TITLE: Why can you find the roots a of polynomial by factoring it?
QUESTION [16 upvotes]: Let $f(x) = x^2 - 9x- 10$
We can state that $f(x) = (x + 1)(x-10)$ since I simply factored it.
The roots of this function is $-1$ and $10$. However, what is the relationship between a factored polynomial and its roots? Why can we assume this?

REPLY [16 votes]: Alfred Yerger explains why factoring gives roots, but that does not explain the converse: why do roots give factors?
Let $p(x)$ be a polynomial with a root $c$.  By long-division of $p(x)$ by $x-c$, we can write $p(x)=q(x)(x-c)+r$, where $r$ is the remainder (a constant since $x-c$ is a linear polynomial).  Now, by substituting $x=c$, we get $p(c)=q(c)(c-c)+r$.  Since $p(c)=0$ and $c-c=0$, we have $0=r$.  That is, there was no remainder after all!  Hence, the result of the long division is that $p(x)=q(x)(x-c)$.  In other words, if we factor out $x-c$ from $p(x)$, we are left with a polynomial $q(x)$.
Something that is easy to miss with this is that in fact, $p(c)=q(c)(c-c)+r$.  That is, $p(c)=r$, whether or not $c$ is actually a root. Thus, the long division will always give $p(x)=q(x)(x-c)+p(c)$.  This is the basis of a computational technique called synthetic division which gives both $q(x)$ and $p(c)$.  And, it takes about as much effort to calculate both as it does to calculate either on its own.