Question Number 92102 by M±th+et+s last updated on 04/May/20
$${how}\:{can}\:{we}\:{factorize}\:\:\:{x}^{\mathrm{5}} −\mathrm{1}\:\:? \\ $$
Commented by mathmax by abdo last updated on 05/May/20
![complex method z^5 =1 with z =re^(iθ) ⇒r^5 e^(i5θ) =e^(i2kπ) ⇒ r=1 and θ_k =((2kπ)/5) k∈[[0,4]] so the roots are z_k =e^(i((2kπ)/5)) z_0 =1 ,z_1 =e^((i2π)/5) , z_2 =e^(i((4π)/5)) , z_3 =e^(i((6π)/5)) , z_4 =e^(i((8π)/5)) we see z_1 ^− =e^(−((k2π)/5)) =z_4 and z_2 ^− =z_3 ⇒ z^5 −1 =Π_(k=0) ^4 (z−z_k ) =(z−1)(z−z_1 )(z−z_1 ^− )(z−z_2 )(z−z_2 ^− ) =(z−1)(z^2 −2Re(z_1 )z +1)(z^2 −2Re(z_2 )z +1) ⇒ x^5 −1 =(x−1)(x^2 −2cos(((2π)/5))x+1)(x^2 −2cos(((4π)/5))x+1) we have cos(((4π)/5)) =−cos((π/5)) =−((1+(√5))/4) cos(((2π)/5)) =2cos^2 ((π/5))−1 =2×(((1+(√5))/4))^2 −1 =(1/8)(6+2(√5))−1 =((6+2(√5)−8)/8) =((−2+2(√5))/8) =((−1+(√5))/4) ⇒ x^5 −1 =(x−1)(x^2 −((−1+(√5))/2)x+1)(x^2 +((1+(√5))/2)x+1) ⇒ x^5 −1=(x−1)(x^2 +((1−(√5))/2)x +1)(x^2 +((1+(√5))/2)x +1)](https://www.tinkutara.com/question/Q92132.png)
$${complex}\:{method}\:{z}^{\mathrm{5}} =\mathrm{1}\:\:\:{with}\:{z}\:={re}^{{i}\theta} \:\Rightarrow{r}^{\mathrm{5}} \:{e}^{{i}\mathrm{5}\theta} \:={e}^{{i}\mathrm{2}{k}\pi} \:\Rightarrow \\ $$$${r}=\mathrm{1}\:{and}\:\theta_{{k}} =\frac{\mathrm{2}{k}\pi}{\mathrm{5}}\:\:\:\:\:{k}\in\left[\left[\mathrm{0},\mathrm{4}\right]\right]\:{so}\:{the}\:{roots}\:{are}\:{z}_{{k}} ={e}^{{i}\frac{\mathrm{2}{k}\pi}{\mathrm{5}}} \\ $$$${z}_{\mathrm{0}} =\mathrm{1}\:\:,{z}_{\mathrm{1}} ={e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{5}}} \:\:\:,\:{z}_{\mathrm{2}} ={e}^{{i}\frac{\mathrm{4}\pi}{\mathrm{5}}} \:,\:\:\:{z}_{\mathrm{3}} ={e}^{{i}\frac{\mathrm{6}\pi}{\mathrm{5}}} \:\:\:\:,\:\:{z}_{\mathrm{4}} ={e}^{{i}\frac{\mathrm{8}\pi}{\mathrm{5}}} \\ $$$${we}\:{see}\:\overset{−} {{z}}_{\mathrm{1}} \:\:={e}^{−\frac{{k}\mathrm{2}\pi}{\mathrm{5}}} \:={z}_{\mathrm{4}} \:\:\:{and}\:\:\overset{−} {{z}}_{\mathrm{2}} \:={z}_{\mathrm{3}} \:\Rightarrow \\ $$$${z}^{\mathrm{5}} −\mathrm{1}\:=\prod_{{k}=\mathrm{0}} ^{\mathrm{4}} \left({z}−{z}_{{k}} \right)\:=\left({z}−\mathrm{1}\right)\left({z}−{z}_{\mathrm{1}} \right)\left({z}−\overset{−} {{z}}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)\left({z}−\overset{−} {{z}}_{\mathrm{2}} \right) \\ $$$$=\left({z}−\mathrm{1}\right)\left({z}^{\mathrm{2}} \:−\mathrm{2}{Re}\left({z}_{\mathrm{1}} \right){z}\:+\mathrm{1}\right)\left({z}^{\mathrm{2}} \:−\mathrm{2}{Re}\left({z}_{\mathrm{2}} \right){z}\:+\mathrm{1}\right)\:\Rightarrow \\ $$$${x}^{\mathrm{5}} −\mathrm{1}\:=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right){x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right){x}+\mathrm{1}\right) \\ $$$${we}\:{have}\:{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\:=−{cos}\left(\frac{\pi}{\mathrm{5}}\right)\:=−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$${cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\:=\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{5}}\right)−\mathrm{1}\:=\mathrm{2}×\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\right)^{\mathrm{2}} \:−\mathrm{1} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\right)−\mathrm{1}\:=\frac{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{8}}{\mathrm{8}}\:=\frac{−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{8}}\:=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\:\Rightarrow \\ $$$${x}^{\mathrm{5}} −\mathrm{1}\:=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} −\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{x}+\mathrm{1}\right)\:\Rightarrow \\ $$$${x}^{\mathrm{5}} −\mathrm{1}=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}{x}\:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{x}\:+\mathrm{1}\right) \\ $$$$ \\ $$
Commented by M±th+et+s last updated on 05/May/20
$${i}\:{am}\:{speachless}\:.\:{god}\:{bless}\:{you}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 06/May/20
$${you}\:{are}\:{welcome}\:{sir}. \\ $$
Answered by niroj last updated on 04/May/20
$$\:\:\mathrm{x}^{\mathrm{6}} −\mathrm{1}=\:\left(\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{2}} −\left(\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:=\:\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{3}} −\mathrm{1}\right) \\ $$$$\:=\:\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right) \\ $$$$\:=\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{x}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}+\mathrm{x}\right) \\ $$$$\:=\:\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\left\{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{x}\right)^{\mathrm{2}} \right\} \\ $$$$=\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{2}} +\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right) \\ $$$$=\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\://. \\ $$
Commented by MJS last updated on 04/May/20
$$\mathrm{right}.\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{better}\:\mathrm{to}\:\mathrm{have}\:\mathrm{square}\:\mathrm{factors} \\ $$$$=\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right) \\ $$
Commented by niroj last updated on 04/May/20
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Commented by john santu last updated on 05/May/20
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Answered by niroj last updated on 04/May/20
$$\:\:\:\mathrm{x}^{\mathrm{5}} −\mathrm{1}=\:\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)\: \\ $$$$\:\:=\:\mathrm{x}\left(\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)−\mathrm{1}\left(\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right) \\ $$$$\:\:=\:\mathrm{x}^{\mathrm{5}} +\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{1} \\ $$$$\:\:\:=\mathrm{x}^{\mathrm{5}} −\mathrm{1}\://. \\ $$
Commented by niroj last updated on 04/May/20
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Commented by M±th+et+s last updated on 04/May/20
$${thank}\:{you}\:{sir} \\ $$