The Difference of Two Squares

Date: 9/21/95 at 20:28:48
From: Pat E Moshfeghian
Subject: Factoring

Dear Dr. Math,

In the following problem, if I factor the problem in Example A's way and then 
factor the same problem with Example B's way, why is it that I do not get the 
same answer?

	Example A:  x^6 - 64  (x to the power of 6 minus 64)
	  	    (x^3 - 8)(x^3 + 8)               factoring binomial
		    (x-2)(x^2+2x+4)(x+2)(x^2-2x+4)   factoring x^3 + y^3

        Example B:   x^6 - 64
                     (x^2) ^3  - 4 ^3
                     (x^2 - 4) ( x^4 + 4x^2 + 16)     
                     Factoring   x ^3 - y ^3                       
                     (x + 2 ) (x - 2 ) ( x ^ 4 + 4x ^2 + 16 )

 after comparing the two answers we can conclude that 

                     ( x^4 + 4x ^ 2 + 16) = (x ^2 + 2x + 4 )( x ^2 - 2x + 4)

      That means the left side of the equation is factorable.
      My question is which law or rule are we using?

Thanks 

Date: 9/21/95 at 21:10:41
From: Doctor Ken
Subject: Re: Factoring

Hello!

This is actually an example of the method "difference of two squares,"
the same method you used the first way you did the problem.

I'll show you why x^4 + 4x^2 + 16 = (x^2 + 2x + 4 )(x^2 - 2x + 4).

The easiest way is to multiply out the right side and see that it equals
the left side: write it as

[(x^2 + 4) + 2x][(x^2 + 4) - 2x]
(x^2 + 4)^2 - (2x)^2
(x^4 + 8x^2 + 16) - (4x^2)
x^4 + 4x^2 + 16, which is what we wanted.

If you had to factor x^4 + 4x^2 + 16, you could write it as

x^4 + 8x^2 + 16 - 4x^2
(x^2 + 4)^2 - (2x)^2
[x^2 + 4 - 2x][x^2 + 4 + 2x], as desired.

Rest assured, this is kind of a tricky problem.

-Doctor Ken, The Geometry Forum

Date: 9/21/95 at 21:35:46
From: Pat E Moshfeghian
Subject: Re: Factoring

Thank you Dr. Math,

The last time I mailed in a question it took about a day before I received 
an answer, which in my book is good.  But wow! This time it was less than 
an hour--This is GREAT!!!

Thank you,

Pat