Why is it so hard to find the roots of polynomial equations?

The idea is basically:

Any monic polynomial can be factored as $f(x) = \prod (x - a_i)$, where $a_{1,\dots,n}$ are the roots of the polynomial.

Now if we expand such a product:

$(x - a_1)(x - a_2) = x^2 - (a_1 + a_2)x + a_1a_2$ $(x - a_1)(x - a_2)(x - a_3) = x^3 - (a_1 + a_2 + a_3)x^2 + (a_1a_2 + a_1a_3 + a_2a_3)x - a_1a_2a_3$

And so on. The pattern should be clear.

This means that finding the roots of a polynomial is in fact equivalent to solving systems like the following:

For a quadratic polynomial $x^2 - px + q$, find $a_1,a_2$, such that

$p = a_1 + a_2$

$q = a_1a_2$

For a cubic polynomial $x^3 - px^2 + qx - r$, find $a_1,a_2,a_3$, such that

$p = a_1 + a_2 + a_3$

$q = a_1a_2 + a_1a_3 + a_2a_3$

$r = a_1 a_2 a_3$

And similarly for higher degree polynomials.

Not surprisingly, the amount of "unfolding" that needs to be done to solve the quadratic system is much less than the amount of "unfolding" needed for the cubic system.

The reason why polynomials of degree 5 or higher are not solvable by radicals, can be thought of as: The structure (symmetries) of the system for such a polynomial just doesn't match any of the structures that can be obtained by combining the structures of the elementary operations (adding subtracting, multiplication, division, and taking roots).

When you try to solve a degree $n$ equation, there are $n$ roots you have to find (in principle) and none of them is favoured over any of the others, which (in some metaphorical sense) means that you have to break an $n$-fold symmetry in order to write down the roots.

Now the symmetry group of the n roots becomes more and more complicated the larger $n$ is. For $n = 2$, it is abelian (and very small!); for $n = 3$ and $4$ it is still solvable (in the technical sense of group theory), which explains the existence of an explicit formula involving radicals (this is due to Galois, and is a part of so-called Galois theory); for $n = 5$ or more this group is non-solvable (in the technical sense of group theory), and this corresponds to the fact that there is no explicit formula involving radicals.

Summary: The complexity of the symmetry group of the $n$ roots leads to a corresponding complexity in explicitly solving the equation.

For a different take, some practical problems are discussed in Wilkinson's classic article The Perfidious Polynomial. If you can't access it, check what Wikipedia has to say on the subject.