From your post it seems you are permitted to use the following:

1. $a\in\mathbb{R}$ is a root of $p(x)$ (i.e. $p(a) = 0$) iff $p(x) = (x-a)q(x)$ for some polynomial $q(x)$.

2. $a\in\mathbb{R}$ is a local extremum of $p(x)$ iff (thanks to Ilya Bogdanov; consider e.g. $p(x) = x^3$) only if $p(x)-p(a) = (x-a)^2q(x)$ for some polynomial $q(x)$.

A proof using these two facts is as follows: write $p(x) = \sum_{k=1}^n c_kx^k$. Then you can write $p(x) - p(a) = (x-a)Q(x,a)$ for some polynomial $Q$ in two variables. But then by those two facts, $a\in\mathbb{R}$ is a local extremum iff $q(a) = 0$, where $q(x) := Q(x,x)$. You can explicitly show (e.g. by computing the $(n-1)$-degree term of $q$) that it is a non-zero polynomial of degree $n-1$.

This can also serve as a motivation to introduce/define the derivative.

From your post it seems you are permitted to use the following:

1. $a\in\mathbb{R}$ is a root of $p(x)$ (i.e. $p(a) = 0$) iff $p(x) = (x-a)q(x)$ for some polynomial $q(x)$.

2. $a\in\mathbb{R}$ is a local extremum of $p(x)$ iff (thanks to Ilya Bogdanov; consider e.g. $p(x) = x^3$) only if $p(x)-p(a) = (x-a)^2q(x)$ for some polynomial $q(x)$.

A proof using these two facts is as follows: write $p(x) = \sum_{k=1}^n c_kx^k$. Then you can write $p(x) - p(a) = (x-a)Q(x,a)$ for some polynomial $Q$ in two variables. But then by those two facts, if $a\in\mathbb{R}$ is a local extremum, then $q(a) = 0$, where $q(x) := Q(x,x)$. You can explicitly show (e.g. by computing the $(n-1)$-degree term of $q$) that it is a non-zero polynomial of degree $n-1$.

This can also serve as a motivation to introduce/define the derivative.

From your post it seems you are permitted to use the following:

1. $a\in\mathbb{R}$ is a root of $p(x)$ (i.e. $p(a) = 0$) iff $p(x) = (x-a)q(x)$ for some polynomial $q(x)$.

2. $a\in\mathbb{R}$ is a local extremum of $p(x)$ iff $p(x)-p(a) = (x-a)^2q(x)$ for some polynomial $q(x)$.

A proof using these two facts is as follows: write $p(x) = \sum_{k=1}^n c_kx^k$. Then you can write $p(x) - p(a) = (x-a)Q(x,a)$ for some polynomial $Q$ in two variables. But then by those two facts, $a\in\mathbb{R}$ is a local extremum iff $q(a) = 0$, where $q(x) := Q(x,x)$. You can explicitly show (e.g. by computing the $(n-1)$-degree term of $q$) that it is a non-zero polynomial of degree $n-1$.

This can also serve as a motivation to introduce/define the derivative.

From your post it seems you are permitted to use the following:

1. $a\in\mathbb{R}$ is a root of $p(x)$ (i.e. $p(a) = 0$) iff $p(x) = (x-a)q(x)$ for some polynomial $q(x)$.

2. $a\in\mathbb{R}$ is a local extremum of $p(x)$ iff (thanks to Ilya Bogdanov; consider e.g. $p(x) = x^3$) only if $p(x)-p(a) = (x-a)^2q(x)$ for some polynomial $q(x)$.

A proof using these two facts is as follows: write $p(x) = \sum_{k=1}^n c_kx^k$. Then you can write $p(x) - p(a) = (x-a)Q(x,a)$ for some polynomial $Q$ in two variables. But then by those two facts, $a\in\mathbb{R}$ is a local extremum iff $q(a) = 0$, where $q(x) := Q(x,x)$. You can explicitly show (e.g. by computing the $(n-1)$-degree term of $q$) that it is a non-zero polynomial of degree $n-1$.

This can also serve as a motivation to introduce/define the derivative.