Factoring a Difference of Squares with Terms That Are Not Monomials

Date: 10/26/2008 at 21:55:53
From: Aaron
Subject: Factoring difference of squares that are not monomials

I need to know how to do this, and I can't do it..like here's an 
example if I'm not making sense.  X power 4 minus (2X - 1) squared.
That's an example out of my text book but I don't understand there 
method of solving.  Could you please give me another example or just
explain it?

Date: 10/26/2008 at 22:58:20
From: Doctor Peterson
Subject: Re: Factoring difference of squares that are not monomials

Hi, Aaron.

First, I assume you understand the general rule: any difference of
squares can be factored as a sum times a difference:

  (a)^2 - (b)^2 = (a + b)(a - b)

The "a" and "b" in this formula can be ANYTHING--a single variable, a
monomial, or a more complicated expression.  The key is to identify
what "a" and "b" are in your expression and write it out to make them
clear before actually factoring.

In your example, you'd have (x^2)^2 - (2x - 1)^2 so that "a" would be
represented by x^2 and "b" would be represented by (2x - 1).

Here's an example that's even more complicated than yours:

  9(x-1)^2 - 4(x+1)^2

First, we see that each term is the square of something; I'll write
them as squares to make that plain:

  [3(x-1)]^2 - [2(x+1)]^2

That is, inside the [] I wrote the square root of each part of a term,
showing what had to be squared to get it: the 9 is 3 squared, and the
(x-1)^2 is already written that way.

Now, in my mind (or maybe on paper, if I'm showing it to someone) I'll
label "a" (the first thing being squared) and "b" (the second thing
being squared):

  [3(x-1)]^2 - [2(x+1)]^2
   \____/       \____/
     a            b

Now, the formula says that the first factor will be a+b, so I write
the sum of those two expressions; and for the second factor I write
the difference of the two:

  (3(x-1) + 2(x+1))(3(x-1) - 2(x+1))
   \____/   \____/  \____/   \____/
  (  a    +   b   )(  a    -   b   )

Now we've factored the expression; but we should do some cleaning up
before we say we're done.  I'll simplify INSIDE each factor by
distributing and combining like terms.  But I'll leave each factor
intact, since that was my purpose.

  (3(x-1) + 2(x+1))(3(x-1) - 2(x+1))

  (3x - 3 + 2x + 2)(3x - 3 - 2x - 2)  <-- watch that last sign!

  (5x - 1)(x - 5)

And we're done!  To check, you might expand both the original ugly
expression, and this factored one (distributing and combining terms),
and see that they both represent the same polynomial.

If you have any further questions, feel free to write back.  The best
thing would be to show your work on a specific problem so I can see
how close you are to getting it right.


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/