Why can you find the roots a of polynomial by factoring it?

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?

Solution 1:

It's not an assumption. It is a general fact that if $ab = 0$, then either $a = 0$ or $b = 0$. The roots of a polynomial are the numbers for which that polynomial evaluates to $0$, so to find the roots, it is enough to find the roots of the factors. So we factor everything as much as we can. In your case, this is a quadratic polynomial, so any factors are linear, and then it is obvious what the roots are.

Solution 2:

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.