Proving a polynomial has x amount of zeros I am new to this thread so sorry if I violate any rules or whatever, but anyway in Calculus right now we are doing stuff related to Fermat's Theorem, Rolle's Theorem, and Intermediate value theorem. I am very confused by Rolles and Fermats and totally don't understand them I try looking online just to be more perplexed than resolved. In our homework it is full of problems like this:

Prove $y = x^3-6x^2+12x-8$ has at most two zeros.
Prove $f(x)=x^5+5x^3+45x+9$ has exactly one zero.

I am just totally lost on what to do and how to begin I really don't understand how to prove these with the theorems given.
Thanks for the help in advance.
 A: The trick is that between any two zeros of your function the derivative must have at least one zero. 
$1)$  Prove $$ y=x^3−6x^2+12x−8$$ has at most two zeros.
If a polynomial has three zeros, then its derivative must have at least two zeros due to Rolle's Theorem. Taking derivative of  $$ y=x^3−6x^2+12x−8$$ We get   $$ y=3x^2−12x+12 =3(x-2)^2$$
Which has only one zero. 
$2)$ Prove $$f(x)=x^5+5x^3+45x+9$$ has exactly one zero.
It has one zero because its degree is odd so it changes sign from negative to positive as as x goes from $- \infty $ to $\infty$
Taking derivative of   $$f(x)=x^5+5x^3+45x+9$$  we get$$5x^4+15x^2+45$$ which is always positive and has no zeros. 
Thus the function is increasing and has only one zero.  
A: i) $f(x) = x^3-6x^2+12x-8$.
Suppose that $f$ has $3$ zeroes in $a<b<c$.
How $f(a)=f(b)=f(c)=0%$ then, by Rolle's Theorem, there is a zero of $f'(x)$ in interval $(a, b)$ and another zero in $(b, c)$. In other words, $f'(x)$ has two zeroes. But $f'(x) = 3x^2-12x+12 = 3(x-2)^2$ has only one zero. Contradiction!
Therefore $f$ has at most two zeroes.
ii) $f(x) = x^5+5x^3+45x+9$
How $f$ has odd degree, then $f$ has at least one zero.
Note that $f'(x) = 5x^4+15x^2+45 > 0$ for all $x\in\mathbb{R}$.
Suppose that $f$ has two zeroes in $a<b$. By Rolle's Theorem, should exists $c$ with $a<c<b$ such that $f'(c)=0$. Contradiction!
Therefore $f$ has exactly one zero.
