Finding a Single Pair of Factors

Date: 02/27/2002 at 20:00:02
From: Lauren
Subject: ax squared - bx - c =

The problem is:

   34x^2 - 41x - 15

I have tried all of the factors of 34, and everything else, but I 
can't get the answer because the only factors of 15 are 1 and 15, and 
3 and 5.

Please help!

Date: 02/27/2002 at 23:17:31
From: Doctor Peterson
Subject: Re: ax squared - bx - c =

Hi, Lauren.

I would just use the quadratic formula to find the zeroes, from which 
I would build the factors. That's the easy way.

But there is a better way to factor this than the method you are 
using, one which I only learned recently. Rather than look at 
combinations of factors of 34 and factors of 15, just look for a pair 
of factors of 34*-15 = -510 whose sum is -41. You can find that 
easily.

I'll demonstrate the work by doing a different problem as a model. 
We'll factor

    21x^2 - 23x - 20

We want to factor ac = 21*-20 = -420, so that the sum of the factors 
is -23. (Note that the factors will have opposite signs, so their sum 
will be the _difference_ of two factors of +420.) We can try 42 and 
10, but their difference is too big; 21 and 20 have too small a 
difference; but 35 and 12 are just right. Writing 23 as 35-12, our 
quadratic is

    21x^2 - 35x + 12x - 20

Taking this in pairs and factoring each pair, we get

    7x(3x - 5) + 4(3x - 5)

Aha! They have a common factor, so we can combine them:

    (7x + 4)(3x - 5)

Notice that if we had reversed the order of the middle terms, it would 
still work:

    21x^2 + 12x - 35x - 20

    3x(7x + 4) - 5(7x + 4)

    (3x - 5)(7x + 4)

Here is where I learned the method:

   Two Methods of Factoring Quadratics
   http://mathforum.org/library/drmath/view/52878.html   

Looking for a single pair of factors this way is much easier than the 
way I learned, which is probably what you are trying to do! Have fun 
finishing your problem.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/