Answer
Proof given below.
Work Step by Step
Proof by contradiction.
f is a polynomial of degree 3 (the leading coefficient of the $x^3$ term is not zero).
Assume the opposite: $f$ has $4$ or more real zeros.
Then, ordering four of the zeros, $ c_{1} \lt c_{2} \lt c_{3}\lt c_{4}$, we have three intervals on which Rolle's theorem applies (polynomials are continuous and differentiable everywhere).
So, $f'$ has three zeros, $d_{1}\in(c_{1},c_{2}),\ d_{2}\in(c_{2},c_{3}),\ d_{3}\in(c_{3},c_{4})$.
Again, with Rolle's theorem applied on $f'$ over two intervals, it follows that
$f''$ has two zeros, $e_{1}\in(d_{1},d_{2}),\ e_{2}\in(d_{2},d_{3})$.
Once again, Rolles theorem gives
$f'''$ has a zero in $(e_{1},e_{2}).$
BUT, if the degree of $f$ is 3, then
the degree of $f'$ is 2,
the degree of $f''$ is 1,
the degree of $f'''$ is 0, a nonzero constant that has no zeros. So, we have a contradiction.
Our assumption was wrong; f can have at most 3 real zeros.
