Teacher2Teacher Q&A #104

Teachers' Lounge Discussion: Factoring trinomials

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From: Jacques

To: Teacher2Teacher Public Discussion
Date: 2005072510:48:52
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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