Proving roots with Mean Value Theorem 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.
 A: 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.
A: Part (a)
Let $f(x)=K_3x^3+K_2x^2+K_1x+K_0$
Since $f(x)$ is a polynomial, it is continuous and differentiable everywhere.
Suppose $f(x)$ has four roots $a_1, a_2, a_3$ such that $f(a_1)=f(a_2)=f(a_3)=f(a_4)=0$.
By Rolle's Theroem there exists;
$b_1$ in $(a_1, a_2)$, $b_2$ in $(a_2, a_3)$, $b_3$ in $(a_3, a_4)$ such that $f'(b_1)=f'(b_2)=f'(b_3)=0$.
So, $f'(x)=3K_3x^2+2K_2x+K_1$ is also a polynomial and thus continuous and differentiable everywhere, by Rolle's Theorem again there exists;
$c_1$ in $(b_1, b_2)$ and $c_2$ in $(b_2, b_3)$ such that $f''(c_1)=f''(c_2)=0$.
So, $f''(x)=6K_3x+K_2$ which is still continuous and differentiable everywhere, by Rolle's Theorem there should exist;
$d_1$ in $(c_1, c_2)$ such that $f'''(x)=0$
however $f'''(x)=6K_3$ which is not zero.
So $f'''(x)$ never equals zero and thus;
$f''(x)$ can only have at most one root,
$f'(x)$ can only have at most two roots, and
$f(x)$ can only have at most three roots.
Part (b)
First, all polynomials are continuous everywhere and differentiable everywhere.
Let $f(x)$ be a polynomial of degree $n$, $f(x)=K_nx^n+K_{n-1}x^{n-1}+...+K_2x^2+K_1x+K_0$.
Suppose that $f(x)$ has $n+1$ roots, $a_{n+1}, a_n, a_{n-1}, ... a_2, a_1$, such that $f(a_n)=f(a_{n-1})=...=f(a_2)=f(a_1)=0$.
Then by Rolle's Theorem there exists some $b_n$ in $(a_{n+1}, a_n)$, $b_{n-1}$ in $(a_n, a_{n-1})$,..., $b_2$ in $(a_3, a_2)$, $b_1$ in $(a_2, a_1)$, such that $f'(b_n)=f'(b_{n-1})=...=f'(b_2)=f'(b_1)=0$.
By applying Rolle's Theorem again there exists some $c_{n-1}$ in $(b_n, b_{n-1})$, $c_{n-2}$ in $(b_{n-1}, b_{n-2})$,..., $c_2$ in $(b_3, b_2)$, $c_1$ in $(b_2, b_1)$ such that $f''(c_{n-1})=f''(c_{n-2})=...=f''(c_2)=f''(c_1)=0$.
If we keep applying Rolle's Theorem we find that eventually there exists two roots $z_2$ in $(y_3, y_2)$ and $z_1$ in $(y_2, y_1)$ such that $f^n(a_2)=f^n(a_1)=0$. Where $y_3, y_2$ and $y_1$ are the roots of $f^{n-1}(x)$.
But $f^n(x)=n!K_n$ and so $f^n(x)>0$ for all $x$ and so the $(n-1)$th derivative of $f(x)$ can only have one root.
Therefore the $(n-2)$th derivative can have at most 2 roots, the $(n-3)$th derivative can have at most 3 roots,..., the 3rd derivative can have at most $(n-3)$ roots, the 2nd derivative can have at most $(n-2)$ roots, the 1st derivative can have at most $(n-1)$ roots and hence $f(x)$, a polynomial of degree $n$, can have at most $n$ roots.
A: I'll do some lower dimensional cases. You can build them up.
A degree-one polynomial has at most 1 real root. This, I take for granted.
Now, we would like to say that a quadratic polynomial as at most 2 real roots. Suppose it has three. Then by the mean value theorem (in fact, just Rolle's Theorem), the derivative of our quadratic has at least 2 real roots. But this is a contradiction, as above I mentioned that a linear polynomial has at most 1 real root. So a quadratic has at most 2 real roots.
Do you see how this continues?
