## How the general solution of polynomials, of any degree greater than 2, is extremely difficult to compute.

There are certain misconceptions that exist about Math, and one of them could be, that if a random system of equations is written down, the Algebraic solution of those equations is at hand. In fact, equations can arise easily, which are known to have numerical answers, but for which the exact, Algebraic (= analytical) answer, is nowhere in sight. And one example where this happens, is with polynomials ‘of a high degree’ . We are taught what the general solution to the quadratic is in High-School. But what I learned was, that the general solution to a cubic is extremely difficult to comprehend, while that of a 4th-degree polynomial – a quartic –  is too large to be printed out. These last two have in fact been computed, but not on consumer-grade hardware or software.

In fact, I was taught that for degrees of polynomials greater than 4, there is no general solution.

This seems to fly in the face of two known facts:

1. Some of those equations have the full number of real roots,
2. Computers have made finding numerical solutions relatively straightforward.

But the second fact is really only a testament, to how computers can perform numerical approximations, as their main contribution to Math and Engineering. In the case of polynomials, the approach used is to find at least one root – closely enough, and then, to use long division or synthetic division, to divide by (1 minus that root), to arrive at a new polynomial, which has been reduced in degree by 1. This is because (1 minus a root) must be a factor of the original polynomial.

Once the polynomial has arrived at a quadratic, computers will eagerly apply its general solution to what remains, thereby perhaps also generating 2 complex roots.

In the case of a cubic, a trick which people use, is first to normalize the cubic, so that the coefficient of its first term is 1. Then, the 4th, constant term is examined. Any one of its factors, or the factors of the terms it is a product of, positively or negatively, could be a root of the cubic. And if one of them works, the equation has been cracked.

In other words, if this constant term is the product of square roots of integers, the corresponding products of the square roots, of the factors of these integers, could lead to roots of the cubic.

(Updated 1/12/2019, 21h05 … )