Here's an alternative to using Rolle's Theorem, which can also be done in a couple of minutes.
Let $x-3=u$. Then the polynomial becomes
so the derivative is
which is quadratic in $u^2$. You can do is either using the quadratic formula,
so simply by noting that $5U^2-15U+4$ is clearly positive for $U\le0$ and negative at $U=1$, so it has two distinct positive roots, hence the quartic derivative has $4$ real roots, all distinct.
It took me more than five minutes to write this up, but the scratchwork and thinking took less than three. The key was to notice that a simple shift makes it easy to expand the factored quintic.
Remark: This approach clearly only works because the roots of the quintic are equally spaced. If the problem started, say with $(x-1)(x-2)(x-4)(x-8)(x-16)$, then Rolle's Theorem would be your best bet. Still, it's a useful alternative to have on tap; suppose, for example, the question had asked for the number of solutions to $f'(x)=1$. Try answering that armed only with Rolle!