factorizar-x-4-1-

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

=(x−x_0 )(x−x_1 )(x−x_2 )(x−x_3 ) =(x−e^(i(Π/4)) )(x−e^((3iΠ)/4) )(x−e^((5iΠ)/4) )(x−e^((7iΠ)/4) ) =(x^2 −x(√)2+1)(x^2 +x(√)2+1)

$$=\left({x}−{x}_{\mathrm{0}} \right)\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)\left({x}−{x}_{\mathrm{3}} \right) \\ $$$$=\left({x}−{e}^{{i}\frac{\Pi}{\mathrm{4}}} \right)\left({x}−{e}^{\frac{\mathrm{3}{i}\Pi}{\mathrm{4}}} \right)\left({x}−{e}^{\frac{\mathrm{5}{i}\Pi}{\mathrm{4}}} \right)\left({x}−{e}^{\frac{\mathrm{7}{i}\Pi}{\mathrm{4}}} \right) \\ $$$$=\left({x}^{\mathrm{2}} −{x}\sqrt{}\mathrm{2}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}\sqrt{}\mathrm{2}+\mathrm{1}\right) \\ $$

Answered by Frix last updated on 04/Mar/24

x^4 +1= =(x^2 −(√2)x+1)(x^2 +(√2)x+1)= =(x−((√2)/2)−((√2)/2)i)(x+((√2)/2)+((√2)/2)i)(x−((√2)/2)+((√2)/2)i)(x+((√2)/2)−((√2)/2)i)

$${x}^{\mathrm{4}} +\mathrm{1}= \\ $$$$=\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)= \\ $$$$=\left({x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}\right)\left({x}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}\right)\left({x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}\right)\left({x}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}\right) \\ $$

Answered by mr W last updated on 04/Mar/24

=x^4 −i^2 =(x^2 +i)(x^2 −i) =(x^2 −(((−(√2)+(√2)i)/2))^2 )(x^2 −((((√2)+i(√2))/2))^2 ) =(x+((−(√2)+(√2)i)/2))(x−((−(√2)+(√2)i)/2))(x+(((√2)+i(√2))/2))(x−(((√2)+(√2)i)/2))

$$={x}^{\mathrm{4}} −{i}^{\mathrm{2}} \\ $$$$=\left({x}^{\mathrm{2}} +{i}\right)\left({x}^{\mathrm{2}} −{i}\right) \\ $$$$=\left({x}^{\mathrm{2}} −\left(\frac{−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}{i}}{\mathrm{2}}\right)^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{2}}+{i}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \right) \\ $$$$=\left({x}+\frac{−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}{i}}{\mathrm{2}}\right)\left({x}−\frac{−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}{i}}{\mathrm{2}}\right)\left({x}+\frac{\sqrt{\mathrm{2}}+{i}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\left({x}−\frac{\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}{i}}{\mathrm{2}}\right) \\ $$

Answered by MATHEMATICSAM last updated on 05/Mar/24

x^4 + 1 = (x^2 )^2 + 2x^2 + 1 − 2x^2 = (x^2 + 1)^2 − ((√2)x)^2 = (x^2 + (√2)x + 1)(x^2 − (√2)x + 1)

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

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