Every polynomial with complex coefficients can be written as the product of linear factors. Enter the linear factors of #P(z)=1+z+⋯+z^6+z^7# any help?

1 Answer
Feb 14, 2018

See below.

Explanation:

#1+z+z^2+cdots+z^7 = (z^8-1)/(z-1)#

but

#(z^8-1)/(z-1)=((x^4+1)(z^4-1))/(z-1) = ((z^4+1)(z^2+1)(z^2-1))/(z-1) = (z^4+1)(z^2+1)(z+1)#

but

#z^2+1 = (z+i)(z-i)# and

#(z^4+1)=(z^2+i)(z^2-i) = (z+sqrti)(z-sqrti)(z+sqrt(-i))(z-sqrt(-i))# and finally

#1+z+z^2+cdots+z^7 = (z+sqrti)(z-sqrti)(z+sqrt(-i))(z-sqrt(-i)) (z+i)(z-i)(z+1)#