p= - 68; q = -24 and the roots of the equation under reference are
-1, -1, -2, -2 and 6.

EXPLANATION:

FIRST METHOD ( On the basis of the solution submitted by Old Original Oskar! and comments by Frederico Kereki)

The sum of the five roots is 0.

Accordingly, the roots are a, a, b, b, and c=-2(a+b).

The sum of the pairwise products of the roots is:

a^{2}+b^{2}+4ab+2(a+b)c = -3(a^{2}+b^{2})-4ab = -23; so that:
3a^{2}+4ba+(3b^{2}-23)= 0.

Solving this equation for a gives a=(-4b+/-sqrt(276-20b^{2}))/6, which produces integer solutions only whenever, b=+/-1 (then a=-2) or b=+/-2 (then a=-1)

Since a and b are symmetric, the feasible solutions are :

(a,b) = (-2, -1); giving, c = -2(a+b) = 6.

Consequently, the required roots are: -2, -2, -1, -1 and 6.

This checks out since:

(x+2)^{2}(x+1)^{2}(x-6)

= x^{5} - 23x^{3} -66x^{2}- 68x - 24; so that p = -68 and q = -24 in terms of the problem under reference.

Comparing the coefficients corresponding to the various exponents of x in both sides of the aforementioned relationship, we obtain:

(i) q/(ab)^2 = 2(a+b)

(ii) (-2q(a+b)/((ab)^2)+ b^2 + 4ab + a^2 = -23

(iii) -2ab(a+b) +(b^2+4ab+a^2)*(q/((ab)^2)) = -66

(iv) (ab)^2 - 2ab(a+b)*(q/((ab)^2)) = 1

By (i); q = 2*(ab)^2 *(a+b)

Substituting the above value of q in (ii) and simplifying, we obtain:
ab = (3*(a+b)^2 - 23)/2 -----------------(#)

Substituting the above value of ab in (iii) and simplifying, we obtain:

5*(a+b)^3 - 23*(a+b) = -66

It can easily be established with a little trial and error that the above equation is satisfied whenever a+b = -3 and substituting this in (#), we obtain ab = 2

Solving for (a+b= -3, ab=2) we obtain (a,b)=(-2,-1),or(-1.-2)
Without any loss of generality, we take (a,b)=(-2,-1) giving:
q = 2*((ab)^2)*(a+b) = -24 and
p = (ab)^2 - 2ab(a+b)*(q/((ab)^2))= -68