To find number of roots in interval $(1,2)$ To find number of roots in interval $(1,2)$
$$f(x)= 3x^2 - 12x + 11 + \frac{x^3-6x^2+11x-6}{5} $$
Now $f(1)f(2)<0$. So it has one root. But how do I know whether it has two or three or whatever. Is there a general rule for these questions where we cannot explicitely find out the roots.
 A: In general, there isn't a lot you can do when you are looking for roots of a function. However, in your case, $f$ is a polynomial, and its derivative is a second-degree polynomial.
That means you can easily find the roots of $f'$, meaning you can plot a rough sketch of $f$. You will discover that one local extrem (the minimum) of $f$ is just slightly right of number $2$. Because $f\to\infty$ as $x\to\infty$, you therefore know that one more root must be to the right of this minimum, therefore also to the right of $2$.
Thhe local maximum is far in the negative. Apply similar logic as before.
A: Edit : My answer is about polynomials only.
You have Descartes'rule of signs (not exactly what you want but a first result) :
https://en.wikipedia.org/wiki/Descartes'_rule_of_signs
And if you want to go further you have Sturm's theorem which allows you to count the number of real roots of a real polynomial in an interval, see :
https://en.wikipedia.org/wiki/Sturm's_theorem
Of course, in your particular case you could explicitely compute the derivative, its sign and then draw the curve by hand to check if you have one or three roots...
A: A general answer is Sturm's algorithm. It is based on a sequence of Euclidean divisions. Here is a short description:
Consider the sequence of polynomials (Sturm's sequence) defined by:


*

*$f_0(x)=f(x)\enspace(\deg f=n),\quad f_1(x)=f'(x),$

*$f_{i-1}(x)=q_i(x)f_i(x)\boldsymbol{\color{red}{-}} f_{i+1}(x)$
and  for all $a\in\mathbf R$ which is not a root of $f(x)$, set 
$$\sigma(a)=\quad \text{number of sign changes in the sequence}\quad
\bigl(f(a),f_1(a),\dots, f_n(a)\bigr)$$
($0$ is not counted as a sign change).
The the number of real roots between $a$ and $b\enspace a<b$ is equal to
$$\sigma(a)-\sigma(b).$$
An elementary solution:
Set $g(x)=x^3-6x^2+11x-6$. Then $f(x)=g'(x)+\frac15g(x)$, hence


*

*$f'(x)=g''(x)+\frac15g'(x)=6x-12+\frac15(3x^2-12x+11)$

*$f''(x)=g'''(x)+\frac15g''(x)=6+\frac15(6x-12)=\frac65(x+3)$
Hence $f'(x)$ has a minimum at $x=-3$ and $f'(-3)=6+\frac15(27+36+11)>0$, which implies the cubic polynomial $f$ has only $1$ real root.
