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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
factorizeinsideR[X]1)X5−12)X6+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)
1)letdecomposeinsideC[x]x5−1x=reiθsox5−1=0⇒r5ei5θ=ei2kπ⇒r=1andθ=2kπ5sotherootsareZk=ei2kπ5withk∈[[0,4]]⇒x5−1=∏k=04(x−Zk)=(x−Z0)(x−Z1)(x−Z2)(x−Z3)(x−Z4)Z0=1Z1=ei2π5,Z2=ei4π5,Z3=ei6π5=Z−2,Z4=ei8π5=Z−1⇒x5−1=(x−1)(x−Z1)(x−Z−1)(x−Z2)(x−Z−2)=(x−1)(x2−2Re(Z1)x+1)(x2−2Re(Z2)x+1)=(x−1)(x2−2cos(2π5)x+1)(x2−2cos(4π5)x+1)wehavecos(π5)=1+54andcos(2π5)=2cos2(π5)−1=2(1+5)216−1=2(6+25)−166=12+45−1616=−4+4516=5−14cos(4π5)=2cos2(2π5)−1=2(5−1)216−1=2(6−25)−1616=12−45−1616=−4−4516=−1+54⇒x5−1=(x−1)(x2−5−12x+1)(x2+1+52x+1)

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