Polynomials : $p(z)=(z-r)q(z)$ Theorem : Let $p$ be a polynomial of degree $n\geq 1$, with leading co-efficient $1$, and let $r$ be a roof of $p$. Then  $p(z)=(z-r)q(z)$, where $q$ is a polynomial of degree $n-1$, with leading co-efficient $1$. The proof is given : Let $p(z)$ have the form $$p(z)=z^n +a_1z^{n-1}+\cdots a_{n-1}z+a_n$$
and let $c$ be any complex number. Then $$p(z)-p(c)=(z^n - c^n)+a_1(z^{n-1}-c^{n-1})+\cdots+a_{n-1}(z-c)=(z-c)q(z)$$
where $q$ is the polynomial given by $$q(z)=z^{n-1} + cz^{n-2}+c^2z^{n-3}+\cdots+c^{n-1}$$$$+a_1(z^{n-2}+cz^{n-3}+\cdots+c^{n-2})+\cdots+a_{n-1}$$
Where is this formula for $q$ derived from? I think with this theorem, one can factor polynomials and solve their roots - provided we are able to guess at least one root $r$. But the problem is we need to find $q$. How can I compute $q$ for, lets say this polynomial : $z^3 -3z^2 +4$?
 A: Have you not learned polynomial long division? It works almost exactly like long division in arithmetic:
$$\require{enclose}\begin{array}{r}
z^2-\phantom 0z-2\\[-3pt]
z-2\enclose{longdiv}{z^3-3z^2 + 0z + 4}\\[-3pt]
\underline{z^3-2z^2\phantom {\;\;+0z+4}}\\[-3pt]
-z^2+0z+4\\[-3pt]
\underline{-z^2+2z\phantom{+4\;\,}}\\[-3pt]
-2z+4\\[-3pt]
\underline{-2z+4}\\[-3pt]
0
\end{array}$$
So $z^3 -3z^2 + 4 = (z-2)(z^2 -z-2)$.

*

*When you write out the division, put in terms for every power of $z$ in the dividend, as I did by adding "$+0z$". You will need that space eventually.

*Take the ratio of the leading term of the dividend to the leading term of the divisor. In this case that is $\frac {z^3}z = z^2$. Write this result on top as the first term of the quotient.

*Multiply the divisor by the just-found quotient term and write the result underneath the dividend, lining up the terms by degree.

*Subtract and write the result below the line. This is the remainder.

*If the remainder is still of equal or higher degree than your divisor, it becomes your new dividend. Repeat the process to get the next term in the quotient.

*Continue until the remainder is of lower degree than the divisor ($0$ is of lower degree than anything other than itself).

This works for any polynomial dividend and divisor. But if the divisor is of degree $1$, then there is a shortcut version called synthetic division that does the same thing, but with much less writing.
As for finding that first root: There is a trick called the Rational Root Theorem. This is great for math homework, because the teacher can carefully construct examples where you can use the rational root theorem to find roots and divide them out, thus giving you good practice in larger degree polynomials and their behavior. It also has some number theory implications.
But for "real world" problems (ones that were not carefully constructed to be easy to solve), the rational root theorem is of practically no value, rarely ever giving you an actual root. There for degrees higher than $4$, you generally have to use approximation techniques to find approximate roots (generally this is easier than using the $3$rd and $4$th degree versions of the quadratic formula as well).
