IMO she has all right to be confused, because this identification-process is actually not as trivial as we perceive it. Maths education doesn't really explain what's really going on there, and perhaps for good reasons (learning by doing is likely the most effective). But to actually get down to the principles, on the risk of making it way more complicated: what we're dealing with here is a problem of inverse modelling.
The general formulation is this:
You are given two functions,
$$\begin{align}
& f : \mathbb{R} \to \mathbb{R}
\\ &f(x) = 8x^2 - 64
\end{align}$$
and
$$\begin{align}
& g : \mathbb{R}^3\times\mathbb{R} \to \mathbb{R}
\\ & g_{(a,b,c)}(x) = ax^2 + bx + c.
\end{align}$$
Now your task is to find values for $a$, $b$ and $c$ that make $g_{(a,b,c)}(x)$ as similar as possible to $f(x)$. In general this would require probing with concrete values and some kind of optimisation procedure (and first of all a notion what we mean by "similar"). We're lucky because we have closed forms for both $f$ and $g$ and can actually make them equal by mere pattern matching. Specifically, she's already determined that $a$ must be $8$. This takes care of the quadratic terms, simplifying the problem to finding $b,c$ such that
$$ \tilde f(x) = -64 $$
and
$$ \tilde g_{(b,c)}(x) = bx + c $$
are equal.
This is where it pays off to write it explicitly as functions, because $\tilde f$ is still more than just a number literal: it is a function of $x$, whose value just happens to not depend on $x$. This highlights that what we need to solve is not just $-64 = bx + c$ (which would be indeterminate as such), but specifically
$$ \forall x\in\mathbb{R} : \quad -64 = bx + c. $$
To have any chance of being true for all numbers, it should certainly be true on any examples, so let's start with the simplest conceivable examples: $x=0$.
$$ -64 = b\cdot0 + c \quad\Rightarrow\quad c = -64, $$
and then $x = 1$,
$$ -64 = b + c = b - 64 \quad\Rightarrow\quad b = 0. $$
So at this point we have determined that $b=0$ and $c=-64$ are the only possible candidate values. Now we still need to show that the expressions are equal for all $x$, but that's just a matter of confirming that the $b=0$ cancels any variations that $x$ could introduce.
Of course, once the student has grokked the principle, she should avoid actually doing it this awkward and slow way and instead just notice that $f$ can be rewritten, adding a term that's equal to zero but pattern-matchable:
$$ f(x) = 8x^2 + 0x - 64 $$
(in German, this is called a "nahrhafte Null", i.e. "nutricious zero").
