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
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.