Question Number 73051 by mathmax by abdo last updated on 05/Nov/19
![factorize inside R[X] 1)X^5 −1 2)X^6 +1](https://www.tinkutara.com/question/Q73051.png)
$${factorize}\:{inside}\:{R}\left[{X}\right] \\ $$$$\left.\mathrm{1}\right){X}^{\mathrm{5}} −\mathrm{1}\:\: \\ $$$$\left.\mathrm{2}\right){X}^{\mathrm{6}} \:+\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 06/Nov/19
![1) let decompose inside C[x] x^5 −1 x=r e^(iθ) so x^5 −1 =0 ⇒r^5 e^(i5θ) =e^(i2kπ) ⇒r=1 and θ =((2kπ)/5) so the roots areZ_k =e^(i((2kπ)/5)) with k∈[[0,4]] ⇒x^5 −1 =Π_(k=0) ^4 (x−Z_k )=(x−Z_0 )(x−Z_1 )(x−Z_2 )(x−Z_3 )(x−Z_4 ) Z_0 =1 Z_1 =e^((i2π)/5) , Z_2 =e^((i4π)/5) , Z_3 =e^((i6π)/5) =Z_2 ^− ,Z_4 =e^((i8π)/5) =Z_1 ^− ⇒ x^5 −1 =(x−1)(x−Z_1 )(x−Z_1 ^− )(x−Z_2 )(x−Z_2 ^− ) =(x−1)(x^2 −2Re(Z_1 )x +1)(x^2 −2Re(Z_2 )x +1) =(x−1)(x^2 −2cos(((2π)/5))x+1)(x^2 −2 cos(((4π)/5))x +1) we have cos((π/5))=((1+(√5))/4) and cos(((2π)/5))=2cos^2 ((π/5))−1 =2(((1+(√5))^2 )/(16)) −1 =((2(6+2(√5))−16)/6) =((12+4(√5)−16)/(16)) =((−4+4(√5))/(16)) =(((√5)−1)/4) cos(((4π)/5)) =2cos^2 (((2π)/5))−1 =2 ((((√5)−1)^2 )/(16)) −1=((2(6−2(√5))−16)/(16)) =((12−4(√5)−16)/(16)) =((−4−4(√5))/(16)) =−((1+(√5))/4) ⇒ x^5 −1 =(x−1)(x^2 −(((√5)−1)/2)x +1)(x^2 +((1+(√5))/2)x +1)](https://www.tinkutara.com/question/Q73108.png)
$$\left.\mathrm{1}\right)\:{let}\:{decompose}\:{inside}\:{C}\left[{x}\right]\:\:{x}^{\mathrm{5}} −\mathrm{1} \\ $$$${x}={r}\:{e}^{{i}\theta} \:\:\:{so}\:{x}^{\mathrm{5}} −\mathrm{1}\:=\mathrm{0}\:\Rightarrow{r}^{\mathrm{5}} \:{e}^{{i}\mathrm{5}\theta} \:={e}^{{i}\mathrm{2}{k}\pi} \:\Rightarrow{r}=\mathrm{1}\:{and}\:\theta\:=\frac{\mathrm{2}{k}\pi}{\mathrm{5}} \\ $$$${so}\:{the}\:{roots}\:{areZ}_{{k}} ={e}^{{i}\frac{\mathrm{2}{k}\pi}{\mathrm{5}}} \:\:{with}\:{k}\in\left[\left[\mathrm{0},\mathrm{4}\right]\right] \\ $$$$\Rightarrow{x}^{\mathrm{5}} −\mathrm{1}\:\:=\prod_{{k}=\mathrm{0}} ^{\mathrm{4}} \left({x}−{Z}_{{k}} \right)=\left({x}−{Z}_{\mathrm{0}} \right)\left({x}−{Z}_{\mathrm{1}} \right)\left({x}−{Z}_{\mathrm{2}} \right)\left({x}−{Z}_{\mathrm{3}} \right)\left({x}−{Z}_{\mathrm{4}} \right) \\ $$$${Z}_{\mathrm{0}} =\mathrm{1}\:\:{Z}_{\mathrm{1}} ={e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{5}}} \:\:,\:{Z}_{\mathrm{2}} ={e}^{\frac{{i}\mathrm{4}\pi}{\mathrm{5}}} \:,\:{Z}_{\mathrm{3}} ={e}^{\frac{{i}\mathrm{6}\pi}{\mathrm{5}}} \:=\overset{−} {{Z}}_{\mathrm{2}} \:\:\:,{Z}_{\mathrm{4}} ={e}^{\frac{{i}\mathrm{8}\pi}{\mathrm{5}}} \:=\overset{−} {{Z}}_{\mathrm{1}} \:\Rightarrow \\ $$$${x}^{\mathrm{5}} −\mathrm{1}\:=\left({x}−\mathrm{1}\right)\left({x}−{Z}_{\mathrm{1}} \right)\left({x}−\overset{−} {{Z}}_{\mathrm{1}} \right)\left({x}−{Z}_{\mathrm{2}} \right)\left({x}−\overset{−} {{Z}}_{\mathrm{2}} \right) \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} \:−\mathrm{2}{Re}\left({Z}_{\mathrm{1}} \right){x}\:+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{Re}\left({Z}_{\mathrm{2}} \right){x}\:+\mathrm{1}\right) \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} \:−\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right){x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}\:{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right){x}\:+\mathrm{1}\right) \\ $$$${we}\:{have}\:{cos}\left(\frac{\pi}{\mathrm{5}}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\:{and}\:{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)=\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{5}}\right)−\mathrm{1} \\ $$$$=\mathrm{2}\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{16}}\:−\mathrm{1}\:=\frac{\mathrm{2}\left(\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\right)−\mathrm{16}}{\mathrm{6}}\:=\frac{\mathrm{12}+\mathrm{4}\sqrt{\mathrm{5}}−\mathrm{16}}{\mathrm{16}}\:=\frac{−\mathrm{4}+\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{16}}\:=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}} \\ $$$${cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\:=\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)−\mathrm{1}\:=\mathrm{2}\:\frac{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{16}}\:−\mathrm{1}=\frac{\mathrm{2}\left(\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}\right)−\mathrm{16}}{\mathrm{16}} \\ $$$$=\frac{\mathrm{12}−\mathrm{4}\sqrt{\mathrm{5}}−\mathrm{16}}{\mathrm{16}}\:=\frac{−\mathrm{4}−\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{16}}\:=−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\:\Rightarrow \\ $$$${x}^{\mathrm{5}} −\mathrm{1}\:=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} −\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}{x}\:+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{x}\:+\mathrm{1}\right) \\ $$$$ \\ $$