Simon's Favorite Factoring Trick

Simon's Favorite Factoring Trick:

A Math Contest Favorite

Have you ever seen a system of equations on a math contest and said to yourself, "Oh no, algebraic manipulation? Here we go again..." ? Chances are, you're not a fan of the good ol' elimination and substitution or the handy-dandy matrix. Well, don't give up yet! Simon's Favorite Factoring Trick might just be what you've been looking for!

Simon's Favorite Factoring Trick tells us this: $$( a + b ) ( x + y ) = ax + ay + bx + by$$

This factoring trick really appears most frequently in these two special cases:

$$( a + 1 ) ( b + 1 ) = ab + a + b + 1$$ $$( a - 1 ) ( b - 1 ) = ab - a - b + 1$$

Ok, seems simple enough. So when is it useful?

Simon's Favorite Factoring Trick was said to have first become popular through the AoPS forums. This factoring trick shows up a lot on contest problems and can be your best friend when it comes to algebraic manipulations.

Let's take a look at an example from a past AMC 12 contest.

(2001 AMC 12, #21) Four positive integers $a, b, c,$ and $d$ have a product of $8!$ and satisfy:

$$ab + a + b = 524$$ $$bc + b + c = 146$$ $$cd + c + d = 104$$

what is $a-d?$

Let's put this trick to use!

The first step is rewriting the system of equations. $$( a + 1 )( b + 1) = 524 + 1 = 525$$ $$( b + 1 )( c + 1) = 146 + 1 = 147$$ $$( c + 1 )( d + 1) = 104 + 1 = 105$$

Since the original equation is missing the $+ 1$ term that is necessary to use Simon's Favorite Factoring Trick, we simply add a $+1$ to the right hand side of the equation.

From here we can factor and rewrite the equations:

Let ($w, x, y,$ and $z) = ( a + 1, b + 1, c + 1, d + 1)$ $$wx = 3\cdot 5\cdot 5\cdot 7$$ $$xy= 3\cdot 7\cdot 7\cdot 7$$ $$yz = 3\cdot 5\cdot 7\cdot 7$$

The Solution

We see that $5^2$ clearly divides $wx$. However, neither $5^2$ nor $5$ can divide $xy$ since $147$ does not have $5$ as a factor. Thus, we can conclude that $w = 25$. Substituting this value back into the equations yields: $(w, x, y, z) = (25, 21, 7, 15)$, which gives us $(a,b, c, d) = (24, 20, 6, 14).$

And surely enough, $a\cdot b \cdot c\cdot d = 8!$ $$a-d = 24 - 14 = \boxed{10}$$

Keep in Mind

We've heard it time and time again, "What is done to one side of the equation must be done to the other!" The same applies when using SFFT.

A lot of the times, SFFT shows up like this, $$xy + x + y = k$$ and we immediately want to do this: $$(x + 1)(y+1) = k$$

But that is wrong! We need to add a 1!

$$( x+ 1)(y+1) = k+1$$

Ah, much better. It is easy to see a form of SFFT, get excited, factor, and forget to balance out the equation. Don't let that happen to you.

Simon's favorite factoring trick which is often abbreviated as SFFT for short, is a useful technique used when factoring. The general form is $ax+ay+bx+by = (a+b)(x+y)$ or $ax-ay-bx+by = (a-b)(x-y)$.

Practice: What are the possible $(x, y)$ pairs such that $x$ and $y$ are positive integers and satisfy $xy-10x-5y=-44$?

Factoring: $x(y-10)-5(y-10)=(x-5)(y-10)=6$

From here, we can solve for $x$ and $y$ based on the divisors of $6$.

Case $1$: $x-5=1$ and $y-10=6$ so $x=6$ and $y=16$.

Case $2$: $x-5=2$ and $y-10=3$ so $x=7$ and $y=13$.

Case $3$: $x-5=3$ and $y-10=2$ so $x=8$ and $y=12$.

Case $4$: $x-5=6$ and $y-10=1$ so $x=11$ and $y=11$.

Thus, the possible $(x, y)$ pairs are $(6, 16)$, $(7, 13)$, $(8, 12)$, and $(11, 11)$.

More information about Simon's favorite factoring trick can be found here