# Simon’s Favorite Factoring Trick

Another elegant factoring technique that will solve hard problems is Simon’s Favorite Factoring Trick or commonly abbreviated as SFFT.  This technique was popularized by Simon Rubinstein-Salzedo, a user of Art of Problem Solving. The idea is adding a term to make the whole expression factorable. Try to do the problem first by yourself before looking at the solution for you to evaluate your skill and maximize your learning.

Sample Problem 1:

Factor completely: $x^4+4$

[toggle title=”Solution:”]

Observe that the two terms are perfect squares. So we can add a term that can make the whole expression a perfect square trinomial.

We can add $4x^2$ to make the expression a perfect square trinomial.

$x^4+4x^2-4$,

However, in the rule of math, we do cannot alter the equation or expression to keep its original form. However adding a zero to a whole expression will not change anything on the original expression.

$x^4+4x^2+4-4x^2$, we add and subtract $4x^2$ or like adding zero.

$(x^2+2)^2-(2x)^2$ observe that this is difference of two squares in the form of $u^2-v^2=(u+v)(u-v)$ where $u=x^2+2$ and $v=2x$

$(x^2+2+2x)(x^2+2-2x)$

Rearranging the expression accordingly,

$(x^2+2x+2)(x^2-2x+2)$ [/toggle]

Sample Problem 2:

Find all positive integral pairs $(x,y)$ that satisfy the equation $xy+x-2y=9$.

[toggle title=”Solution:”]

Look at the left hand side of equation. Think of a number that you can add or subtract to make it factorable. In this equation we subtract the left side by 2.

$xy+x-2y=9$ $xy+x-2y-2=9-2$ $x(y+1)-2(y+1)=7$ $(y+1)(x-2)=7$

7 is a prime number which means that the factor of 7 is only 1 and itself. By trial and error,

Case 1:

$y+1=7$   and   $x-2=1$

$y=6$       and   $x=3$

One pair is $(3,6)$

Case 2:

$y+1=1$   and   $x-2=7$

$y=0$       and   $x=9$

Since we are ask for positive integral pairs this is not a solution since y=0.

Therefore the only solution is $(3,6)$ [/toggle]