Proofs from Complex polynomials

Polynomial division gives

with a constant remainder $r$. At $z=\alpha$, the first term vanishes, leaving $p(\alpha)=r$.

If $p(\alpha)=0$, polynomial division gives

so $z-\alpha$ is a factor. Conversely, if $p(z)=(z-\alpha)q(z)$, then $p(\alpha)=0$.

Proceed by induction on the degree. A degree-one polynomial already has the required form. Suppose the result holds for degree $m$. If $p$ has degree $m+1$, the fundamental theorem of algebra supplies a root $\alpha_1$. The factor theorem gives

for a polynomial $q$ of degree $m$. Apply the induction hypothesis to $q$ and multiply the factors.

Write $p(z)=\sum_{j=0}^n a_jz^j$ with $a_j\in\mathbb R$. If $p(\alpha)=0$, then

For multiplicity, write $p(z)=(z-\alpha)^kq(z)$ with $q(\alpha)\ne0$. Since $p$ has real coefficients, conjugating this factorisation gives $p(z)=(z-\overline\alpha)^k\widetilde q(z)$, where $\widetilde q(z)=\overline{q(\overline z)}$ and $\widetilde q(\overline\alpha)=\overline{q(\alpha)}\ne0$. Thus the conjugate root has exactly the same multiplicity.