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Question Number 29169 by abdo imad last updated on 04/Feb/18

factorize inside R[x] the polynomial p(x)= x^8 −1.

$${factorize}\:{inside}\:{R}\left[{x}\right]\:{the}\:{polynomial} \\ $$$${p}\left({x}\right)=\:{x}^{\mathrm{8}} −\mathrm{1}. \\ $$

Commented by abdo imad last updated on 06/Feb/18

p(x)=(x^4 )^2 −1=(x^4 −1)(x^4 +1)=(x^2 −1)(x^2 +1)(x^4 +1) =(x−1)(x+1)(x^2 +1) (x^4 +1) but x^4 +1=(x^2 +1)^2 −2x^2 =(x^2 +1−(√)2x)(x^2 +1+(√)2x)so p(x)=(x−1)(x+1)(x^2 +1)(x^2 −(√2)x+1)(x^2 +(√2) x +1).

$${p}\left({x}\right)=\left({x}^{\mathrm{4}} \right)^{\mathrm{2}} \:−\mathrm{1}=\left({x}^{\mathrm{4}} −\mathrm{1}\right)\left({x}^{\mathrm{4}} +\mathrm{1}\right)=\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{4}} +\mathrm{1}\right) \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)\:\left({x}^{\mathrm{4}} +\mathrm{1}\right)\:{but} \\ $$$${x}^{\mathrm{4}} +\mathrm{1}=\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{2}{x}^{\mathrm{2}} =\left({x}^{\mathrm{2}} +\mathrm{1}−\sqrt{}\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{1}+\sqrt{}\mathrm{2}{x}\right){so} \\ $$$${p}\left({x}\right)=\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:{x}\:+\mathrm{1}\right). \\ $$$$ \\ $$

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