Q&A #104

Teachers' Lounge Discussion: Factoring trinomials

T2T || FAQ || Ask T2T || Teachers' Lounge || Browse || Search || Thanks || About T2T

View entire discussion
[ next >>]

From: Sandie Walser <walser@arn.net>
To: Teacher2Teacher Public Discussion
Date: 1998061010:28:27
Subject: trinomials

I have had the best success with teaching polynomials using the
grouping method.  I no longer teach all of the special cases.  The
better students pick them up and discover them, while the grouping
method always works and is not confusing for the weaker students.
(Sorry about the exponent, but I can't get my computer to cooperate.)

3x2(squared) + 8x + 4	 Multiply a and c       12

since the last term is positive, list all factors of 12 and then add
the factors.  The pair of factors that are the same as the middle term
are substituted placing the smaller one first.  Factor the first two
terms of the new expression and then factor the next two terms.  This
should create a binomial that can be factored from the phrase.

3x2(squared) + 8x + 4        12      +
                            1, 12   13
                            2,  6    8
                            3,  4    7
3x2(squared) + 2x + 6x + 4
     x(3x + 2) + 2(3x + 2)
     (3x + 2)(x + 2)

The same theory works with c as negative, but you will subtract the
factors to find the middle term.

It is also fantastic for finding primes.   If it is prime over
integers, the middle term will not be listed.  I do have to stress
over integers, as I have had students successfully factor over

Post a reply to this message
Post a related public discussion message
Ask Teacher2Teacher a new question

[Privacy Policy] [Terms of Use]

Math Forum Home || The Math Library || Quick Reference || Math Forum Search

Teacher2Teacher - T2T ®
© 1994- Drexel University. All rights reserved.
The Math Forum is a research and educational enterprise of the Drexel School of Education.The Math Forum is a research and educational enterprise of the Drexel University School of Education.