solving polynominals of higher degrees than 2

Galois proved that it can't be done for degree greater than four.

http://mathworld.wolfram.com/CubicFormula.html

http://mathworld.wolfram.com/QuarticEquation.html

pretty complex from the looks of the links
what if you were to take out an x as a common factor so that:
x( ax^2 + bx + c/x) = 0
i realise that according to the above one of the roots would be 0 which is impossible if c/x is to be valid, but frankly i was hoping there would be something more simple that can be done perhaps by bringing the equation to a quadratic state or two

Synthetic Division
You should look up synthetic division. It is not a direct solution like the quadratic formula, but it is a systematic way to solve polynomials with rational roots.

There is no good consistant way to do it symbolically. There are standard forms of these equations and if you can work it into one of these standard forms, you can solve a 3rd or higher polynomial. It is a pain to try to do this in software (I know from experience).

Are you doing this with a computer? Have you considered trying a numeric algorithm (i.e. changing the numbers until you find the right one)?

Equations of degree >2
As Markr commented, there is no general formula for polynomial equations of degree >2.
You should also realize that the quadratic equation is the result of the general case of the method of completing the square.
One method of solving these equations is to determine the overall behavior of the graph using differential calculus.
Rolle's Theorem in particular addresses this. If the value of a continuous function changes sign, a zero exits between the boundries. Newton's method is also helpful.

Equations of degree >2
As Markr commented, there is no general formula for polynomial equations of degree >2.
You should also realize that the quadratic equation is the result of the general case of the method of completing the square.
One method of solving these equations is to determine the overall behavior of the graph using differential calculus.
Rolle's Theorem in particular addresses this. If the value of a continuous function changes sign, a zero exits between the boundries. Newton's method is also helpful.

I solved some 3rd or 4th order equations once using matlab and eiggenvalues. I forgot what I did but it is worth a try. Also I think there is a free equivalent of matlab on the web.