Question Number 33743 by prof Abdo imad last updated on 23/Apr/18
![let p(x)=(1+x^2 )(1+x^4 )....(1+x^2^n ) with n integr 1) find the roots of p(x) 2) factorize p(x) inside C[x]](Q33743.png)
$${let}\:{p}\left({x}\right)=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)....\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \right)\:{with}\:{n}\:{integr} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{factorize}\:{p}\left({x}\right)\:{inside}\:{C}\left[{x}\right] \\ $$
Commented by caravan msup abdo. last updated on 24/Apr/18
![we have proved that p(x)=((1−x^2^(n+1) )/(1−x^2 )) if x^2 ≠1 so p(x)=0 ⇔ x^2^(n+1) = e^(i2kπ) if x=r e^(iθ) we get r=1 and 2^(n+1) θ =2kπ ⇒ θ_k = ((2kπ)/2^(n+1) ) = ((kπ)/2^n ) and k∈[[1,2^(n+1) −1]] the roots of p(x) are the complex z_k =e^(i((kπ)/(2^n ))) with 1≤k≤ 2^(n+1) −1 and k≠2^n 2) p(x) =Π_(k=1 and k≠2^n ) ^(2^(n+1) −1) (z−e^(i((kπ)/2^n )) )](Q33779.png)
$${we}\:{have}\:{proved}\:{that}\:{p}\left({x}\right)=\frac{\mathrm{1}−{x}^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${if}\:{x}^{\mathrm{2}} \neq\mathrm{1}\:{so}\:{p}\left({x}\right)=\mathrm{0}\:\Leftrightarrow\:{x}^{\mathrm{2}^{{n}+\mathrm{1}} } =\:{e}^{{i}\mathrm{2}{k}\pi} \\ $$$${if}\:{x}={r}\:{e}^{{i}\theta} \:{we}\:{get}\:{r}=\mathrm{1}\:{and}\:\mathrm{2}^{{n}+\mathrm{1}} \theta\:=\mathrm{2}{k}\pi \\ $$$$\Rightarrow\:\theta_{{k}} \:=\:\frac{\mathrm{2}{k}\pi}{\mathrm{2}^{{n}+\mathrm{1}} }\:=\:\frac{{k}\pi}{\mathrm{2}^{{n}} }\:{and}\:{k}\in\left[\left[\mathrm{1},\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}\right]\right] \\ $$$${the}\:{roots}\:{of}\:{p}\left({x}\right)\:{are}\:{the}\:{complex} \\ $$$${z}_{{k}} \:\:={e}^{{i}\frac{{k}\pi}{\mathrm{2}^{{n}} \:\:\:}} \:{with}\:\mathrm{1}\leqslant{k}\leqslant\:\mathrm{2}^{{n}+\mathrm{1}} \:−\mathrm{1}\:\:{and}\:{k}\neq\mathrm{2}^{{n}} \\ $$$$\left.\mathrm{2}\right)\:{p}\left({x}\right)\:=\prod_{{k}=\mathrm{1}\:\:{and}\:{k}\neq\mathrm{2}^{{n}} } ^{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}} \:\:\left({z}−{e}^{{i}\frac{{k}\pi}{\mathrm{2}^{{n}} }} \right) \\ $$