Factoring Trinomials With Prime Factors

Date: 05/16/2002 at 21:39:26
From: Scott
Subject: Factoring Trinomials

Factor each trinomial, if possible.  If the trinomial cannot be 
factored using integers, write PRIME.

1)  48x^2 + 22x - 15

I've multiplied -15 and 48 and I usually get -720 and I can't find a 
number that goes into -720 and adds up to 22.

2)  -4y^2 + 19y - 21

I've multiplied -21 and -4 and I get 84, but I can't find any 
factors of 84, that equal 19.

3)  6a^2 - 7a + 18

I've multiplied 18 and 6 and I get 108, but I can't find any factors 
of 108 that equal -7.

Help!

Date: 05/17/2002 at 12:30:32
From: Doctor Peterson
Subject: Re: Factoring Trinomials

Hi, Scott.

If I don't see the appropriate pair of factors on sight, I make 
a systematic search by first finding the prime factorization. I'll 
demonstrate on the first, 

    720 = 2*2*2*2*3*3*5.

Since the product is -720, we will have a positive and a negative 
factor, and it is the _difference_ of two factors of 720 that has to 
equal 22. Taking a first guess, let's just divide the prime factors 
evenly (so the factors will be fairly close together), and try

    2*2*2*2 * 3*3*5

      16    *  45

The difference is 45-16 = 29. That's a bit too big, so we have to make 
the factors closer. Maybe if we swap a 2 for a 3 it will make the 16 
big enough:

    2*2*2*3 * 2*3*5

       24   *   30

The difference is now 30-24 = 6. That's too small. How about instead 
swapping a 4 for a 5:

    2*2*5 * 2*2*3*3

     20   *    36

The difference is now 36-20 = 16. We want to make the smaller one 
smaller, so let's start here and swap a 10 for a 9:

    2*3*3 * 2*2*2*5

      18  *   40

The difference is now 40-18 = 22. We found it!

It would be a lot harder if this were in fact prime. To show that 
there is no way to factor it, you would have to make a list of ALL 
pairs of factors and show that none work. In this case, there are 30 
factors, in 15 pairs, so that wouldn't be terribly hard, if you know 
how to make such a list.

In some cases, however, it's obvious that you can't factor it: If the 
desired sum of two positive factors is less than twice the square of 
the product, then no pair of factors can give a sum so low. Can you 
see why?

But ultimately, the most efficient way to see whether a quadratic can 
be factored is to use the quadratic formula to find its zeros. 
Factoring is a waste of time in hard cases.

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