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Question Number 30592 by abdo imad last updated on 23/Feb/18
let p(x)=x^(2n)  −2cosα x^n  +1  1) find roots lf p(x)  2)factorize p(x) inside C[x]  3)factorize p(x) inside R[x].
letp(x)=x2n2cosαxn+11)findrootslfp(x)2)factorizep(x)insideC[x]3)factorizep(x)insideR[x].
Answered by sma3l2996 last updated on 23/Feb/18
1)let say  p(x)=0  x^(2n) −2cos(α)x^n +1=(x^n )^2 −2cos(α)x^n +cos^2 (α)+sin^2 (α)  =(x^n −cosα)^2 +sin^2 (α)=0  x^n −cosα=isinα  or  x^n −cosα=−isinα  x_1 ^n =sinα+icosα=e^(iα)  ; x_2 ^n =cosα−isinα=e^(−iα)   so roots of p(x) are  :  x_(1,k) =e^(i(α+2kπ)/n)  ; x_(2,k) =e^(−i(α+2kπ)/n)    \k=(0,1,2,...,n−1)  2)  p(x)=Π_(k=0) ^(n−1) (x−e^(i(α+2kπ)/n) )×Π_(k=0) ^(n−1) (x−e^(−i(α+2kπ)/n) )  p(x)=Π_(k=0) ^(n−1) (x−e^(i(α+2kπ)/n) )(x−e^(−i(α+2kπ)/n) )  3)  p(x)=Π_(k=0) ^(n−1) (x^2 −(e^(i(α+2kπ)/n) +e^(−i(α+2kπ)/n) )x+e^(i(α+2kπ−α−2kπ)/n) )  p(x)=Π_(k=0) ^(n−1) (x^2 −2cos(((α+2kπ)/n))x+1)
1)letsayp(x)=0x2n2cos(α)xn+1=(xn)22cos(α)xn+cos2(α)+sin2(α)=(xncosα)2+sin2(α)=0xncosα=isinαorxncosα=isinαx1n=sinα+icosα=eiα;x2n=cosαisinα=eiαsorootsofp(x)are:x1,k=ei(α+2kπ)/n;x2,k=ei(α+2kπ)/nk=(0,1,2,,n1)2)p(x)=k=0n1(xei(α+2kπ)/n)×k=0n1(xei(α+2kπ)/n)p(x)=k=0n1(xei(α+2kπ)/n)(xei(α+2kπ)/n)3)p(x)=k=0n1(x2(ei(α+2kπ)/n+ei(α+2kπ)/n)x+ei(α+2kπα2kπ)/n)p(x)=k=0n1(x22cos(α+2kπn)x+1)

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