Quintic polynomial with only positive roots

Question

The Abel-Ruffini theorem states that the solutions to a general polynomial equation of the form$$x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0$$ have no algebraic expression in terms of the coefficients of the polynomial. Suppose that we know beforehand that such a polynomial with integer coefficients has only real roots that are all positive and that the polynomial is irreducible. Could there be some way to determine an algebraic expression for the roots in terms of the coefficients?

Answer 1

As @eyeballfrog points out in the comments, the equivalent question asks whether a monic quintic with integer coefficients and real roots is solvable using the four fundamental operations as well as roots. The answer is no. Consider the following counterexample: $$f(x) = x^5-6x^3+6x-2$$

• We have $f'(x) = 5x^4-18x^2+6$ which is a biquadratic polynomial. One can explicitly find the $4$ roots which correspond to critical points of $f$. Evaluating $f$ at these points yields alternating signs of output, proving that $f$ indeed has $5$ real roots. (You may verify this in a graphing calculator)
• The polynomial is irreducible by Eisenstein's criterion. Thus, the Galois group $G$ of the splitting field of $f$ is a transitive subgroup of $S_5$.
• The discriminant of $f$ is equal to $126032$, which is not a perfect square. Thus, $G$ is not contained in $A_5$. This means that $G$ is either $S_5$ or a subgroup of $F_5$.
• We know that $G$ is contained in $F_5$ iff the sextic resolvent has rational roots. In our case, the sextic resolvent is given by: $$g(y) = y^6 + 48y^5 + 144y^4 - 20280y^3 -191232y^2 + 2014960y + 21159072$$ which has no rational roots by the rational root test. It follows that the Galois group is $S_5$.
• Since $S_5$ is not a solvable group, we cannot solve for the roots of $f$ using the four fundamental operations and $n$th root operations.