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Q&A #104

Teachers' Lounge Discussion: Factoring trinomials

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From: Jacques <jfeuzeu@hotmail.com>
To: Teacher2Teacher Public Discussion
Date: 2005072511:48:52
Subject: factoring trinomials using "illegal move"

	
I have tried this "illegal move" technique(see next paragraphs) as
mention by Kelly Stone on Jun 03, 1998. When the leading coefficient
is negative (for example -6x^2+7x+3), or when a common factor is not
factor out (for example -12x^2+14x+6), the technique doesn't seem to
work; unless there is something I'm doing wrong. If someone knows how
to use this "illegal move" technique, for a trinomial with a negative
leading coefficient or with a common factor without factoring it out,
let me know.
Thanks.
    


"ILLEGAL MOVE" TECHNIQUE

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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