<s>[INST] Solving $4k^3 + 17k^2 - 228k -1116 = 0$ The equation given to me is $$4x^4 + 16x^3 - 17x^2 - 102x -45 = 0$$
I'm asked to find it's resolvent cubic which is not so difficult to find. But the problem is that the question further asks to find the solution of resolvent cubic.
I have found resolvent cubic using Ferrari's method. The resolvent cubic came out to be $$4k^3 + 17k^2 - 228k -1116 = 0$$
The trouble begins now when I'm trying to solve this cubic it's getting very exhaustive and long calculations. But since the question asks to find the solution of the cubic. 
Now can someone help me solve this cubic easily or there's no way out to this problem. I want to escape the tedious calculations while solving this cubic. 
Kindly help me if you can.
 [/INST] Let's use Bill Dubuque's AC method on this cubic. Start by factoring $228 = 2^3 \cdot 3 \cdot 19$ and $1116 = 2^2 \cdot 3^2 \cdot 31$. Their greatest common divisor is $2 \cdot 3 = 6$, so it is an ideal candidate to reach the form $f(6x) = a(6x)^3 + b(6x)^2 + c(6x) + d$.
To do this, transform the roots $k \mapsto 1/x$ and multiply by $-1$ to get:
$$\begin{align} (2^2 \cdot 3^2 \cdot 31)x^{3}+(2^2 \cdot 3 \cdot 19)x^{2}-17x-4 = 0 \\
\implies (2^3 \cdot 3^3 \cdot 31)x^{3}+(2^3 \cdot 3^2 \cdot 19)x^{2}-17 \cdot 6x-24 = 0 \end{align}$$
and now since this is in the form $f(6x)$, let $6x = u$:
$$31u^3 + 38u^2 - 17u - 24 = 0$$
After searching for candidates using the rational root theorem, one might stumble across $-1$ as a root. Synthetic division yields:
$$\begin{array}{c|rrr}&31&38&-17&-24\\-1&&-31&-7&24\\\hline\ &31&7&-24&0\\\end{array}$$
thus this polynomial can be factored as $(u+1)(31u^2+7u-24)$ and here $u = -1$ is again a root of the quadratic. Therefore $(u+1)^2 (31u - 24) = 0$ since the constant term and $u^3$ term must match, or that: $$k = \frac{1}{1/6}, \frac{1}{(24/31)/6} = 6, \frac{31}{4}.$$
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