# How to solve a polynomial $P(x) = 0$

At the moment we just learnt the factor theorem of polynomials and how if $x-a$ is a factor of $P(x)$, then $P(a) = 0$. We're then taught to find the roots of a polynomial its best to check the factors of it's constant and see what happens when it's factors are plugged into the polynomial. If you get 0, good job and divide the polynomial, and repeat the process. I feel like this is a bit like being taught factorisation for quadratics, its a good and all, until you learn the quadratic formula and it can solve all quadratics!

How would I find the roots of any non constant, single variable polynomial?

I'm guessing there's unlikely to be an equation for such a task, and I'm uncertain of how it would be done. If it can be done at all.

Thanks

• It is not possible. – Vim Apr 2 '16 at 7:20
• In general, there is no closed form solution for polynomials of degree >2. Of course there are some special cases with closed form solutions but those are few and far in between. – Mark Apr 2 '16 at 7:23

There is not a General method to solve an arbitrary polynomial. There are some cryteria however.

First: Not every polynomial seen in an arbitrary field has roots in that field, e.g. $x^2-2$ seen in $\mathbb{Q}$ has no roots $x^2+1$ seen in $\mathbb{R}$ has no roots. But in the complex numbers this is true, every polynomial of degree n has exactly n roots.

The polynomials which don't have any roots can be however reduced in multiplication of polymomial of lower degree. If a polynomial cannot be split is called irreducible. There are good criteria to say if a pol. Is irredicible in a certain field.

Now let's talk about resolutive formulas to find roots of a pol. There are such formulas For polynomial of degree 1,2,3 and 4, but not from 5,6,... Why there aren't such formulas? Well, it can be proved that cannot exiat such formulas for high degree polynomial, and this can be seen using some group theory, and in particolar studying some properties of the Group $A_n$ which, from n=5,... Is non solvable.

It cannot be done at all. (Abel-Ruffini theorem) It can only be done on polynomial with degree less than 5, but at degree 3 (Cubic function) the formula is very lengthy already, so maybe "trial and error" is better.

Sometimes, "casus irreducibilis" happen with polynomial with degree 3.

Or, if you have a computer, you can use Newton's method to approximate a root.

• Instead of there being an equation to solve all polynomials, could there be a certain method? What I mean is with calculus, to find a gradient at a point of a function there is not an equation per say to find it, but there is a method by taking the derivative to get it. – frog1944 Apr 2 '16 at 8:53
• @frog1944 You see, some polynomial doesn't have root that can be represented by elementary functions, so there is not any certain method. I think what students have to do have the free coefficient quite small, so rational root theorem is enough. Some roots can be easily found by simple tricks: If free coefficient = 0, then it has a root 0. If the sum of coefficients = 0, then it has root 1. If alternative sum of coefficient = 0, then it has root -1. – user202729 Apr 2 '16 at 9:05

First of all there are formulas to compute the roots of polynomials up to degree $4$ there are no formulas that work across the board for polynomials of degree $5$ and beyond. The most common approach is numerical and there are fairly good algorithms.

Here is a small trick I like to use and it works for me most of the time. Suppose you are given a polynomial $a_0+a_1x+\ldots+a_ix^i+\ldots+a_nx^n$. Then suppose it has a rational root $a/b$ with $gcd(a,b)=1$ then $a$ divides $a_n$ and $b$ divides $a_0$. This trick can be used to find rational roots if at all you are stuck with doing this problem by hand.