TITLE: Proving roots with Mean Value Theorem
QUESTION [7 upvotes]: Stewart wants me to prove stuff but I have no idea how to.
a) Show that a polynomial of degree 3 has at most three real roots.
b) Show that a polynomial of degree n has at most n real roots.

REPLY [4 votes]: According to Rolles theorem and the polynomial theorem any third degree polynomial sports at most 1 to  3 roots. Therefore, any polynomial of degree n possesses at least one root, whether real or complex. Moreover, a polynomial with a degree n bears at most n roots. To back this argument with Rolles theorem assume that there are three roots such that $a>b>c$. According to Rolles theorem there must be a number $m$ such that $f'(m) = 0$ between $a$ and $b$. Likewise there must be a value $n$ such that $f'(n) = 0$ between $b$ and $c$. This implies that $m$ and $n$ are minimums or maximums. Since there are two solutions for $f'(x)$ because a third degree polynomial's derivative is a degree two function. Therefore the third degree polynomial acquires at most three real roots. If the function is twice differentiated there must be such a "$c$" between $m$ and $n$ such that $f"(c) = 0$ thus revealing that indeed that a third degree function bears at least one root. Moreover if you took a quadratic function and analyze between $x$- intercepts there is either a max or min value in between the two thus asserting the fact that when the derivative equals to zero there is indeed two zeros per critical point. As a result, there must be two c.p s to have three real roots.