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 March 25th, 2014, 12:56 PM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 factor factor x^2y^3-x^3y^2+y^2z^3-y^3z^2+x^3z^2-z^3x^2
 March 25th, 2014, 01:03 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: factor $(y - x)(x - z)(y - z)(xy + xz + yz)$
 March 25th, 2014, 01:15 PM #3 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 Re: factor can you show steps
March 25th, 2014, 03:42 PM   #4
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: factor

Hello, mared!

I'll try to de-mystify the procedure.
I sort of "invented" the method many years ago.

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

Arrange in terms of $x^3,\,x^2,\,\cdots$

$\;\;\;x^3z^2\,-\,x^3y^2 \,+\, x^2y^3\,-\,x^2z^3 \,+\, y^2z^3\,-\,y^3z^2$

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

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

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

$\text{W\!e have: }\:(z\,-\,y)\,\big[x^3z\,+\,x^3y \,-\,x^2z^2\,-\,x^2yz\,-\,x^2y^2\,+\,y^2z^2\big]$

Arrange in terms of $y^2,\,y,\,\cdots$

$\;\;\;(z\,-\,y)\,\big[y^2z^2\,-\,x^2y^2 \,+\,x^3y\,-\,x^2yz\,+\,x^3z\,-\,x^2z^2\big]$

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

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

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

$\text{W\!e have: }\:(z\,-\,y)(x\,-\,z)\big[-xy^2\,-\,y^2z\,+\,x^2y\,+\,x^2z\big]$

Arrange in terms of $z,\,\cdots$

$\;\;\;(z\,-\,y)(x\,-\,z)\big[x^2z\,-\,y^2z \,+\,x^2y\,-\,xy^2\big]$

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

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

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

$\text{Therefore: }\:(x\,-\,y)(y\,-\,z)(z\,-\,x)(xy\,+\,yz\,+\,zx)$

 March 25th, 2014, 05:09 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,757 Thanks: 2138 The factors (x - y), (y - z), and (z - x) exist by the factor theorem. The final factor is then easily found.

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