# Talk:Quintic function

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## Details

It is important to have good, detailed examples, this page lacks that. Please add more details on the computation of the Galois groups here. —Preceding unsigned comment added by 188.26.49.57 (talk) 06:38, 3 April 2010 (UTC)

Agreed. To talk about F(5) without showing how to get it, leaves a novice reader clueless. 71.132.139.87 (talk) 23:43, 28 June 2011 (UTC)

## The page is messy

I agree. In my opinion an encyclopedia article should not contain a cartoon, especially one that is only loosely connected with the subject. Thinking that something is funny is POV anyway. Also I am disturbed by the repeated mentioning of computer algebra systems. I know they are powerful. I know they may very well be mentioned somewhere in the article. But the way they are in right now simply breaks the flow of the text and obscures the core idea I wish to learn from the article. Pure mathematics and practical applications should really be in separate sections.  Pt (T) 22:13, 20 Apr 2005 (UTC)
I disagree with the statement that an encyclopedia article shouldn't contain a cartoon. In some cases, and not just those about humor or cartoonists, a cartoon is very appropriate. I do agree though that this particular cartoon is only loosely connected with the subject (as a pun, of course, the joke has no substantive contribution to the content of the article). As the contributor of the cartoon, however, I would like to note that this type of cartoon might be seen in a techincal (even encyclopedic) article in certain types of mathematical, scientific, and engineering magazines, and even serious academia need not be completely humorless. Perhaps very few people besides me find the cartoon appropriate (or funny for that matter), in which case it should go, but if this is an issue, there should be some discussion to determine if this is the case. CyborgTosser (Only half the battle) 16:32, 30 May 2005 (UTC)
The cartoon is great. It actually serves as a pedagogical tool. I'm thinking about even putting it on my door so students passing by can look at it. --C S (Talk) 17:54, 19 November 2005 (UTC)

did you actually make that cartoon? i just cant get over it (no im serious) good work :)

btw is this the easiest approach? im finding the reduction a bit difficult

Is it true that Galois was the one who proved insolvability for equations of degree higher than 5? I think I saw somewhere that Abel's proof was valid for n>5 too.

I'd dispute that the Brin Radical solution counts as an Algebraic solution. The formula for a Brin Radical includes an infinite sum, and hence is not strictly speeking algebraic (although I'd prefer someone with a bit more knowledge on this to verify this).

From Galois Theory 2nd Edition, I Stewart, Chapman and Hall, ISBN 0-412-34550-1 Therorem 15.7 (p144) If K is a Field of characteristic zero and n >= 5, then the general polynomial of degree n over K is not soluble by radicals. --Pfafrich 18:59, 18 November 2005 (UTC)

I've split the Brin Radical material into a seperate page. I think the previous treatment gave too much emphesis to the Brin Radicals, and downplayed the insovability by radicals which is a very deep and beautiful result, which gavce rise to the foundations of much of modern group theory. --Pfafrich 20:57, 18 November 2005 (UTC)

Shouldn't the general formula be more like:

${\displaystyle ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f=0}$

${\displaystyle ax^{5}+bx^{4}+cx^{3}+dx+e=0}$

Or is this a different type of quintic than the article is talking about? --Psiphiorg 21:07, 18 November 2005 (GMT)

I think it's the same thing, ie. you get rid of the quadratic term with a substitution of variables like the one for the cubic. Phr 11:52, 25 February 2006 (UTC)
I think this was my mistake when originally typing in. ${\displaystyle ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f=0}$ is the general form. Phr may be correct in that it can be simplified, but I've not checked this. --Salix alba (talk) 15:47, 25 February 2006 (UTC)

## The degree after a quintic equation...

What comes after a quintic equation? How do you solve and graph it? And what comes after that? How do you solve polynomials where the highest degree is over 1 million, or progressive? I have been looking around and it seems that every one likes to say they cannot be solved. — Preceding unsigned comment added by 24.141.44.123 (talkcontribs) 02:46, 21 September 2006‎

For polynomial equation of degree >4, there does not exist a general solution. It is not a case of "they haven't found it yet!". A "formula" giving the solutions has been proven not to exist. --Veddan (talk) 19:50, 15 May 2008 (UTC)
This is false. There exist many general solutions for finding the roots of polynomials of arbitrarily high degree. If you're interested in numerical solutions only, you can use Laguerre's method, the Jenkins-Traub method, or the Durand-Kerner method to obtain the roots. If you want analytical solutions, you can use Hiroshi Umemura's method based on Siegel modular forms (see Tata Lectures in Theta II, Solution of algebraic equations in terms of theta constants) or the Birkeland-Meyr method based on multivariate hypergeometric functions (see here for starters). These are generalizations of the various methods described on the Bring radical page for solving the quintic. The only thing that has been proven regarding the solutions of equations of the fifth degree and higher is that their solutions need not involve radicals. This is the content of the Abel-Ruffini theorem. All of these other solutions I've mentioned go beyond radicals. Stormwyrm (talk) 03:03, 29 January 2009 (UTC)
After the quintic equations, there come sextic equations, septic equations, octic equations, nonic equations and decic equations. You may find this amusing: the name of a 100th degree polynomial is a hectic equation. —Preceding unsigned comment added by Protactinium-231 (talkcontribs) 07:40, 16 December 2009 (UTC)

## The caption in the image is truncated and therefore misleading.

The caption of the image "Polynomialdeg5.png" appears in my browser as:

 Polynomial of degree 5: f(x) = (x+4)(x+2)(x+1)(x-1)(x-3)/20


without the final "+2".

Can the size of the image be adjusted to account for the size of the caption? Or perhaps the caption can be shrunk to fit? —Preceding unsigned comment added by Ndokos (talkcontribs) 13:38, November 22, 2006

I have added spaces between the factors and around the plus sign so that the caption can word wrap. How does it look now in your browser?
--Psiphiorg 00:39, 23 November 2006 (UTC)
Looks good! Thanks.
--Ndokos 21:18, 30 November 2006 (UTC)

"this is known as the Abel-Ruffini theorem, first published in 1824, which was one of the first applications of group theory in algebra"

This is untrue, due to a confusion between Abels work which has no mention of groups, with the slightly later work of Galois, who developed finite group theory to prove the inslovability (by the way Galois proof reads more like a sketch of a proof, rather than a proof itself)

Galois didn't have too much time to write, because of the duel, isn't it? Albmont 17:29, 28 February 2007 (UTC)

## David Dummit,Sigeru Kobayashi and Hiroshi Nakagawa

The Quintic is solvable by radicals,it was demonstrated by David Dummit and (independently), Sigeru Kobayashi and Hiroshi Nakagawa in 1991-92.How was it done?.Is Abel-Ruffini Theorem fallacious? Is Galois' great theorem fallacious?

Arkapravo Bhaumik 06:23, 25 January 2007 (UTC)

I tried some research, but much of it is over my head. I'm pretty sure that Kobayashi and Dummit just demonstrated that some quintics are solvable by radicals, not all of them, though (note "David Dummit and (independently) Sigeru Kobayashi and Hiroshi Nakagawa give methods for finding the roots of a general solvable quintic in radicals."). I tried solving a quintic with radicals in Mathematica, and got [1] as the solution; therefore I think that it might be a little complex for this article :-). — 15:13, 25 January 2007 (UTC)
I liked that huge expression; I think it deserves a page. Something like "Examples of Solvable Quintics". I will show it to my daughters to scare them off the computer late at night <evil grin>. Albmont 17:32, 28 February 2007 (UTC)
The solution by radicals of any solvable quintics is given by Maple function solve, if the variable _EnvExplicit has been set to true. This works also for solvable quintics depending on parameters. When substituting the solutions in the quintic, Maple is able to simplify to zero, when the coefficients are numeric. But not when they depend on parameters. Is Mathematica able to do this simplification for the three parametric examples of the page? It is not clear for me if this kind of information has its place in this page.D.Lazard (talk) 09:00, 3 June 2010 (UTC)

We ought to have a page on the "Solution Of the Qunitic by David Dummit,Sigeru Kobayashi and Hiroshi Nakagawa" this would deal with elliptic functions and help us to structure the solution. Arkapravo Bhaumik 05:59, 8 February 2007 (UTC)

## [itex] Tags

I switched almost all the equations and variable mentions to use [itex] formatting. This eliminated the mixed appearance of variables forced to italics using double apostrophes and superscripts done via sup tags, and the formatting produced by putting everything between the math tags. I did not touch the image caption that gave trouble earlier.

I did this most immediately because of the poor result of using ''t''<sup>5</sup> (namely, t5). The spacing is off such that the ${\displaystyle t}$ leans forward into the superscript 5, which left me guessing at whether I was staring at some rendering error, a ${\displaystyle t^{6}}$, an ${\displaystyle l^{5}}$, or even an ${\displaystyle I^{5}}$, before settling on ${\displaystyle t^{5}}$, and thank goodness for context. The spacing is also off enough with commas and periods that immediately follow double-apostrophe italicized variables to make the page unsightly without use of the math tag (though if it's any more difficult to read, it's only because the spacing isn't what I'd expect after too much time with TeX output).

If I've unknowingly committed some great faux pas, please inform me: I'm new here.

Jay Uv. 02:29, 26 January 2007 (UTC)

## Requested Move

The following discussion is an archived discussion of the proposal. Please do not modify it. Subsequent comments should be made in a new section on the talk page. No further edits should be made to this section.

The result of the proposal was no consensus to move the page, per the discussion below. Dekimasuよ! 03:07, 11 December 2007 (UTC)

I suggest moving this page to Quintic function because it makes more sense to derive an equation from a function than the other way around. —Celtic Minstrel (talkcontribs) 20:58, 6 December 2007 (UTC)

• Oppose The existing title is idiom; and "quintic function" covers a much broader class of objects. Septentrionalis PMAnderson 19:11, 8 December 2007 (UTC)
The above discussion is preserved as an archive of the proposal. Please do not modify it. Subsequent comments should be made in a new section on this talk page. No further edits should be made to this section.

## Linear algebraic methods

hi, im currently in year twelve studying maths b and c. and i have come across the need to find an exact value for a quintic 64x^5 + 16x^3 + x - 4. i read this sentence "The quintic equation can be solved by creating a companion matrix of the quintic equation and calculating the eigenvalues of said matrix." and thought - yippee i can solve it as i have a good knowledge of matrices and eigenvalues, but i did it and the result was not equal to what three computed answers gave me, so I'm thinking this statement is untrue. can someone confirm this for me?

cheers

I may be incorrect, but I believe that solving a quintic by finding the eigenvalues of a it's comparison matrix would be trivially true. Given a fifth degree monic polynomial p(t) and it's comparison matrix C(p) then, det(\lambda*I - C(p)) = p(t) Thus we'll still need to factor p(t).

Davepntr (talk) 18:00, 12 October 2008 (UTC)

After some more research I believe this section is referring to more sophisticated algorithms designed for finding eigenvalues. However these algorithms are designed for computers, and not to be done by hand. I believe that this section requires expantion/clarification and should perhaps be considered for moving to a more general topic on polynomials as this is not specific to quintics.

Davepntr (talk) 06:11, 13 October 2008 (UTC)

Agreed, this section has nothing to do with quintics in particular. Perhaps it is alluding to the interpretation of the Jenkins-Traub algorithm as linear algebra operations, though that is not the normal or efficient way of implementing it. I will remove this section. --Macrakis (talk) 18:05, 23 October 2010 (UTC)

## Real or complex

The article says: Solving linear, quadratic, cubic and quartic equations by factorization into radicals is fairly straightforward when the roots are rational and real but I thought the various formulae worked for extracting complex roots as easily as real roots. Certainly so for the quadratic, and I thought so for the cubic and quartic as well, which are just elaborations on the quadratic. (Now I think about it, what is the problem with irrational roots?) --Michael C. Price talk 12:14, 1 April 2009 (UTC)

## Factorisation

The article says

Other quintics like ${\displaystyle x^{5}-x+1=0}$ cannot be factorized

but if ${\displaystyle \alpha =-1.1673039782614...}$ is a root (which it is) we can write

${\displaystyle x^{5}-x+1=(x-\alpha )(x^{4}+\alpha x^{3}+\alpha ^{2}x^{2}+\alpha ^{3}x+\alpha ^{4}-1)-(\alpha ^{5}-\alpha +1)}$

But since the ${\displaystyle (\alpha ^{5}-\alpha +1)}$ term vanishes we can solve for the remaining roots and factorise ${\displaystyle x^{4}+\alpha x^{3}+\alpha ^{2}x^{2}+\alpha ^{3}x+\alpha ^{4}-1}$ using the quartic equation. Where am I going wrong? --Michael C. Price talk 10:25, 7 April 2009 (UTC)

Okay, I think I see the answer. Obtaining ${\displaystyle \alpha =-1.1673039782614...}$ requires an infinite number of steps. Perhaps I'll add this to the article since it is not explained very well. --Michael C. Price talk 10:45, 7 April 2009 (UTC)

## Introductory (graphical) characterization

This my be splitting hairs, but the introduction seems to poorly characterize the appearance of graphs of quintics. It states that "Because they have an odd degree, normal quintic functions appear similar to normal cubic functions when graphed, except they may possess an additional local maximum and local minimum each." The thing is, you can easily get a cubic-looking graph with an extra "wiggle" (i.e., pair of inflection points) by integrating a quartic which has only three unique roots. Consider f(x) = x^5 +x^3. It has two local extrema but cannot be confused for a cubic. Does it matter?

Would it be better to have "...except that it may possess an additional pair of inflection points"? Koyarpm (talk) 17:34, 1 October 2010 (UTC)

## Hideous tables

If I had plenty of time right now, I'd rid this article of the hideous html tables. Michael Hardy (talk) 03:27, 16 March 2011 (UTC)

## φ

In this edit, the constants in the quintic root are expressed as:

φ = (1+sqrt(5))/2
a=sqrt(2/φ)
b=sqrt(2*φ)
c=5^(1/4)

a=sqrt(5-sqrt(5))
b=sqrt(5+sqrt(5))
c=sqrt(5)

What is the advantage of introducing φ into this? The φ-less form seems simpler, and I don't see any motivation for introducing φ here. --Macrakis (talk) 18:23, 17 August 2011 (UTC)

Dear Macrakis; I was the one who gave that exact solution in the first place. The golden ratio naturally appears in the 5th roots of unity; if you plot the roots in the complex plane, they form the vertices of a pentagon. This then explains why you can express a lot of trigonometric angles pi*n/5 in terms of the golden ratio. For example, cos18 = (1/4)sqrt[2(5+sqrt(5))] = 5^(1/4)/2 sqrt[φ]. When you solve a quintic equation, naturally the fifth roots of unity are involved, hence why the radical solution of the quintic x^5-5x+12 = 0, which has a discriminant D that is divisible only by 2 and 5, or D = 2^12*5^2, can be expressed in terms of the golden ratio and the square root of 2. I hope it is clearer now.Titus III (talk) 15:12, 20 August 2011 (UTC)
Using the expression golden ratio to describe the solutions is like using the expression circumference to describe the normal distribution.--Lefschetz (talk) 09:41, 21 August 2011 (UTC)
Like Macrakis and Lefschetz, I prefer the formulation without golden ratio. Introducing golden ratio here is useful only for people accustomed with it, that is very few people. Moreover this quintic is only an example and the fact that ${\displaystyle {\sqrt {5}}}$ (and thus φ) appears in the solution is very specific to this example. Introducing φ may give the false idea that the golden ration appears in the solution of all solvable quintics. Thus I vote for the formulation without golden ratio. D.Lazard (talk) 09:55, 22 August 2011 (UTC)
Whatever we do should by guided by Reliable Sources, so I did a Google search on ["golden ratio" quintic]. I found one book on the quintic which did mention the golden ratio: Jerry Michael Shurman, Geometry of the quintic. One mention is in constructing an icosahedral golden configuration, two are for writing a solution. So a reputable mathematician does mention the golden ratio in this connection (though he actually defines it as 0.618 = 1/1.618 = 1.618 - 1). But I'm not sure that makes it worth mentioning in the WP article. --Macrakis (talk) 16:43, 22 August 2011 (UTC)

Hello all. I'll summarize my answer in three points:

Point 1. Here is a cut-away view of an icosahedron,

If you guys remember, Felix Klein wrote "Lectures on the Icosahedron and the Solution of the Fifth Degree". Turns out the Cartesian coordinates of the icosahedron's vertices can be defined in terms of the golden ratio φ = (1+√5)/2. Centered at the origin, and with edge-length 2, then,

{\displaystyle {\begin{aligned}&(0,\,\pm 1,\,\pm \phi )\\&(\pm 1,\,\pm \phi ,\,0)\\&(\pm \phi ,\,0,\,\pm 1)\\\end{aligned}}}

So unless Klein has been refuted, please tell me again that the golden ratio, via the icosahedron, is not implicit in the solution of fifth degree equations, eh?

Point 2. Here are the five 5th roots of unity xi,

As one can see, they form the vertices of a regular pentagon. The diagonals D of a regular pentagon can be calculated based upon the golden ratio φ and the known side T,

${\displaystyle {\frac {D}{T}}=\phi ={\frac {1+{\sqrt {5}}}{2}}}$

Thus, consistent with the icosahedron (of course), these vertices can also be simply expresed in terms of φ as,

{\displaystyle {\begin{aligned}x_{1}&=\phi ^{0}=1\\2x_{2}&=\phi ^{-1}+5^{1/4}i{\sqrt {\phi }}\\2x_{3}&=\phi ^{-1}-5^{1/4}i{\sqrt {\phi }}\\2x_{4}&=-\phi +5^{1/4}i{\sqrt {\phi ^{-1}}}\\2x_{5}&=-\phi -5^{1/4}i{\sqrt {\phi ^{-1}}}\\\end{aligned}}}

Point 3. Finally, by Lagrange, all five roots of a solvable quintic can be given as,

${\displaystyle y_{k+1}=\omega ^{k}u_{1}^{1/5}+\omega ^{2k}u_{2}^{1/5}+\omega ^{3k}u_{3}^{1/5}+\omega ^{4k}u_{4}^{1/5}}$

for k = {0,1,2,3,4}, the four ui are the roots of the quartic resolvent, and ω is any complex 5th root of unity. Thus, if the roots are given in full, then ALL solvable quintics, via the 5th roots of unity, have the golden ratio naturally imbedded in the roots.

Conclusion: Thus, for the golden ratio to appear in the solution of quintics is not arbitrary at all but, by the established results of Lagrange and especially Klein's work on the icosahedron, is simply a natural consequence. Notice that, say, it is not the plastic constant that pops out, but it had to be φ. And why φ? By points 1,2,3, one can have an inkling of the reason. (I hope.) Titus III (talk) 13:10, 23 August 2011 (UTC)

Thanks for the explication. But the way the article is currently written, φ comes in as a deus ex machina, with no motivation. I don't think we should expand the Quintic function article to include the discussion of the icosahedral symmetries (too long and too technical for most readers), but it may be worthwhile to start an article on say the Icosahedral symmetries of the quintic (based on Klein and Shurman), to which we refer to motivate φ... or at least cross-reference Icosahedral_symmetry#Related_geometries in the solutions section. --Macrakis (talk) 14:36, 23 August 2011 (UTC)
Dear Macrakis: You are much welcome. (P.S. I made some minor changes to the 5th roots of unity.) Titus III (talk) 15:51, 23 August 2011 (UTC)
Dear Titus III: Your long contribution is not convincing at all. In fact, it is proved in Lazard (2002) paper (see the article for an exact reference) that, if L is the minimal radical extension which contains a root of a solvable quintic, then L contains the golden ratio if and only if the discriminant is the product of 5 by a square. Thus, for almost all solvable quintics, the golden ration does not appears in the expression of the first root. It appears in the expression of the other roots, because they involve the fifth roots of unit. By the way, the correct expression of the fifth roots of unit in term of the golden ratio is
{\displaystyle {\begin{aligned}x_{1}&=1\\2x_{2}&=\phi ^{-1}+{\sqrt {-\phi -2}}\\2x_{3}&=\phi ^{-1}-{\sqrt {-\phi -2}}\\2x_{4}&=-\phi +{\sqrt {\phi ^{-1}-2}}\\2x_{5}&=-\phi -{\sqrt {\phi ^{-1}-2}}\\\end{aligned}}}
Conclusion: Golden ratio does not appear naturally when solving quintics, but is strongly related to the fifth roots of units. Thus there is no reason to involve it in the example. However a section about the fifth roots of units (roots of the quintic ${\displaystyle x^{5}-1}$) would be useful to show that they may be expressed by radicals and in term of the golden ratio. D.Lazard (talk) 12:59, 24 August 2011 (UTC)

Dear Daniel: 1. Hm, the "correct expression" for the 5th roots of unity? Given the equality,

${\displaystyle 2x_{2}=\phi ^{-1}+5^{1/4}i{\sqrt {\phi }}=\phi ^{-1}+{\sqrt {-\phi -2}}}$

surely you are not serious when you say that only one is correct, while the other is incorrect? They are equal. One may be simpler perhaps, but not "more" correct.

2. I'm curious about your result that, "...L contains the golden ratio if and only if the discriminant is the product of 5 by a square." This is a family of solvable quintics:

${\displaystyle x^{5}+10x^{3}+5(n^{2}+3n+18)x^{2}-5(n^{3}+n^{2}+15n-14)x+(n^{4}-n^{3}+37n^{2}+441)=0\,}$

with solution,

${\displaystyle x_{i}=z_{1}^{1/5}+z_{2}^{1/5}+z_{3}^{1/5}+z_{4}^{1/5}}$

where z_i are the roots of,

${\displaystyle z^{4}+(n^{4}-n^{3}+37n^{2}+441)z^{3}+(n^{7}+9n^{6}+70n^{5}+312n^{4}+1123n^{3}+2623n^{2}+4326n+3530)z^{2}-(n^{4}-n^{3}+37n^{2}+441)z+1=0}$

The discriminant D of the quintic, using the formula in the article is,

${\displaystyle D=5^{5}(-21+n)^{2}(17+2n+n^{2})^{3}(225+20n+26n^{2}+3n^{3}+n^{4})^{2}\,}$

It is easy to make ${\displaystyle 17+2n+n^{2}=y^{2}}$. However, I am having trouble expressing the roots in terms of the golden ratio if n is rational. Is your result valid only if the coefficients of the quintic are integers?Titus III (talk) 22:26, 24 August 2011 (UTC)

You are using f for both the function and a variable, with same appearance.84.229.210.231 (talk) 07:27, 26 November 2011 (UTC)

Fixed. D.Lazard (talk) 08:51, 26 November 2011 (UTC)

## Function and equation

Anon 24.240.63.6 made this change and I made a additional little change. User D.Lazard then reverted per "An equation is not a function.". Perhaps this revert and the edit summary were inspired by the fact that the sentence happened to immediately follow the sentence about the equation g(x)=0. I think that—as this article is about the quintic function and not about the quintic equation—this was a good call by anon. So I brought it back to the functions version, but to avoid possible frowning and/or confusion, I moved the equation sentence two lines down toward the end of the lead: [2] - DVdm (talk) 14:55, 3 February 2014 (UTC)

I agree with DVdm edit and comment. Just after my edit, I had something like that in mind, but I was too lazy to do it myself :-) D.Lazard (talk) 15:52, 3 February 2014 (UTC)

## Solution through elliptic functions

Recently, Footmath has added a section about the solution through elliptic integrals, based on a recent and badly cited article. It seems that the author of this recent article has rediscovered old Hermite's method (without an exact reference, I am unable to know if Hermite is cited). In any case, this new section essentially duplicates Bring radical, and is not better written. I have therefore replaced this section by a explicit link to this page and I have added explicit credit to the authors of 19th century. Some cleanup is yet needed for avoiding duplicates with the almost unreadable section "Beyond radicals" and for avoiding double citations. D.Lazard (talk) 18:32, 21 May 2014 (UTC)

## Auxiliary equation

Auxiliary equation is mentioned three times, it needs explaining! 86.166.163.239 (talk) 08:26, 19 May 2015 (UTC)

I have edited the article for clarifying this. D.Lazard (talk) 09:56, 19 May 2015 (UTC)

## Tagged assertion

Before my recent edits, the article contained the following assertion, which I quote verbatim for the record:

It is then a necessary (but not sufficient) condition that the irreducible solvable quintic
${\displaystyle z^{5}+a\mu ^{4}z+b\mu ^{5}=0\,}$
with rational coefficients must satisfy the simple quadratic curve
${\displaystyle y^{2}=(20-a)(5+a)\,}$
for some rational ${\displaystyle a,y}$.

As it is, it is a mathematical non-sense. I have tried to give it a meaning by replacing y by b, and rewording the sentence, but this results in a wrong assertion (at least if there is no typo in the formula for Cayley's resolvent). As no source is provided, which allows to correct this sentence, I'll remove it as [[WP:OR]. D.Lazard (talk) 09:59, 20 May 2015 (UTC)

There is another option:
${\displaystyle {x}^{5}+k_{2}\,x+k_{1}=0}$
${\displaystyle k_{2}={\frac {5\,{b}^{2}\,\left({b}^{2}\,g-{c}^{3}\right)\,\left({b}^{4}\,{g}^{2}-{b}^{2}\,{c}^{3}\,g-{c}^{6}\right)\,\left({b}^{4}\,{g}^{2}+4\,{b}^{2}\,{c}^{3}\,g-{c}^{6}\right)}{c\,{g}^{3}\,{\left({b}^{2}\,g+{c}^{3}\right)}^{3}}}}$
${\displaystyle k_{1}={\frac {{b}^{2}\,\left({b}^{2}\,g-{c}^{3}\right)\,\left({b}^{4}\,{g}^{2}+{c}^{6}\right)\,\left({b}^{8}\,{g}^{4}-22\,{b}^{6}\,{c}^{3}\,{g}^{3}-6\,{b}^{4}\,{c}^{6}\,{g}^{2}+22\,{b}^{2}\,{c}^{9}\,g+{c}^{12}\right)}{{c}^{2}\,{g}^{4}\,{\left({b}^{2}\,g+{c}^{3}\right)}^{4}}}}$
${\displaystyle m={\frac {{b}^{3}\,c\,g-b\,{c}^{4}}{{b}^{2}\,{g}^{3}+{c}^{3}\,{g}^{2}}}}$
solution:
${\displaystyle g\,{m}^{\frac {4}{5}}-{\frac {b\,g\,{m}^{\frac {3}{5}}}{c}}+c\,{m}^{\frac {2}{5}}+b\,{m}^{\frac {1}{5}}}$
Sorry, I'm probably writing in the wrong section, but my additions were removed (I'm new). — Preceding unsigned comment added by A.Samokrutov (talkcontribs) 16:35, 21 May 2015‎
To A.Samokrutov: Please sign your posts in the talk pages with four tildes (~~~~).
This formula seems a new parameterization of solvable quintics in Bring–Jerrard form. The article gives already three different such parameterizations, which are all simpler than yours. As Wikipedia is not a media for publishing original research (see WP:OR), this new parameterization cannot be inserted in Wikipedia if it has not been published in a referred and notable academic journal. Ever if it has been reliably published, it is not necessary worth to mention it in Wikipedia: this would need that it would be cited in secondary sources (see WP:SECONDARY), or, at least that it is evident that would improve the article. In any case, because of your conflict of interest (see WP:COI), you are misplaced to judge if this would improve the article. D.Lazard (talk) 20:56, 21 May 2015 (UTC)
Please look quintic, sextic, septic maybe something fit for a wiki? A.Samokrutov (talk) 16:30, 22 May 2015 (UTC)
Again, without a referred article explaining the meaning of these formulas, how there were obtained, and why they are interesting, these formulas may not be inserted in Wikipedia.
PS to my preceding post: There is no evidence that above parameterized set of quintics is solvable. A proof of solvability could be an expression of b, c, g in terms of k1 and k2 or of the rational root of the Cayley's resolvent in terms of b, c, g. D.Lazard (talk) 17:19, 22 May 2015 (UTC)

## Depressing the quintic

Article reads: "the Tschirnhaus transformation ...which depresses the quintic". I bet most readers have no idea what it means to "depress" a quintic. Some more explanation would be helpful (or maybe even an article on depressing polynomials?). --104.132.34.86 (talk) 23:36, 15 February 2016 (UTC)

Fixed. D.Lazard (talk) 09:32, 16 February 2016 (UTC)

## Quintic or irreducible quintic?

The section Quintic function#Roots of a solvable quintic (paragraph 2) says

If the quintic is solvable, one of the solutions may be represented by an algebraic expression involving a fifth root and at most two square roots, generally nested.

Should this start out as "If an irreducible quintic is solvable..."? I ask because a quintic that can be factored into a cubic and a quadratic can be solved without fifth roots—can it also be solved using a fifth root? Loraof (talk) 21:13, 13 June 2016 (UTC)

Every equation may be solved using a fifth root. It suffice to replace any number appearing in the solution by the fifth power of its fifth root. This may seem a joke, but Galois theory allows showing that fifth roots (if any) may always be eliminated from any solution of an equation of lower degree. The beginning of the section already said that "solvable quintics" refer to irreducible quintics solving by radicals. I have clarified this by adding that only irreducible quintics are considered in the section. D.Lazard (talk) 22:29, 13 June 2016 (UTC)

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