# Solutions of Polynomials

For a general quadratic equation $ax^{2}+bx+c=0$ the solutions are given by

$$x=\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}$$ and

$$x=\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}$$

– it may not necessarily be the most efficient method of calculating the solutions to a quadratic since in some cases factorisation is much faster, but it gets the job done and if you’re feeling a little bit lazy and can’t be bothered to factorise a quadratic then this equation practically does the job for you.

Naturally at some point in my mathematics studies (probably during my A Levels) I asked myself the question – does a formula exist that I can use to solve cubic equations? Naively (I later realised), I tried to find a formula using similar techniques to how the quadratic formula is derived. Of course I failed (but it’s amazing how much you can learn even when you fail). However, formulae do exist that give the solutions to a general cubic equation (equation of general cubic).

It is worth noting that any cubic $ax^{3}+bx^{2}+cx+d=0$ can be transformed into a cubic of the form $x^{3}+px+q=0$ by using the transformation $x=y-\dfrac{b}{3a}$. The solutions to this new cubic (given here without derivation) are

$$x=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}$$

$$x=j\left(\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\right)+j^{2}\left(\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\right)$$

and

$$x=j^{2}\left(\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\right)+j\left(\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\right)$$

where $j=\mathrm{e}^{\frac{2\pi i}{3}}$

The big problems here are that these formulae are much bigger than the formulae for the solutions to a quadratic, it certainly doesn’t roll off the tongue quite as easily, and also it is very cumbersome to work with. Nonetheless we have our equations, whether or not they are easy to work with may not be relevant – the fact that they exist is what matters here.

So what about quartic equations $ax^{4}+bx^{3}+cx^{2}+dx+e=0$ Are there any formulae for the solutions of a quartic? Yes there are … but the expressions are extremely long; you can see them at the wikipedia page on quartic equations. Again, we have the formulae and we at least know that they exist so we can be satisfied with that for now. Since general quadratic, cubic and quartic polynomials have closed formulae to describe their solutions they are said to be “solvable by radicals”

But things stop there. There are no such formulae for quintic (degree 5) polynomials or indeed any polynomial of degree greater than or equal to 5. It’s not that the formulae have yet to be found – they just don’t exist. The reason why they don’t exist takes us into Galois Theory where it can be proved that general polynomials of degree 5 and greater are not solvable by radicals – in other words, no formulae exist that give the solutions to general polynomials of degree 5 and above.

This is not to say that these polynomials cannot be solved at all, nor that their solutions don’t exist – it is quite easy to construct a quintic polynomial where all of the solutions can easily be found by (educated or uneducated) guesswork – but not using a systematic method or any series of ‘standard formulae’. To solve a general quintic (or higher degree) polynomial we will usually have to resort to numerical methods unless we have some additional knowledge about the polynomial or unless we are lucky enough to be able to guess some solutions!

Comments closed.