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!
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$\begingroup$In suspect you forgot $z^1$. So, $$ p(z) = \frac{1-z^9}{1-z},\quad z\neq 1, $$ i.e., $p(z) = 0$ iff $z^9 = 1$, $z\neq 1$. This should help.
$\endgroup$ $\begingroup$Hint:
Note that you can reorder the terms as: $$ (1+z^3+z^6)+(z+z^4+z^7)+(z^2+z^5+z^8) $$ so you have $$ 1(1+z^3+z^6)+z(1+z^3+z^6)+z^2(1+z^3+z^6)=(1+z^3+z^6)(1+z+z^2) $$ now, to find an irreducible factorization in $\mathbb{C}$, you have to factorize the two factors.
$\endgroup$ $\begingroup$Since $p(1) \neq 0$, $(z-1)$ will not be a factor.
Are you sure you're not missing a term of $z$?
Hint: For a polynomial with real coefficients, if a complex number $a+bi$ is a root, then so is its conjugate $a-bi$.
$\endgroup$ 2 $\begingroup$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: $$p(z)=\big(z+(-1)^{\frac{1}{9}}\big)\big(z-(-1)^{\frac{2}{9}}\big)\big(z+(-1)^{\frac{3}{9}}\big)\big(z-(-1)^{\frac{4}{9}}\big)\big(z+(-1)^{\frac{5}{9}}\big)\big(z-(-1)^{\frac{6}{9}}\big)\big(z+(-1)^{\frac{7}{9}}\big)\big(z-(-1)^{\frac{8}{9}}\big)$$
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