# why can we factor a polynomial using its solutions

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"

• Familiar with the Fundamental theorem of algebra? Jul 16, 2020 at 17:16
• Have you looked at the Factor Theorem? Jul 16, 2020 at 17:16
• "and where does the coefficient $a$ in the factored form come from?" It is the same $a$ as in $f(x)=ax^n+bx^{n-1}+\dots$ Jul 16, 2020 at 17:21
• 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. Jul 16, 2020 at 17:29

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).