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Q&A #104


Factoring trinomials

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From: Kelly Stone
To: Teacher2Teacher Service
Date: Jun 03, 1998 at 16:16:51
Subject: Re: Factoring trinomials

That method for factoring trinomials was difficult to follow on the computer,
but I think I got it.  

Have you ever heard of the "illegal move"?  It is as follows, and my students 
like to use it:  (It only needs to be used when a<>1.)

Example:  6x^2-7x-3

This is difficult to factor because the coefficient of the squared term is
not 1. Therefore I remove the 6 by multiplying it with the c term. My new
trinomial is:

          x^2-7x-18

Now this trinomial is easily factored into (x-9)(x+2).

I did an "illegal move," and I now need to "undo" it. Since I multiplied by 6 
in the first step, to "undo" it I now divide each constant by 6.

         (x-9/6)(x+2/6)

I now have a factored form with fractions. That is not acceptable so I first 
reduce the fractions to lowest terms.

         (x-3/2)(x+1/3)

The binomials still have fractions that cannot be reduced, so I simply take the 
denominator of the fraction and squeeze it in front of the x in that binomial, 
making the denominator the coefficient of x.

         (2x-3)(3x+1)

It works every time!

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