**Continued Examination of the Roots of f(x)=x^5-2*x^3-8*x-2, a Quintic Not Solvable by Radicals.**

How could we prove that there are exactly 3 real roots?

Well, first take the derivative of f(x)

`> `
**fprime:=diff(f(x),x);**

Let's solve for critical points:

`> `
**solve(fprime=0,x);**

So, we have two real critical points. Then f(x) has one relative maximum

and one relative minimum. Notice from below that the values of f(x) change sign between -2 and -1, between -1 and 0, and between 2 and 3.

`> `
**f(-2);f(-1);**

`> `
**f(-1);f(0);**

`> `
**f(0);f(3);**

Since the values change sign's 3 times, and there is one max and one min,

there must be exactly 3 roots, so let's find them:

`> `
**fsolve(f(x)=0,x=-2..-1);**

`> `
**fsolve(f(x)=0,x=-1..0);**

`> `
**fsolve(f(x)=0,x=2..3);**

These roots certainly don't look like nice numbers which could be expressed as combinations of the coefficients of f(x) (which were integers), but that is not a proof, especially since we are allowed to take square roots, cube roots, etc. of the coefficients.

Abel began to examine which polynomial's have solutions by radicals, but did not complete his work before his death. Galois was the first to determine a condition for a polynomial to be solvable by radicals.

8.4.6. Theorem (Galois). Let f(x) be a polynomial over a field K of characteristic zero. The equation f(x) = 0 is solvable by radicals if and only if the Galois group of f(x) over K is solvable.

While this theorem is beyond the scope of the course, we can still get an idea of how you

would finish off the above proof that our f is not solvable by radicals (just read the following for the big picture idea - not for details):

By Theorem 8.3.10 there exists a splitting field F for f(x) with F subset C. The polynomial f(x) is irreducible by Eisenstein's criterion, and so adjoining a root of f(x) gives an extension of degree 5. By the fundamental theorem of Galois theory, the Galois group of f(x) over Q must contain a subgroup of index 5, so since its order is divisible by 5, it follows from Cauchy's theorem that it must contain an element of order 5. By Proposition 8.1.4, every element of the Galois group of f(x) gives a permutation of the roots, and so the Galois group is easily seen to be isomorphic to a subgroup of S_5 . This subgroup must contain an element of order 5, and it must also contain the transposition that corresponds to the element of the Galois group defined by complex conjugation. Therefore, by the previous lemma, the Galois group must be isomorphic to S_5 . Applying Theorem 8.4.6 completes the proof, since S is not a solvable group.