# Polynomials - using Newton's method, or not?

I have to find a root of polynomial of degree $$n\ge2$$. I need to write code to calculate the root for different values of $$n$$. Only 1 real positive solution is needed.

I can use general Newton's method for all degrees, or only for 5th+, and for 2,3 and 4 I can use algebraic formulas that solve quadratic, cubic and quartic equations. Since algebraic formulas need to use sqrt, acos and similar functions - does it make sense at all to use algebraic formulas - or is it better to use Newton's method for all degrees? I guess the latter will actually be faster?

This is my equation:

$$\frac{1}{(b_1x + 1)} + \frac{1}{(b_2x + 1)} + ... + \frac{1}{(b_nx + 1)} -k = 0$$

and constraints:

$$1 \leq k0$$

e.g. for $$n=3$$:

$$b_1 b_2 b_3 (-k)x^3 +\left(b_1 b_2+b_3 b_2+b_1 b_3- b_1 b_2 k-b_3 b_2 k-b_1 b_3 k\right)x^2 + \left(2 b_1+2 b_2+2 b_3-b_1 k-b_2 k-b_3 k\right)x + 3-k =0$$

which can be written as

$$\sum_{i=0}^n(n-i-k)\binom{\{b_1,...,b_n\}}{i}x^i=0.$$

• Is the question how to find the roots of a polynomial the fastest? – Reinstate Monica Jul 23 '19 at 18:40
• Yes - is it faster to use Newton's method for 2nd, 3rd and 4th degree polynomial or is it faster to use quadratic, cubic and quartic equations? – Bojan Vukasovic Jul 23 '19 at 18:42
• @Apass.Jack I updated equation to be equal to 0 – Bojan Vukasovic Jul 24 '19 at 12:44
• These are all integral polynomials, might be counting something, it possible you have rational roots in which case use the rational roots theorem and be done precisely and quickly. – Algeboy Jul 24 '19 at 12:44
• @Algeboy I need only one positive real solution. I will update question. – Bojan Vukasovic Jul 24 '19 at 12:45

The sum goes to infinity at x = -1 / $$b_i$$. Between $$b_i$$ and $$b_{I+1}$$ it is continuous. So each of these intervals contains a point wher the sum of fractions equals k. Go from there.