How do I choose the correct range of numbers to find the zeros...
Trial and error, We know from Descartes rule of signs that there is exactly one positive root, and 0 or 2 negative roots. Also we know that for large positive x f(x) is positive, and "large" negative x f(x) is negative from the sign of the x^3 term.
So evaluate f(0) it should be -ve, take positive steps untill f changes sign (in this case between 1 and 2). Then as the sum of the roots is equal to -the coefficient of x^2, we conclude that the negative roots if they exist, are >=-5.
There are almost certainly other methods of establishing a suitable range.
RonL
A rather simple method (that computer programs use) is the "bisection method".
First find an interval that has only one zero and work in that interval.
Example,
$\displaystyle y=x^3+x-1$
We can to find the zero of this. That is to solve,
$\displaystyle x^3+x-1=0$
You can draw a graph and see where it intersects only once and work in that interval. Look below.
So work on interval,
$\displaystyle [0,2]$
Now, chose the midpoint $\displaystyle x=1$ does it work?
$\displaystyle 1^3+1-1=1>0$ too large.
Meaning the solution must be in the interval.
$\displaystyle [0,1]$.
Now chose the midpoint there $\displaystyle x=.5$ does it work?
$\displaystyle (.5)^3+(.5)-1=-.375$ too little.
Meaning the solution must be in the interval.
$\displaystyle [.5,1]$
Choose the midpoint there $\displaystyle x=.75$ does it work?
$\displaystyle (.75)^3+(.75)-1=.1719$ too large.
Thus, the solution is on the interval
$\displaystyle [.75,1]$
And so on.