Question Number 204979 by Simurdiera last updated on 04/Mar/24
$${factorizar} \\ $$$${x}^{\mathrm{4}} \:+\:\mathrm{1} \\ $$
Answered by Skabetix last updated on 04/Mar/24
$$=\left({x}−{x}_{\mathrm{0}} \right)\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)\left({x}−{x}_{\mathrm{3}} \right) \\ $$$$=\left({x}−{e}^{{i}\frac{\Pi}{\mathrm{4}}} \right)\left({x}−{e}^{\frac{\mathrm{3}{i}\Pi}{\mathrm{4}}} \right)\left({x}−{e}^{\frac{\mathrm{5}{i}\Pi}{\mathrm{4}}} \right)\left({x}−{e}^{\frac{\mathrm{7}{i}\Pi}{\mathrm{4}}} \right) \\ $$$$=\left({x}^{\mathrm{2}} −{x}\sqrt{}\mathrm{2}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}\sqrt{}\mathrm{2}+\mathrm{1}\right) \\ $$
Answered by Frix last updated on 04/Mar/24
$${x}^{\mathrm{4}} +\mathrm{1}= \\ $$$$=\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)= \\ $$$$=\left({x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}\right)\left({x}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}\right)\left({x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}\right)\left({x}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}\right) \\ $$
Answered by mr W last updated on 04/Mar/24
$$={x}^{\mathrm{4}} −{i}^{\mathrm{2}} \\ $$$$=\left({x}^{\mathrm{2}} +{i}\right)\left({x}^{\mathrm{2}} −{i}\right) \\ $$$$=\left({x}^{\mathrm{2}} −\left(\frac{−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}{i}}{\mathrm{2}}\right)^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{2}}+{i}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \right) \\ $$$$=\left({x}+\frac{−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}{i}}{\mathrm{2}}\right)\left({x}−\frac{−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}{i}}{\mathrm{2}}\right)\left({x}+\frac{\sqrt{\mathrm{2}}+{i}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\left({x}−\frac{\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}{i}}{\mathrm{2}}\right) \\ $$
Answered by MATHEMATICSAM last updated on 05/Mar/24
$${x}^{\mathrm{4}} \:+\:\mathrm{1} \\ $$$$=\:\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{1}\:−\:\mathrm{2}{x}^{\mathrm{2}} \\ $$$$=\:\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)^{\mathrm{2}} \:−\:\left(\sqrt{\mathrm{2}}{x}\right)^{\mathrm{2}} \\ $$$$=\:\left({x}^{\mathrm{2}} \:+\:\sqrt{\mathrm{2}}{x}\:+\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:−\:\sqrt{\mathrm{2}}{x}\:+\:\mathrm{1}\right) \\ $$