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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.

Blue
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Heisenberg
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  • Here's a hint to get you going: $-6$ is a root. – Macavity Mar 12 '16 at 12:08
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    If one notices that the discriminant of this cubic is zero, one can conclude that the cubic has a double root and so shares a root with its derivative, but that (quadratic) derivative can be readily factored. – Travis Willse Mar 12 '16 at 12:12
  • @Macavity you mean first I should find any one of the roots of the cubic by hit and trial and further use it to find the other two roots. – Heisenberg Mar 12 '16 at 12:13
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    You can use your favourite approach for the cubic. Mine usually involves testing for integer roots first. – Macavity Mar 12 '16 at 12:15
  • @Travis don't you think that finding discriminant of this cubic will involve long calculations. – Heisenberg Mar 12 '16 at 12:23
  • @Macavity I also use the same approach but in this case the calculations of discriminant were also really long. – Heisenberg Mar 12 '16 at 12:25
  • I just tested some small factors of $1116$ - usually worthwhile, especially in textbook settings. – Macavity Mar 12 '16 at 12:26
  • @Macavity yeah! because the product of roots will always be equal to $-1116$. – Heisenberg Mar 12 '16 at 12:30
  • To be precise the product will be $\dfrac{1116}4=279$, but that's a longer story. – Macavity Mar 12 '16 at 12:45
  • @Macavity oh! yeah the product will be $\frac{constant term}{leading coefficient}$ I made a typo in my previous comment. But I'm talking about the discriminant calculation in this. That's long too. – Heisenberg Mar 12 '16 at 12:52
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    @user109256 Yes, certainly, which is why I posted my remark as a comment and not an answer. – Travis Willse Mar 12 '16 at 13:04
  • In this case, one method is to look for the roots of the original polynomial---since $4 = 2^2$ and $45 = 3^2 \cdot 5$, there are $2(3)(3)(2) = 36$ rational roots to check, compared with $108$ (!) for the resolvent, and use these to reconstruct the roots of the resolvent, and hence the resolvent itself. This arguably contravenes the spirit of the problem (i.e., the usual reason for invoking the resolvent in the first place). – Travis Willse Mar 12 '16 at 13:11
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    @Travis is there a way to escape this extremely long calculation of cubic. – Heisenberg Mar 12 '16 at 13:11
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    @user109256 Yes, see my previous remark, but note that it doesn't apply in all situations, it only works here because the original quartic turns out to factor nicely, which is not something one usually knows a priori! In short, I don't think there's a nice way to handle this sort of problem without knowing anything more about the polynomial at the outset than we probably do, because the polynomial doesn't have any apparent special form and because the constant term, $1116$, has many factors. – Travis Willse Mar 12 '16 at 13:15
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    @Travis sadly we still don't have easy way to handle cubics with large coefficients. But I think these types of problems shouldn't be asked in exams. – Heisenberg Mar 12 '16 at 13:20
  • Look at my deleted answer. – Piquito Dec 21 '21 at 01:37

3 Answers3

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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. :-)

Oscar Lanzi
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    Applying the Rational Root Theorem directly might not be a practical method to carry out by hand unless one gets lucky with the order they choose. Since $1116 = 2^2 \cdot 3^2 \cdot 31$, there are $2(3)(3)(2) = 36$ integers to check, and $3 \cdot 36 = 108$ rational numbers to check. – Travis Willse Mar 12 '16 at 13:08
  • That's why I spent some time describing the procedure you would use. You're supposed to use Descartes' rule of signs. If there can be only one positive root and we find the remainder changing signs between x=6 and x=8, then there can be no positive whole number roots and any positive fractional root must lie between 6 and 8 -- you should be able to find that the latter constraint admits $only$ $one$ positive fractional candidate. Among negative roots, you're allowed two sign changes unless you hit on a zero first. You really end up needing relatively few trials ... to get $all$ the roots! – Oscar Lanzi Mar 12 '16 at 13:38
  • Again, this really depends on the order one chooses to test the roots/other numbers in the first place. Why, for example, would we test $8$, a priori? Descartes' Rule of Signs does give some information here, but it turns out not to be so useful: After all, once we find one rational root, we can apply polynomial long division to reduce the root-finding problem to the computationally much shorter problem of solving the resulting quadratic equation. – Travis Willse Mar 12 '16 at 13:47
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    In the case that there are three distinct rational roots, naively one expects that it takes an average of $\tfrac{1}{6} \cdot 108 = 18$ trials to find a solution, and Descartes' Rule of Signs /does/ tell us that in this case there are two negative roots (counting multiplicity), so if we try the negative candidate rational roots first, we can expect an average of $13.5$ trials. This is a modest improvement, but it turns out not to help so much here, as the roots are not all distinct. – Travis Willse Mar 12 '16 at 13:51
  • Which we indeed find. If we try positive whole numbers, then negative integers in order of increasing absolute value, and breaking when the rule of signs demands it, we find a root in 11 trials. Then we need no trials to solve the residual quadratic (but we do need to recognize a four-digit perfect square), and the presence of a double root emerges. – Oscar Lanzi Mar 12 '16 at 14:07
  • The point is not so much what happens in this particular case, which one can only know after the fact, but rather how much work one should expect at the beginning of the process, or early on anyway (say, after observing the sign changes and working out the prime factorizations of the first and last coefficients). I suppose it is true that one really has a good practical reason to compute with integers first, as computing with integers is generally less computationally intensive than computing with noninteger rationals of similar magnitude. – Travis Willse Mar 12 '16 at 14:23
  • I wonder if in practice it's better to carry out a binary search among the integer candidate roots of a given sign. In practice this might let us halt using Descartes earlier on average. This might be a worthwhile question in its own right... – Travis Willse Mar 12 '16 at 14:27
  • To speed things up: all coefficients of the cubic are divisible by $4$ except $17$. Hence if there is an integer root, it's divisible by $2$. Assume this and reduce coefficients, this leaves $3^2\times31$ for the constant coefficient, hence the first to try is $\pm3$, etc. – Jean-Claude Arbaut Dec 20 '21 at 23:45
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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: $$k = \frac{1}{1/6}, \frac{1}{(24/31)/6} = 6, \frac{31}{4}.$$

Toby Mak
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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 $x-\frac{31}{4}$.

orangeskid
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