To elaborate on SmileyCraft's answer (they give the answer but not the underlying reason why, which I feel is more important)...
The Roots:
On first contending with the roots issue, which the comments hint you had trouble with: we know $-2i$ and $5$ are roots. We know that since $-2i$ is a root, $+2i$ is also a root because it's the conjugate of $-2i$. Then, since $5, -2i, +2i$ are all roots, we know that the polynomial can be factored as
$$f(x) = a(x - 5)(x - 2i)(x + 2i)$$
up to a leading coefficient $a$ that I'll talk about later. (You assumed $a=1$, and I'll explain why that's wrong in the next section.) It should be clear that this form is correct otherwise, since plugging in any of our roots gives $f(x) = 0$.
Since you wanted real coefficients and such, we note that $(x-2i)(x+2i)=x^2 + 4$, thus giving us the form you posed in the question (again, up to that constant $a$).
Finding $a$:
So now you know that the polynomial has the form
$$f(x) = a(x-5)(x^2 + 4)$$
The reason we include this $a$ is because the leading coefficient of the polynomial doesn't actually affect the roots: it does affect certain aspects of its behavior, but not that one.
(Note of import: that is only true when the polynomial is factored in this form. For example, in the other standard form of a cubic, $f(x) = ax^3 + bx^2 + cx + d$, the coefficient $a$ does affect the zeroes. It's only if in this factored form that the leading coefficient does not affect the roots. Play around on a graphing calculator, shouldn't be hard to convince yourself of these facts.)
We also know that the $y$-intercept of the graph is $(0,25)$. Thus, plug in $x = 0$ and $f(x) = 25$ in the equation above, and then you'll be able to figure out what $a$ actually is.
Plugging in these values, then, we see
$$25 = a(-5)(4) \;\;\; \Rightarrow \;\;\; 25 = a(-20) \;\;\; \Rightarrow \;\;\; a = \frac{25}{-20} = -\frac{5}{4}$$
Thus, the polynomial in its factored form is
$$f(x) = \left( -\frac{5}{4} \right)(x-5)(x^2 + 4)$$
There is a solution using generalized hypergeometric series.
Writing $x = -41^{1/5} t$, the equation becomes
$$ t^5 - s t - 1 = 0 \tag 1$$
where $s = 41^{-4/5}$. There is a series for the root of (1) near $t=1$ in powers of $s$:
$$ \eqalign{t &= \sum _{k=0}^{\infty }{\frac {{3125}^{-k}{16}^{k}\Gamma \left( 2\,k-1/
10 \right) \Gamma \left( 2\,k+2/5 \right) \Gamma \left( 4/5 \right)
\Gamma \left( 3/5 \right) {s}^{5\,k}}{\Gamma \left( -1/10 \right)
\Gamma \left( 4/5+k \right) \Gamma \left( 3/5+k \right) \Gamma \left(
2/5+k \right) k!}}\cr &+1/5\,{\frac {{3125}^{-k}{16}^{k}\Gamma \left( 2\,k+
4/5 \right) \Gamma \left( 2\,k+3/10 \right) \Gamma \left( 6/5 \right)
\Gamma \left( 3/5 \right) {s}^{5\,k+1}}{\Gamma \left( 3/10 \right)
\Gamma \left( k+6/5 \right) \Gamma \left( 4/5+k \right) \Gamma \left(
3/5+k \right) k!}}\cr &-1/25\,{\frac {{3125}^{-k}{16}^{k}\Gamma \left( 2\,k
+6/5 \right) \Gamma \left( 7/5 \right) \Gamma \left( 4/5 \right) {s}^{
5\,k+2}}{\Gamma \left( {\frac{7}{10}} \right) \Gamma \left( k+7/5
\right) \Gamma \left( k+6/5 \right) \Gamma \left( 4/5+k \right) k!}
\Gamma \left( 2\,k+{\frac{7}{10}} \right) }\cr &+{\frac {{3125}^{-k}{16}^{k
}\Gamma \left( 2\,k+8/5 \right) \Gamma \left( 7/5 \right) \Gamma
\left( 6/5 \right) {s}^{5\,k+3}}{125\,\Gamma \left( {\frac{11}{10}}
\right) \Gamma \left( k+8/5 \right) \Gamma \left( k+7/5 \right)
\Gamma \left( k+6/5 \right) k!}\Gamma \left( 2\,k+{\frac{11}{10}}
\right) }}
$$
Best Answer
If you ask me, you did the good thing ignoring the hint and solving it your way. However, there is an "algorithm" for solving these type of problems that the hint is probably referring to. Notice that sum of coefficients is $0$, so $x = 1$ is root. Dividing by $x - 1$ leaves $$6x^4 + 11x^3 - 40x^2 +11x +6 = 0$$ and noticing that $0$ isn't a root and dividing by $x^2$ you get $$6(x^2 + \frac{1}{x^2}) + 11(x + \frac{1}{x}) - 40 = 0$$ Now substitute $y = x + \frac{1}{x}$, and notice $x^2 + \frac{1}{x^2} = y^2 - 2$, you are left with two quadratic equations to solve.