Linear Factorization of Complex Polynomials I am trying to find a linear factorization of the polynomial $$p(z) = 1 +z+z^2 +z^3 +z^4 +z^5 + z^6 +z^7 +z^8$$
I know what it means by linear factorization in the sense of non-complex polynomials, but i'm not sure where to begin for a complex polynomial of degree 8. I tried some trial and error by factoring out $(z-1)$ and $i$ but didn't seem to have much luck! Could someone help me get on the right track please!

The polynomial factorizes, over the reals as ($z\neq 1$)
$$p(z) = 1 +z+z^2 +z^3 +z^4 +z^5 + z^6 +z^7 +z^8=\frac{1-z^9}{1-z}=(1+z+z^2)(1+z^3+z^6)$$
and its roots are the (complex) $9$-th roots of unity (except $1$ of course):
$$-(-1)^{\frac{1}{9}},(-1)^{\frac{2}{9}},-(-1)^{\frac{3}{9}}, (-1)^{\frac{4}{9}},-(-1)^{\frac{5}{9}}, (-1)^{\frac{6}{9}}, 
-(-1)^{\frac{7}{9}},(-1)^{\frac{8}{9}}$$
which should make the $8$ linear factors clear. So:
$$p(z)=\big(z+(-1)^{\frac{1}{9}}\big)\big(z-(-1)^{\frac{2}{9}}\big)\big(z+(-1)^{\frac{3}{9}}\big)\big(z-(-1)^{\frac{4}{9}}\big)\big(z+(-1)^{\frac{5}{9}}\big)\big(z-(-1)^{\frac{6}{9}}\big)\big(z+(-1)^{\frac{7}{9}}\big)\big(z-(-1)^{\frac{8}{9}}\big)$$ 
