Difference between revisions of "Simon's Favorite Factoring Trick"

m (Rub'i'nstein)
Line 9: Line 9:
== Examples ==
== Examples ==
([[AIME]] 1987/5) <math>m</math> and <math>n</math> are integers such that <math>m^2 + 3m^2n^2 = 30n^2 + 517</math>. Find <math>3m^2n^2</math>.
'''Outline Solution:''' Rearrange to <math>m^2 + 3m^2n^2 -30n^2= 517</math>. The key step is changing the equation to <math>m^2 + 3m^2n^2 -30n^2-10= 507</math>, where the equation factors to <math>(3n^2 + 1)(m^2 - 10) = 507 = 3\cdot 13^2</math>.  This makes things much simpler.  The rest of the problem is left as an exercise to the reader.
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=454191#p454191 AIME 1987/5]

Revision as of 14:12, 21 June 2006


Simon's Favorite Factoring Trick (abbreviated SFFT) is a special factorization first popularized by AoPS user Simon Rubinstein-Salzedo. This appears to be the thread where Simon's favorite factoring trick was first introduced.

Statement of the factorization

The general statement of SFFT is: ${xy}+{xk}+{yj}+{jk}=(x+j)(y+k)$. More oftenly however SFFT is introduced as $xy + x + y + 1 = (x+1)(y+1)$ or $xy - x - y +1 = (x-1)(y-1)$.


This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually ${x}$ and ${y}$ are variables and $j,k$ are known constants. Also it is typically necessary to add the ${j}{k}$ term to both sides to perform the factorization.


Invalid username
Login to AoPS