Finding number of roots using Rolle's Theorem, and depending on parameter I need to count the number of real solutions for $ f(x) = 0 $ but I have an $m$ in there.
$$ f(x) = x^3+3x^2-mx+5 $$
I know I need to study $m$ to get the number of roots, but I don't know where to begin. Any suggestions?
Taking the derivative of the function: $$f'(x)=3x^2+6x-m$$
If the function $f$ has two roots with $x=a$ and $x=b$, then $$f(a)=0=f(b)$$ 
Since $f(x)$ is continuous and differentiable, by Rolle's theorem, we have that there exists at least one  $c \in (a,b)$ such that $$f'(c)=0$$
Namely, $$f'(c)=3c^2+6c-m=0$$
And here is where you begin to consider how the roots exist depend on the parameter.
You can use the discriminant of the cubic function to find the nature of the roots. The general cubic equation is

$$ax^3+bx^2+cx+d=0$$

so in your case

$$a=1,\ b=3,\ c=-m,\ d=5$$

In general, the discriminant is

$$\Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$

You can substitute in your values and get a cubic expression in $m$. Then your equation has


three distinct real roots, if $\Delta&gt;0$.
one or two distinct real roots, if $\Delta=0$.
one real roots (and two complex roots), if $\Delta&lt;0$.


To really finish your answer, you would need to distinguish one or two real roots in the second case. You may also want so solve for $m$ in each (in)equality, to give simpler conditions on $m$. Since you asked for "suggestions" and "where to begin," I'll leave it at that.

Note that this did not need Rolle's theorem. Do you really need to use it?