Algebra Pre-Algebra and Basic Algebra Math Forum

 August 24th, 2009, 01:27 AM #1 Newbie   Joined: Aug 2009 Posts: 1 Thanks: 0 Factor x^4 + 1 Hi, How can I factor x^4 + 1 using two polynomial of grade 2 with real coefficients? I managed to do with complex numbers but I can't find non-complex polynomials is there a method? Thanks August 24th, 2009, 06:39 AM #2 Senior Member   Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0 Re: Factor x^4 + 1 Here's how I would do it: x^4 + 1 = (x^2 + Bx + C)(x^2 + Dx + E) = x^4 + (B + D)x^3 + (BD + C + E)x^2 + (BE + CD)x + CE Equating by power: B + D = 0 BD + C + E = 0 BE + CD = 0 CE = 1 So B = -D BE + CD = 0 -> -DE + CD = 0 -> D(C - E) = 0 -> C = E Since CE = 1 and C = E, either C = 1 or C = -1. Assume C = 1. Then BD + C + E = 0 -> BD = -2 -> B = sqrt(2). This gives: x^4 + 1 = (x^2 + sqrt(2) x + 1)(x^2 - sqrt(2) x + 1) which is the factorization you want. August 25th, 2009, 09:19 AM #3 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: Factor x^4 + 1 Easier to do as a difference of squares, but a rose is a rose is a rose ... if there are complex numbers then there are complex numbers ..period. August 25th, 2009, 10:20 AM   #4
Global Moderator

Joined: Dec 2006

Posts: 21,103
Thanks: 2321

Quote:
 Originally Posted by profetas I managed to do with complex numbers but I can't find non-complex polynomials . . .
If complex numbers are used at all, you should know that the complex linear factors come in conjugate pairs whose products are the real quadratic factors.
For example, (x + (1 + i)/?2)(x + (1- i)/?2) = x + ?2x + 1. Tags factor Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Albert.Teng Algebra 7 June 16th, 2012 05:30 AM daigo Algebra 3 June 14th, 2012 06:38 PM HellBunny Algebra 3 February 18th, 2012 11:31 AM Eminem_Recovery Algebra 11 June 19th, 2011 09:50 PM haebin Calculus 2 September 14th, 2009 10:25 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      