I'll use an example recurrence but my question is meant to be generalized.
Let's say we had some recurrence, such as:
$$F(n) = -8F(n-1) + 9F(n-2) + 92F(n-3) - 140F(n-4)$$
where we already know the first few base constants $F(0), F(1), F(2), F(3)$ so the entire recurrence is defined for all integers $n \geq 0$.
Normally we convert this to some kind of characteristic polynomial:
$$x^n = -8 x^{n-1} + 9 x^{n-2} + 92 x^{n-3} - 140x^{n-4}$$
Divide everything by $x^{n-4}$ and put everything on one side:
$$x^4 +8 x^3 - 9 x^2 - 92 x + 140 = 0$$
This polynomial can be factored:
$$(x - 2)^2 (x + 5) (x + 7) = 0$$
And now we know that the roots are $2, -5, -7$. The $2$ root has multiplicity $2$, whereas the $-5$ and $-7$ roots each have multiplicity $1$.
From this we can say that:
$$F(n) = a \cdot (2)^n + b \cdot n \cdot (2)^n + c \cdot (-5)^n + d \cdot (-7)^n$$
And then we use the original four values of $F$ that we do know to solve a short system of equations and solve for $a, b, c, d$ to finish up the closed form.
The short version of my question is basically "Why does this work?"
Why can we use a "characteristic polynomial" (what is this, exactly) instead of a recurrence?
Why does that polynomial's roots directly correspond to the closed-form of that recurrence?
Why do we need to add an additional term with another power of $n$ for roots of multiplicity $>1$?