BadNewsBears
Diamond Member
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!
from teacher2teacher. I take no credit.
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!
from teacher2teacher. I take no credit.
