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 HW Helper P: 3,540 It depends what you want, do you want that 4th degree factored among the reals or factored into its linear roots? What you should do is find all the complex roots as Mark has done, so what you have is $$x_1=1$$ $$x_2=cis(2\pi/5)$$ $$x_3=cis(4\pi/5)$$ $$x_4=cis(-2\pi/5)$$ $$x_5=cis(-4\pi/5)$$ So since these are all its roots, to factorize it into its linear roots it's simple. $$x^5-1=(x-1)(x-cis(2\pi/5))(x-cis(4\pi/5))(x-cis(-2\pi/5))(x-cis(-4\pi/5))$$ Now if you want only real factors, notice that $$(x-cis(\theta))(x-cis(-\theta))=x^2-x(cis(\theta)+cis(-\theta))+cis(\theta)cis(-\theta))$$ Can you simplify this to get rid of any imaginary numbers?