Given $f: \Bbb C \to \Bbb C$ s.t. $f(z) = a_nz^n + \cdots + a_1z + a_0$ with non-zero coefficients, I want to prove there are exactly $n+1$ complex numbers with $f(z) = w_0(z-w_1) \cdots (z-w_n)$.
I'm given that:
- Any nonzero degree polynomial equation with nonzero coefficients has at least one root
- Any polynomial equation with at least one root $w$ has a form $f(z) = (z-w)g(z)$, where $g$ is a polynomial function with non-zero coefficients of $n-1$ degrees
Is there something like induction done the other way around, i.e., from n to zero? I ask because, if you give me a small $n$ I can do this proof manually repeatedly applying (1) then (2), but doing this for an arbitrary starting $n$ doesn't seem like it would follow the usual induction method, and would instead require something like induction in the opposite direction.
Suppose the statement were false. Then there would be some polynomial $f(z)$ with smallest possible degree $n+1, n \geq 1$ where the statement is false. We know the degree has the form $n+1, n \geq 1$ because any linear polynomial $az+b$ has the form $a(z- (-\frac ba))$.
We know from $(1)$ that $f(z)$ has some root $w_{n+1}$. Therefore, we know from $(2)$ that $f(z)=(z-w_{n+1})g(z)$, where $\deg g \lt \deg f$. But since $f$ has the smallest possible degree of any polynomial that can't be written in the required form, we know that $g$ can be written in the required form.
So $g(z)=w_0(z-w_1) \ldots (z-w_n)$, which means $f(z)=w_0(z-w_1) \ldots (z-w_n)(z-w_{n+1})$. This contradiction proves that our hypothesis, the existence of $f$ that can't be written in the required form, is false. That proves the theorem.
Restating this in the form of infinite descent, if $f=(z-w_{n+1})g$ can't be written in the required form, then $g$, which has smaller degree, also can't be written in the required form. But that means we could create an infinite chain of polynomials with decreasing degrees, which is impossible.