Factoring a Sum of SquaresDate: 10/15/2005 at 17:59:18 From: talin Subject: sum of squares vs. factoring perfect square trinomials I do not understand why you cannot factor a sum of two squares, but you can factor a perfect square trinomial. The perfect square trinomial can be factored like this: (a+b)square = (a squared - ab + b squared) (a+b)squared is a sum of squares which can't be factored Date: 10/15/2005 at 23:11:45 From: Doctor Peterson Subject: Re: sum of squares vs. factoring perfect square trinomials Hi, Talin. Let's correct what you wrote: a^2 + 2ab + b^2 (a trinomial) factors as (a + b)^2 a^2 - b^2 (a difference of squares) factors as (a + b)(a - b) a^2 + b^2 (a sum of squares) can't be factored Now, that last fact isn't really quite true! Properly, we should say "it can't be factored _over the real numbers_". If you allowed complex numbers, you could write a^2 + b^2 = (a + bi)(a - bi) If you know about imaginary numbers, try expanding the right side and see what I mean! One way to see why this happens is to consider one of the reasons we factor, to solve a polynomial equation. Take the equation x^2 - 1 = 0 To solve it, we can factor and then set each factor equal to 0: (x + 1)(x - 1) = 0 x + 1 = 0, or x - 1 = 0 x = -1, or x = 1 So each factor corresponds to a solution of the equation. Now look at this equation: x^2 + 1 = 0 If we could factor this sum of squares (over the reals!), we would have a real solution of the equation. But here are the solutions: x^2 = -1 x = +- sqrt(-1) x = i, or x = -i Those aren't real numbers; so we'd better not be able to factor it! In general, we can just say that x^2 + b^2 is positive for all x, so x^2 + b^2 = 0 can't have any real solutions; that means it can't be factored with real coefficients. Does that help make it clear why some polynomials can't be factored, and this kind in particular? If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/