# How do computer programs find roots of high-degree polynomials?

My question is motivated by curiosity about the optimization of high-degree polynomial functions.

Let's say your experiment data are modeled by a non-trivial 15th degree polynomial. Taking the derivative of that function would leave you with a 14th degree polynomial. As far as I know, there is simply no way to "manually" find the roots (and therefore critical points) of such a polynomial (please correct me if this is wrong).

However, it seems (Computing roots of high degree polynomial numerically.) that certain programming languages (MATLAB, Mathematica, etc...) have a pretty easy time finding roots of polynomials of much higher degree (up to degree 2000 in less than a minute, apparently).

How do they find those roots, and are they exact values or approximations?

• What do you mean by "exact values or approximations"? Computers don't usually output exact values (and the most meaningful way to refer to many number is "the root of this polynomial in this interval"). Commented Jan 14, 2016 at 22:25
• I think he means symbolic computation. Commented Jan 14, 2016 at 22:28
• Since there is no closed form solution for determining the roots of polynomials of degree 5 or higher, MATLAB and other programs will give only approximations of the roots of a polynomial unless it fits into special classes where the roots are known (e.g. $z^5-1 = 0.$). Read about Abel's Impossibility Theorem at mathworld.wolfram.com/AbelsImpossibilityTheorem.html for reference.
– Joel
Commented Jan 14, 2016 at 22:46
• How does Matlab do it numerically (i.e., approximately)? Look at the code: edit roots. Commented Jan 14, 2016 at 23:25
• @Joel: nonetheless, computation in the field of algebraic real numbers is decidable. Abel's result says that you can't give a closed solution for polynomials of degree greater than 5 using radicals, but that doesn't mean you can't give precise computable descriptions of roots. Google for "root isolation" to learn more. (Or see mathoverflow.net/questions/116069/…) Commented Jan 14, 2016 at 23:28