Teacher2Teacher Q&A #104

Factoring trinomials

T2T || FAQ || Teachers' Lounge || Browse || Search || T2T Associates || About T2T

View entire discussion
[<<prev]

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!

Teacher2Teacher - T2T ®