Roots for f' - polynomial function with degree 3

Question

For this function: $f: \mathbb{R} \rightarrow \mathbb{R}$, $f(x)=(x-1)\cdot(x-2)\cdot(x-3)\cdot(x-4)$ , proof that $f'(x)$ (the first derivate) have 3 distinct real roots.

How can I solve this without brute force? (open parentheses). $f$ is polynomial function with degree 4. After derivation, it will have degree 3 so $f'$ it will have 3 roots (because $f'$ have degree 3), but no 100% distincts or real.

Also, I could use Rolle theorem on $f'(x)$ but in this way I need to open parentheses and then would be a lot of work. Is there an easy way to solve this problem?

Answer 1

HINT

Since $f(1) = 0 = f(2)$, by Rolle's Theorem, $f'$ must have a root on $(1,2)$. Similarly, argue $f'$ must have a root on $(2,3)$ and $(3,4)$. Since the intervals are disjoint, so are the roots distinct.

Answer 2

You could find the critical points $z_1,z_2$ such that $f''(z_i) = 0$ and then use the second derivative test $\left( f'''(z_i) > 0\text{ or }f'''(z_i) <0\right)$ and evaluation (of $f'$) towards infinity to find that you have three roots, one less than $z_1$, one between $z_1,z_2$, and one greater than $z_2$, where I've assumed $z_1<z_2$

Although that seems like an awful lot just to find an 'elegant' way to do it.

You could also use Sturm's Theorem