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I am having an issue factoring x^4-x^3+x-1=0.

We are using imaginary numbers and the answer the book provide still alludes me on how to get there.
 Sep 19, 2012
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You have to divide the factors of -1(independent term), by the factors of 1(leading coefficient).
That´ll give you two possible factors: x+1, x-1. divide each by your original polynomial.
Synthetic division:
1, -1, 0, 1, -1 by 1.
1, 0, 0, 1, 0. = x^3+1(x-1). That would be factoring to a third-degree polynomial.
1, 0, 0, 1 by -1.
1,-1, 1, 0. = x^2-x+1(x-1)(x+1) That would be factored to a second-degree polynomial, which you can then factor as: [x^2-(x+1)](x-1)(x+1). Where x^2-x+1 is a difference of squares a^2-b^2.
a^2 = x^2. and b^2 = x+1.
Imaginary number = i = square root of -1.
i^2 is equal to -1. -1 times -x-1 is equal to x+1.
x+1 is equal to -x-1(i^2)
Result:
(x-square root of -x-1i)(x+square root of -x-1i)(x-1)(x+1)
I think that's the correct answer.
Hope this could help.
 Sep 19, 2012

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