Can someone please explain why we are able to factor an $n$ degree polynomial function using only it roots?
What I mean is this:
Lets say we have a function defined like so: $$f(x) = ax^4 + bx^3 +\dots$$ It can supposedly be factored like so: $$f(x) = a(x−p)(x−q)(x−r)\dots$$ Where $p, q, r$ etc. are the solutions of the function being equal to $0$. Is there a simple proof for why this is valid, and where does the coefficient $a$ in the factored form come from? (I don't want some lame answer for $a$ like :"if $a$ wasn't there the factored form wouldn't equal the original form"
Consider your polynomial $p(x)$, with zeros $z_1, z_2, \ldots, z_n$. Take:
$\begin{align*} p(x) &= q(x) (x - z_i) + r(x) \end{align*}$
(plain polynomial division, $q$ is quotient, $r$ remainder). You know that the degree of $r$ must be less than the degree of $x - z_i$, i.e., it is a constant. Now:
$\begin{align*} p(z_i) &= q(z_i) \cdot 0 + r(z_i) \end{align*}$
so you see that $r(z_i) = 0$, but $r(x)$ is a constant. Thus you conclude:
$\begin{align*} p(x) &= q(x) (x - z_i) \\ &\vdots \\ &= a (x - z_1) (x - z_2) \dotsm (x - z_n) \end{align*}$
The $a$ is just the leading coefficient of $p(x)$, the coefficient of the highest power of $x$ (if you multiply out the rest, the leading coefficient is 1, a monic polynomial).