Let $f(x) =3x^5 - 15x +5$. By Eisenstein’s Criterion we can show that $f(x)$ irreducible over $\mathbb{Q}$(Since $5 \nmid 3, 5 \vert -15,5$ and $5^2 \nmid 15$). Since $g$ is continuous and $ f( -2 ) = -61, f( -1 ) = 17 ,f( 0 ) = 5,f( 1 ) = -7,f( 2 ) = 71$. So by intermediate value theorem we can say that $f$ has three real roots. Clearly $f'(x) > 0 $ for all $x > 2$ and $f'(x) < 0$ for all $x < -2$. So $f(x)$ is monotone and so $f$ does not have zeroes in these regions. Suppose $f(x)$ has $4$ real zeroes then by using Rolle's theorem we can say that $f'(x)$ should contain at least $3$ zeroes between the roots of $f$. But $f'(x) = 15(x^4 -1)$ does not have $3$ real zeroes. So $f$ has two other complex roots. Let $K$ be the smallest subfield of complex numbers containing $\mathbb{Q}$ and $5$ roots of $f(x)$. Then using fundamental theorem of Galois theory we can say that $Gal(K/\mathbb{Q}) \approx S_5$, the symmetric group of five letters. Since $S_5$ is not solvable, by a theorem of Galois we can conclude that $f(x)$ is not solvable by radicals.
That is each zero of the polynomial $f(x)$ cannot be written as an expression involving elements of $\mathbb{Q}$ combined by the operations of addition, subtraction, multiplication, division, and extraction of roots.
How do the roots of $f$ look like? We have information about the location of real roots but I think that information may not help in finding some expression for roots. Precisely, my question is that that does there exist a series, continued fractions, or some integral which represent the roots of $f(x)$?.
It turns out one of the roots is $$x_0=\frac13 \ _4F_3\left(\frac15,\frac25,\frac35,\frac45,\frac12, \frac34,\frac54,\frac{625}{20736}\right)= 0.334166718886923432353275582781735638464962574956948610…$$ using “bring radical” link’s hypergeometric definition. Here are properties which will give you an integral, summation, and continued fraction representations here. You will have to either look up the other roots or solve for them, and derive your own values using the link. The integral representations are integrals of progressively simpler integrands of the Hypergeometric function like this:
$$x_0=\frac1{3Γ\left(\frac45\right)}\int_0^\infty \frac{\ _3F_3\left(\frac15,\frac25,\frac35,\frac12,\frac34,\frac54,\frac{625t}{20736}\right) }{e^t\sqrt[5]t}dt=\frac13\sum_0^\infty \frac{Γ(5n+1)}{405^n Γ(2(2n+1))n!}$$
sum proof.
Here are the other solutions of which I am unsure how to derive. The complex solution is also the conjugate of the one posted. Notice the similar hypergeometric arguments:
$$\mathop =_{\text{conjugate}}^{\text{with}}-.080931889…\pm 1.5063232344…i$$
Other real solutions:
Top=$-1.56912279…$, Bottom=$1.39681985…$
I would write out the rest of the roots the same way and using the continued fraction formula in the Wolfram functions site link, but these are too complicated. If someone else wants to write them out, be my guest. The same goes for the nome representation. I hope this helps. Please correct me and give me feedback!