Why can't repeated differentiation be used to solve polynomials? Take
$$x^2+-3x+2=0$$
This has solutions $x=1$ and $x=2$.
However, I can do anything I want to both sides of an equation and yield a new valid equation as long as I do the same thing to both sides, right?
So
$$
\frac{\mathrm{d}}{\mathrm{d}x} (x^2-3x+2) = \frac{\mathrm{d}}{\mathrm{d}x} 0
$$
$$
2x - 3 = 0
$$
$$
x = \frac{3}{2}
$$
Why is this not a valid method of solving polynomials?
 A: The rule "you can do anything you like to both sides of an equation" should be clarified here: you can apply any function you like to both sides of an equation (though even this has some limitations, which I will get to in a moment). For example, if you know that $2x = y + 4$, you could, say, perform the function $f(x) = x - 4$ to both sides, and get $2x - 4 = y$. Or you could apply the function $g(x) = x/2$ to both sides to get $x = \frac{y + 4}{2}$. Or even, I could square both sides, to obtain the true statement $4x^2 = (y + 4)^2$. These are all permissible.
While this is not strictly on point, there are some limitations to this.

*

*If the function has full domain, generally speaking, the resulting equation tends to be a consequence of the previous equation, but not the other way around. For example, $4x^2 = (y + 4)^2$ is certainly true if $2x = y + 4$ is, but if $4x^2 = (y + 4)^2$ is true, then maybe $2x = y + 4$ or maybe $2x = -(y + 4)$. If the function you're applying is not injective (which squaring is not!), then you may introduce erroneous solutions.

*Also, dually, if your function has a restricted domain, that restriction could eliminate solutions! I could take the square root of both sides of $2x = y + 4$ to get $\sqrt{2x} = \sqrt{y + 4}$, but this will instantly eliminate solutions like $(x, y) = (-2, -8)$, which were present in the original equation!

So, my point is, without some common sense, you can gain or lose solutions. But as I said, this is not strictly on point.
Now, differentiation is not a function. Well, actually, it is a function, but it's not a real function, i.e. a function that takes real numbers and spits out real numbers. It's a function that acts on functions! It's a rule for taking certain real functions (differentiable functions), and producing other real functions.
It can't act on real numbers. If I ask you, say, for the derivative of $11$, the best you can do is interpret $11$ as the constant map that maps every real number to $11$, and give me the answer $0$ (which, again, is a constant map, not a real number).
When you see the equation $x^2 - 3x + 2 = 0$, this is an equation of real numbers. Yes, the left hand side and right hand side can be viewed as functions, but it's not an equality of functions. The left hand side is a different function to the right hand side. We are looking for the few points $x \in \Bbb{R}$ where the function values of these two wildly different functions happen to coincide.
Taking the derivative of both sides yields nothing of interest, because we didn't have equality of functions in the first place.
