Solving a Perfect Square Trinomial
Date: 05/11/98 at 14:42:09 From: Joe Subject: Solutions for perfect square trinomial How do I solve this kind of problem? m^2 - 4m + 4 - n^2 + 6n - 9 This is as far as I have gotten: (m^2 - 4m + 4) - (n^2 - 6n + 9) ( )^2 - ( )^2 Please help.
Date: 05/11/98 at 17:04:51 From: Doctor Sam Subject: Re: Solutions for perfect square trinomial Joe, You began by grouping together the terms that sort of seem to go together: (m^2 - 4m + 4) - (n^2 - 6n + 9) The next step is to notice that the trinomials in each parentheses are perfect squares. That is: m^2 - 4m + 4 = (m-2)^2 and n^2 - 6n + 9 = (n-3)^2 If you don't see this you can always try factoring the expressions using trinomial factoring: m^2 - 4m + 4 = (m - ?)(m -?) and you will see that both factors must be the same. So now we have changed the original problem into (m-2)^2 - (n-3)^2. What next? Well, I cannot help but notice that this is a difference of squares. I assume that you are familiar with the way something like A^2 - B^2 factors? It is a difference of two squares, and its factors are: (A - B)(A + B) This is a pattern that you can use NO MATTER WHAT A and B stand for. In your problem A = m - 2 and B = n - 3. So: (n-2)^2 - (m-3)^2 = ([m-2]-[n-3]) ([m-2]+[n-3]) = (m - n + 1) (m + n - 5) I hope that helps. -Doctor Sam, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum