How does one determine whether a polynomial has a double root? Given the following function:
$$x^{3}+x^{2}+2x+8=0$$
The roots are $-3$, and $-1$.
However, $-3$ is counted as a double root, all three roots are $-3, -3, -1.$
My question is, how is it determined that $-3$, but not $-1$, is the double root?
 A: This equation does not have $-3$ or $-1$ as solutions, so I'll instead focus on one which does:
$$x^3 + 7 x^2 + 15 x + 9 = 0$$
Here, $-1$ is a "single" nonrepeated root, and $-3$ is a double root.

Why? It's quite simple: it's because you can write
$$x^3 + 7 x^2 + 15 x + 9 = (x-(-3))(x-(-3))(x-(-1))$$
In general, if a polynomial $f$ has roots $r_1,\cdots,r_n$ and leading coefficient $a$, you can write
$$f(x) = a (x-r_1)(x-r_2) \cdots (x-r_n)$$
and if the factor $(x-r)$ appears $m$ times, you say $r$ is a root of multiplicity $m$. ($m=2$ being double roots, $m=3$ being triple roots, and so on.)
Another way to look at it, it is because $(x+3)^2$ evenly divides $x^3 + 7 x^2 + 15 x + 9$, resulting in another polynomial, but $(x+3)^3$ will not.
Yet another way, it's because you can "factor out" or divide $x^3 + 7 x^2 + 15 x + 9$ by $x+3$, get a polynomial, and do it again. (In general if a root appeared $n$ times, then you could divide by the corresponding linear factor $n$ times but not $n+1$.)

An additional way is via calculus. If $r$ is a double root, you can write
$$f(x) = (x-r)^2 g(x)$$
Then by differentiation you see
$$f'(x) = 2(x-r) g(x) + (x-r)^2 g'(x)$$
Hence $f'(r) = 0$.
