Q&A #18119

"Bottoms up" as a method to factor trinomials

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

View entire discussion
[<<prev] [next>>]

From: Pat Ballew (for Teacher2Teacher Service)
Date: Jan 01, 2007 at 16:40:33
Subject: Re: "Bottoms up" as a method to factor trinomials

Well, the secret is that 8 = 24/3... If you consider that the solutions of x^2 + bx +c = 0 are the same as the solutions of 2x^2 + 2bx + 2c etc... then you are a step closer to understanding the solution.... If we take 3x^2 + 14x + 8 = 0 and (x=u/3) and substitute we get (3(u/3)^2 + 14 (u/3) + 24/3) = 0 and now if we simplify the first term we get u^2/3 + 14 u/3 + 24/3 = 0 now if we multiply all terms by 3 we get u^2 + 14u + 24... and solve to get the two solutions you had, u=12 and u=2, but remember that we wanted x, not u, and x=u/3 thus the final solution... A similar mathod you might have seen that works on the same method involves multiplying each term by the a coefficient to make it a perfect square... Using your second example, we rewrite 12x^2 + 13x-35 as 144x^2 + 156 x - 420 = 0 and write factors with 12x in each lead (12x + ) (12x - ) and use the ordinary method, but with larger numbers to produce the same result.... As a third approach.. you can simply divide everything by the a term... x^2 + 13/12 x - 35/12 = 0 and then look for two numbers that multiply to make -35/12 (which of course is -420/144) and add up to 13/12... if we assume denominators of 12, then the choice is the same as two numbers that have a product of -420 and a sum of 13... Hope that is all clear... Good luck -Pat Ballew, for the T2T service

Post a 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- The Math Forum at NCTM. All rights reserved.