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August 24th, 2009, 12:27 AM   #1
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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
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August 24th, 2009, 05:39 AM   #2
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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.
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August 25th, 2009, 08:19 AM   #3
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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.
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August 25th, 2009, 09:20 AM   #4
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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.
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