Expert: Ahmed Salami - 4/27/2010

Question
x^3+2x^2+10x-20=0

Answer
Hi N,
We could use the Newton's method to obtain a solution. The discriminant test shows that the equation has one real root and two complex (conjugate) roots.
Given a function f(x) and its derivative f'(x), we begin with a first guess x(0). Provided the function is reasonably well-behaved a better approximation x(1) is given by
x(1) = x(0) - f[x(0)]/f'[x(0)]
Now,
f(x) = x^3 + 2x^2 + 10x - 20
f'(x) = 3x^2 + 4x + 10
Starting with, say, x(0) = 1
f[x(0)] = 1^3 + 2(1^2) + 10(1) - 20
       = 1 + 2 + 10 - 20
       = -7
f'[x(0)] = 3(1^2) + 4(1) + 10
        = 3 + 4 + 10
        = 17
x(1) = x(0) - f[x(0)]/f'[x(0)]
    = 1 - (-7/17)
    = 1 + 7/17
    = 1 + 0.41
    = 1.41

You continue in this manner using
x(2) = x(1) - f[x(1)]/f'[x(1)]
and so on until f[x(n)] is as close to zero as you desire. This would also be when you see no noticeable difference (at least according to your desire) between x(n-1) and x(n).
The solution is actually 1.36881 to five decimal places.

Regards