Consider $x^2 -3x + 2 = 0$ It factorizes as $(x-1)(x-2) = 0$ and so has two clearly distinct roots $x = 1$ and $x = 2$.
Now consider $x^2 -2x +1 = 0$. It factorizes as $(x-1)(x-1) = 0$ In this case there are two factors which are zero if $x = 1$, so the root $x=1$ is considered to repeat (with "multiplicity" 2).
Now look at $x^2 = 0$. It factorizes as $x.x = 0$, or if it makes it clearer, $(x-0)(x-0) = 0$. Here the root $x = 0$ occurs twice, i.e. with multiplicity 2.
FtoA says that for a polynomial of degree n, there are n (some, possibly complex) roots if you include the multiplicity. This is perhaps more clearly expressed that a polynomial $p_n(x)$ will factorize as $(x-c_1)(x-c_2)....(x-c_n)$ where the $c_i$'s can be repeated and can be zero.