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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"

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  • $\begingroup$ Read about Bezout's theorem. $\endgroup$
    – Michael Rozenberg
    Jul 16, 2020 at 17:16
  • 2
    $\begingroup$ Familiar with the Fundamental theorem of algebra? $\endgroup$
    – Alexey Burdin
    Jul 16, 2020 at 17:16
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    $\begingroup$ Have you looked at the Factor Theorem? $\endgroup$
    – highbeta
    Jul 16, 2020 at 17:16
  • $\begingroup$ "and where does the coefficient $a$ in the factored form come from?" It is the same $a$ as in $f(x)=ax^n+bx^{n-1}+\dots$ $\endgroup$
    – JMoravitz
    Jul 16, 2020 at 17:21
  • $\begingroup$ As for a "simple proof"... the fundamental theorem of algebra is one of the classic results in an undergraduate course in Algebra (abstract/modern algebra, a course on group theory, ring theory, etc... not just a pre-calculus how to perform arithmetic with letters). To fully understand and appreciate it, you'll probably want to get a lot more under your belt with regards to Groups, Rings, Euclidean Domains, Unique Factorization Domains, etc... Any decent Algebra textbook should cover this in plenty of detail. Else, quotient-remainder or degrees might not make full sense. $\endgroup$
    – JMoravitz
    Jul 16, 2020 at 17:29

1 Answer 1

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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).

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