Question Number 29169 by abdo imad last updated on 04/Feb/18
![factorize inside R[x] the polynomial p(x)= x^8 −1.](https://www.tinkutara.com/question/Q29169.png)
$${factorize}\:{inside}\:{R}\left[{x}\right]\:{the}\:{polynomial} \\ $$$${p}\left({x}\right)=\:{x}^{\mathrm{8}} −\mathrm{1}. \\ $$
Commented by abdo imad last updated on 06/Feb/18
$${p}\left({x}\right)=\left({x}^{\mathrm{4}} \right)^{\mathrm{2}} \:−\mathrm{1}=\left({x}^{\mathrm{4}} −\mathrm{1}\right)\left({x}^{\mathrm{4}} +\mathrm{1}\right)=\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{4}} +\mathrm{1}\right) \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)\:\left({x}^{\mathrm{4}} +\mathrm{1}\right)\:{but} \\ $$$${x}^{\mathrm{4}} +\mathrm{1}=\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{2}{x}^{\mathrm{2}} =\left({x}^{\mathrm{2}} +\mathrm{1}−\sqrt{}\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{1}+\sqrt{}\mathrm{2}{x}\right){so} \\ $$$${p}\left({x}\right)=\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:{x}\:+\mathrm{1}\right). \\ $$$$ \\ $$