \,+\,x^2y^2\,+\,x^2yz \,+\,x^2z^2\,-\,y^2z^2\bigg]$

$\text{Re-order: }\;(y\,-\,z)\bigg[-y^2z^2\,+\,z^2y^2\,+\,x^2yz \,-\,x^3y \,+\,x^2z^2\,-\,x^3z\bigg]$

$\text{Factor: }\;(y\,-\,z)\bigg[-y^2(z^2\,-\,x^2)\,+\,x^2y(z\,-\,x)\,+\,x^2z(z\,-\,x)\bigg]$

$\text{Factor: }\;(y\,-\,z)\bigg[-y^2(z^2\,-\,x^2)\,+\,x^2y(z\,-\,x)\,+\,x^2z(z\,-\,x)\bigg]$

$\text{Factor: }\;(y\,-\,z)\bigg[-y^2(z\,-\,x)(z\,+\,x)\,+\,x^2y(z\,-\,x) \,+\,x^2z(z\,-\,x)\bigg]$

$\text{Factor: }\;(y\,-\,z)(z\,-\,x)\bigg[-y^2(z\,+\,x) \,+\,x^2y\,+\,x^2z\bigg]$

$\text{Expand: }\;(y\,-\,z)(z\,-\,x)\bigg[-y^2z\,-\,xy^2\,+\,x^2y \,+\,x^2z\bigg]$

$\text{Re-order: }\;(y\,-\,z)(z\,-\,z)\bigg[x^2z\,-\,y^2z\,+\,x^2y\,-\,xy^2\bigg]$

$\text{Factor: }\;(y\,-\,z)(z\,-\,x)\bigg[z(x^2\,-\,y^2)\,+\,xy(x\,-\,y)\bigg]$

$\text{Factor: }\;(y\,-\,z)(z\,-\,x)\bigg[z(x\,-\,y)(x\,+\,y) \,+\,xy(x\,-\,y)\bigg]$

$\text{Factor: }\;(y\,-\,z)(z\,-\,x)(x\,-\,y)\bigg[z(x\,+\,y)\,+\,xy\bigg]$

$\text{Expand: }\;(y\,-\,z)(z\,-\,x)(x\,-\,y)\bigg[xz\,+\,yz\,+\,xy\bigg]$

$\text{Re-order: }\;(x\,-\,y)(y\,-\,z)(z\,-\,x)(xy\,+\,yz\,+\,xz)$

Great! None of my classmates can do it.
Quote:
 Originally Posted by johnny soroban is the second smartest mathematician in this forum.
Then who's the smartest? I guess it's Mark

June 8th, 2011, 04:50 AM   #6
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Joined: Sep 2010

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Re: Factor

Quote:
Originally Posted by Eminem_Recovery
Quote:
 Originally Posted by johnny soroban is the second smartest mathematician in this forum.
Then who's the smartest? I guess it's Mark
huh plenty of them... Dont forget CrG and especially ZardoZ!

 June 8th, 2011, 02:51 PM #7 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 skipjack
June 9th, 2011, 03:10 AM   #8
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Re:

Quote:
 Originally Posted by johnny skipjack
And Johnny is the third smartest person.

June 15th, 2011, 06:41 AM   #9
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Quote:
 Originally Posted by Eminem_Recovery None of my classmates can do it.
Then what method did they try? The factor theorem tells you immediately that (x - y), (y - z) and (z - x) are factors.

June 15th, 2011, 06:54 AM   #10
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Joined: Nov 2009
From: Northwest Arkansas

Posts: 2,766
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Re:

Quote:
 Originally Posted by johnny skipjack
At least he's consistent...

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