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.
 A: Maube the resolvent cubic has a rational root?  Try to find one before resorting to the Ferrari method.
I would try integers first, selecting factors of 1116 to match the Rational Root Theorem.  If you try positive integers Descartes' rule of signs says there can be only one positive root, so if you use increasing choices and the remainder goes positive, you know there are no positive integer roots.
With negative integers, the rule of signs allows two sign changes before you have to give up.
If you did not find an integer root, then you can try fractions.  Such a fraction must be one-half or one-fourth an integer (why?).  You can pinpoint the best candidates from the sign changes you found in the remainders for the integer candidates.
Only if you find no rational roots are you justified in using the full (and yes, unwiledy) Ferrari treatment.
I now hint shamelessly at the answer.  I was "not positive", but I had a "sixth" sense that the resolving cubic would have a rational root.   :-)
A: 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: $$k = \frac{1}{1/6}, \frac{1}{(24/31)/6} = 6, \frac{31}{4}.$$
A: Ferrari's method: If $a$, $b$, $c$, $d$ are the roots of the quartic, form the equation with roots
$$a b + c d, a c + b d, a d + b c$$
The coefficients of this equation are symmetric in $a$, $b$, $c$, $d$ ( since we have $a b + c d$ and all its symmetric transforms).
After we solve the cubic, the quadric is solved using the equalities
$$(a+b+c+d)^2 - 4 ( a b + c d) - 4 ( a d + b c) = (a- b + c - d)^2$$
Now, with $u = a b + c d$, $v = a c + b d$, $w= a d + b c$ we have
$(u-v)(u-w)(v-w) = (a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$$
so the discriminant of the cubic resolvent equals the discriminant of the quadric.
If the quartic is
$$x^ 4 + s x^3 + p x^2 + q x + r$$
then the Ferrari cubic resolvent is
$$x^3- p x^2 + (q s - 4 r)x + (4 p r - r s^2 + 4 p r)$$
So, if the particular quartic is
$$4(x^4 + 4x^3 - \frac{17}{4}x^2 - \frac{51}{2}x -\frac{45}{4})$$
then the Ferrari resolvent is
$$x^3 + \frac{17}{4}x^2 - 57 x - 279$$
or
$$4 x^3 + 17 x^2 - 228 x - 1116$$
Calculating the discriminant of $4 x^3 + 17 x^2 - 228 x - 1116$  we get $0$. That means that the cubic has multiple roots. To find a multiple root, calculate the GCD of the cubic and its derivative
$$\textrm{gcd}(x^3 + \frac{17}{4}x^2 - 57 x - 279,  3 x^2 + \frac{17}{2}x - 57) = x+6$$
Now, divide the cubic $x^3 + \frac{17}{4}x^2 - 57 x - 279$ by $(x+6)^2$ and find $x-\frac{31}{4}$.
