Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

No Difference of Squares


Date: 02/08/97 at 18:55:27
From: Jeff & Beth Johnson
Subject: algebra problem: factoring

I am in Algebra I and we are factoring. Almost the entire class has 
been having problems with this problem and no one understands it. 
Could you try to help?

The original problem:  180x^2y - 108xy^2 -75x^3

Working on it, I took out the greatest common factor:

             3x(60xy - 36y^2 - 25x^2)

Next I broke down the -36y^2 and the -25x^2 because they are the 
difference of two perfect squares:

                3x[60xy(-6y-5x)(6y+5x)]

Is this right?  If it is right, is it prime?  

Thank you for any help you might give,  
Jennie J.  


Date: 02/08/97 at 20:08:37
From: Doctor Toby
Subject: Re: algebra problem: factoring

Dear Jennie,

Your first step is correct:

   3x (60 x y - 36 y^2 - 25 x^2)

Next you looked at -36 y^2 - 25 x^2 and tried to factor it as a 
difference of squares. But this is *not* a difference of squares,
because -36 y^2 is not a square (unless you're getting into imaginary 
numbers, which probably isn't the case here). 36 y^2 is a square, 
however. Just check; (-6y - 5x) (6y + 5x) = -36 y^2 - 60 x y - 25 x^2,
which is not the same as -36 y^2 - 25 x^2.

As it turns out, factoring -36 y^2 - 25 x^2 doesn't help you that 
much. Suppose, for the sake of argument, that factoring 
-36 y^2 - 25 x^2 = (-6y - 5x) (6y + 5x) were correct. Then you still 
wouldn't have 3x (60xy (-6y - 5x) (6y + 5x)).  Instead, you would have 
3x (60xy + (-6y - 5x) (6y + 5x)).  This doesn't help much since you 
need a factoring that includes the 60xy.

But you are on the right track! It *is* important that 36y^2 and 25x^2 
are both squares. Since there is no difference of squares (and you 
need to involve the 60xy anyway), you can't use the rule for 
difference of squares. What you need to learn to factor is something 
which is *itself* a square.

Look at these two products:

   (A + B)^2 = A^2 + 2 A B + B^2
   (A - B)^2 = A^2 - 2 A B + B^2 

(You can check these for yourself by multiplying.)

Notice that there are squares in these formulas also, as well as the 
*cross term* 2AB.

You have 3x (60 x y - 36 y^2 - 25 x^2). You can't use your squares 
yet, because they're negative squares right now. So factor out -1 to 
get -3x (-60 x y + 36 y^2 + 25 x^2), or -3x (36 y^2 - 60 x y + 25 x^).
This fits one of the patterns in the last paragraph, so you should be 
able to factor the expression now.

I hoped I helped!

-Doctor Toby,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
Middle School Factoring Expressions

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

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

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/