Is there a non-calculus intuitive explanation for why an $n^{th}$ degree polynomial has at most $n-1$ turning points?

Question

Is there a non-calculus intuitive explanation for why an $n^{th}$ degree polynomial has at most $n-1$ turning points?

The calculus explanation is clear enough, but is there a "classical" explanation?

Answer 1

You can show by pure algebra that the turning points occur at arguments that correspond to what we often call the derivative being zero.

For definiteness consider a cubic polynomial

$y=ax^3+bx^2+cx+d$.

Say it has a turning point at $x=x_0, y=y_0$. For that to work $y-y_0$ must have at least a double root at $x=x_0$. Thus the quotient

$\frac{y-y_0}{(x-x_0)^2}$

must give a zero remainder when the long division is carried out.

Do this division with $y=ax^3+bx^2+cx+d$ as above and you get

$ax^3+bx^2+cx+d-y_0=Q(x-x_0)^2+R$

where the quotient $Q$ and the remainder $R$ are given by:

$Q=ax+(2ax_0+b)$

$R=(3a{x_0}^2+2bx_0+c)x+(a{x_0}^3+b{x_0}^2+cx_0+d-y_0)-(3a{x_0}^2+2bx_0+c)x_0$

The second term in the remainder is automatically zero because $y_0$ is defined as $ax^3+bx^2+cx+d$ evaluated at $x=x_0$, but getting the rest of the remainder to also equal zero requires satisfying a quadratic relation

$3a{x_0}^2+2bx_0+c=0$

which is the "zero derivative" condition for a cubic polynomial. Likewise for any degree $n$ the condition for a turning point corresponds to a degree $n-1$ relation corresponding to a "zero derivative".