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
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.