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).
Galois Theory assigns, to each polynomial, a mathematical structure called a group. A polynomial is solvable in radicals (that is, you can write down its roots in terms of its coefficients, the 4 arithmetical operations, and square roots, cube roots, etc.) if and only if the corresponding group is a "solvable" group. The definition of solvable group won't mean much to you if you haven't done a course in group theory; there should be a sequence of groups, starting with the trivial group and ending with the group corresponding to the polynomial, such that each group in the sequence is a "normal" subgroup of the next group, and the "quotient" of each group by the previous group is "commutative".
What's more, if the group corresponding to the polynomial is solvable, then this sequence of in-between groups, together with the quotient groups, can be used to construct the formula for the roots of the polynomial.
To expand this into something you could actually use to determine whether a polynomial is solvable in radicals, and to solve it if it is, takes a semester of advanced undergraduate level mathematics. Get yourself a good text on Galois Theory (assuming you have already done courses on Linear Algebra and an introductory Abstract Algebra course - if not, you'll have to study those first), read it, and enjoy.
Best Answer
If $p(x)=x^5+Ax^4+Bx^3+Cx^2+Dx+E$ has root $a$, then
$$p(x)=(x-a)(x^4+A'x^3+B'x^2+C'x+D')$$
where
$$\begin{align} A&=A'-a\\ B&=B'-aA'\\ C&=C'-aB'\\ D&=D'-aC'\\ E&=-aD' \end{align}$$
We can easily unwind the first two equations to
$$\begin{align} A'&=A+a\\ B'&=B+aA'=B+aA+a^2 \end{align}$$
and the last two (provided $a\not=0$) to
$$\begin{align} D'&=-E/a\\ C'&=(D'-D)/a=-(E/a^2+D/a) \end{align}$$
Alternatively you could keep going in the "forward" direction with
$$\begin{align} C'&=C+aB'=C+aB+a^2A+a^3\\ D'&=D+aC'=D+aC+a^2B+a^3A+a^4 \end{align}$$
Either way, if you have a (good) numerical approximation to $a$, you can compute (approximately) the coefficients $A'$, $B'$, $C'$, and $D'$ of a quartic and then apply your favorite root-finding algorithm to that polynomial. If you succeed in getting a second root, you can repeat the basic idea outlined here, to factor that root out of the quartic leaving a cubic, etc. (Once you get things down to a quadratic, the quadratic formula will finish things off.)
Note, however, that a quintic may have only one real root, so some caution is called for if you're trying to make this mindless and mechanical; in particular, Newton-Raphson is not going to help with a quartic that has no real roots.