finding the zeros of a polynomial that have irrational zeros $x^3-5x^2+x+8=0$
I know that the zeros are approx. $-1.07$, $1.72$, and $4.34$ by looking by using a graphing calculator, but how do I find the zeros without? Rational roots theorem does not work here!
 A: Use the bisection-method or the newton-method.
Also possible is the regula falsi.
Idea of the bisection-method :
Let a and b be numbers, such that f(a) < 0 and f(b) > 0 or vice versa.
Then, take m := $\frac{a+b}{2}$.
If f(m)=0, you are done. Otherwise choose a or b, such that the signs differ 
and continue the process.
The newton-method iterates as follows :
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\ ,$$ 
with some starting value $x_0$
Regula falsi is similar, but uses 
$$\frac{f(b)-f(a)}{b-a}$$
instead of f '(x).
Here a,b are numbers with f(a) < 0 and f(b) > 0 or vice versa.
Regula falsi does not need the derivates, but converges slightly slower than
the newton-method.
A: As you note, there is no rational root. I'm afraid you have to either use a numerical approximation or use the general solution of cubic. There are many methods to find the general solution in closed form. Take a look at http://en.wikipedia.org/wiki/Cubic_function
A: To count the number of roots in a given interval, we can use the Sturm Sequence.
The Sturm Sequence for $f(x)=x^3-5x^2+x+8$ is the sequence of polynomals
$$
x^3-5x^2+x+8\\
3x^2-10x+1\\
\frac{44}{9}x-\frac{77}{9}\\
\frac{117}{16}
$$
If $N(x)$ is the number of sign changes in the Sturm Sequence at $x$, then for $a<b$, $N(a)-N(b)$ is the number of real roots of $f$ in $(a,b)$.
As $x\to-\infty$, the signs are "$-,+,-,+$"; that is, three sign changes: $N(-\infty)=3$.
At $x=0$, the signs are "$+,+,-,+$"; that is, two sign changes: $N(0)=2$.
As $x\to+\infty$, the signs are "$+,+,+,+$"; that is, no sign changes: $N(+\infty)=0$.
Thus, there are three real roots of $f$, one negatve and two positive.
We can use the Sturm Sequence to find intervals with a single root, then use bisection or Newton to find that root.
