Q&A #104

Teachers' Lounge Discussion: Factoring trinomials

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From: Jacques

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.


Have you ever heard of the "illegal move"?  It is as follows, and my
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
trinomial is:


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.


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


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
making the denominator the coefficient of x.


It works every time!

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