A Quintic Which is Solvable By Radicals
Let's look at x^5-1. Notice that this polynomial with a=1, f=-1, and all the rest of the quintic coefficients 0 has a real root at x=1.
Notice that this root is easily expressed in terms of a and/or f - for example
x1=a represents x1=1. We'd like to look at the other roots and see that they are also
expressable in terms of the coefficients a=1 and f=-1.
> g:=(x^2+((1+sqrt(5))/2 )*x +1)*(x^2+((1-sqrt(5))/2)*x+1);
I figured out how to write f as a product of two quadratic equations.
I did this via algebra: assume that
= f(x) = (x^2 +ax +1)(x^2+bx +1),
multiply everything out, and then solve for a and b.
You should check this for yourself!
Now I'm going to find the roots of the first factor of g using the quadratic formula (Maple will make this easy)
> solve(x^2+((1+sqrt(5))/2 )*x +1=0,x);
Let's check to make sure these are roots:
Good, these are roots.
Now find a radical expression of the first root (one) in terms of a=1 and/or
Use Maple commands to find the roots of the second factor of g. Define these roots as three and four:
Now list all 5 roots, which are all expressible in terms of the coefficients: