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So I came across the unsolvable quintic polynomial noticing that solutions can be found by connections with ellipses and such here.

But more importantly, I was considering methods we use (or at least I know and understand) the application of different types of functions on a polynomial when I took notice of a few things.

Let us say we have a polynomial $P(x)$ in the form:$$P(x)=x^5+ax^4+bx^3+cx^2+dx+f$$

And we were trying to solve $P(x)=0$ i.e the roots of this polynomial.

A good try for luck involves the rational roots theorem, where $x$ may have the solution $x=f'$, where $f'=$ the factors of $f$, if and only if $f$ is an integer. If it is rational and non-integer, then $x$ may equal any given factor of the numerator of $f$ divided any possible factor of the denominator of $f$. Does not work for irrational $f$.

If we do find one such, we can now reduce this quintic polynomial into a quartic polynomial, which is always factorable.

But this won't be the case for our problem, as that is too easy.

I note that solving this problem is equivalent to solving:$$e^{x^5}e^{ax^4}e^{bx^3}e^{cx^2}e^{dx}e^f=1$$

Another transformation:$$\ln(x-r_1)+\ln(x-r_2)+\ln(x-r_3)+\ln(x-r_4)+\ln(x-r_5)=\ln(0)$$Where $r_n$ is the $n^{th}$ root. This obviously won't work, however, because we can't define $\ln(0)$.

Another interesting idea is to try to evaluate$$\cos(P(x))=\cos(0)=1$$or maybe using another trigonometric function. We can use identities to expand this, though we must note the problems of a trigonometric function's periods.

I also thought about the ideas of substitution like $x=t-\frac{a}{5}$ to remove the $x^4$ term, creating a depressed quintic, but that still doesn't help us much.

And I don't believe moving terms to the other side will help.

But moving terms to the other side (mainly $f$) allows us to take the log of everything.

However, this means we are no longer solving the same quintic.

Perhaps there is a sort of thing we can try to do:$$g(P(x))=g(0)$$Find some function $g$ that is hopefully one-to-one that can simplify the problem. or possibly $x=q(z)$ as a sort of substitution?

We could also use $x=cos(u)$, but I'm not sure that will go far.

I also note that possibly Lagrange inversion theorem might help (not something I can do)?

And what can integrals/derivatives/calculus do for us?

Perhaps we could use $$\int_{r_n}^{r_m}P(x)dx=0$$Where we realize that if the area is $0$, then the root/graph must cross the $x-axis$ around $r_n,r_m$.

I can't think of some derivatives can do for us, but finding relative maxima/minimum help set ranges for possible roots.

One can also try to use $Q(x)$ and $C(x)$ where $Q$ is a quadratic and $C$ is cubic, but I have noted that not all quintic polynomials are composite.

My question goes:

  1. Do any of these ideas appear to work when trying to solve the quintic polynomial?

  2. How are quintic polynomials generally attempted?

  3. Does Calculus play a role, as I have yet to see that occur.

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Maybe off topic. For polynomial:

$${x}^{5}+q\,{x}^{4}+p\,{x}^{3}+t\,{x}^{2}+k\,x-m=0$$

with root R, characteristic polinomial of matrix:

$$\small\begin{pmatrix}a & -c\,t-b\,p+g\,m-d\,k & -b\,t+d\,m-c\,k & c\,m-b\,k & b\,m\cr b & b\,q+a & -c\,t+g\,m-d\,k & d\,m-c\,k & c\,m\cr c & c\,q+b & b\,q+c\,p+a & g\,m-d\,k & d\,m\cr d & d\,q+c & c\,q+d\,p+b & d\,t+b\,q+c\,p+a & g\,m\cr g & g\,q+d & d\,q+g\,p+c & g\,t+c\,q+d\,p+b & d\,t+b\,q+c\,p+g\,k+a\end{pmatrix}$$

can have a root:

$$\small{x=g\,{R}^{4}+\left( g\,q+d\right) \,{R}^{3}+\left( d\,q+g\,p+c\right) \,{R}^{2}+\left( g\,t+c\,q+d\,p+b\right) \,R+d\,t+b\,q+c\,p+g\,k+a}$$

For example:$${x}^{5}-m=0$$

characteristic polinomial of matrix:

$$\begin{pmatrix}a & g\,m & d\,m & c\,m & b\,m\cr b & a & g\,m & d\,m & c\,m\cr c & b & a & g\,m & d\,m\cr d & c & b & a & g\,m\cr g & d & c & b & a\end{pmatrix}$$

has a root:

$$x=g\,{m}^{\frac{4}{5}}+d\,{m}^{\frac{3}{5}}+c\,{m}^{\frac{2}{5}}+b\,{m}^{\frac{1}{5}}+a$$

For example:

$${x}^{5}+{x}^{4}-4\,{x}^{3}-3\,{x}^{2}+3\,x+1=0$$

has a root:

$$x=2\,\mathrm{cos}\left( \frac{2\,\pi \,k}{11}\right)$$

if a=1,b=1,c=1,d=1,g=1 characteristic polinomial

$$-{x}^{5}-6\,{x}^{4}+{x}^{3}+10\,{x}^{2}+6\,x+1$$

has a root:

$$x=16\,{\mathrm{cos}\left( \frac{2\,\pi \,k}{11}\right) }^{4}+16\,{\mathrm{cos}\left( \frac{2\,\pi \,k}{11}\right) }^{3}-8\,{\mathrm{cos}\left( \frac{2\,\pi \,k}{11}\right) }^{2}-10\,\mathrm{cos}\left( \frac{2\,\pi \,k}{11}\right) -2$$

Sorry for my bad English.

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  • $\begingroup$ This is a very nice observation $\endgroup$ – Yuriy S Jul 3 '16 at 18:51
  • $\begingroup$ How did you come by this observation? Can any root of the original polynomial be used? $\endgroup$ – Yuriy S Jul 3 '16 at 19:45
  • $\begingroup$ dxdy.ru/topic87105.html (in Russian) $\endgroup$ – AlexSam Jul 3 '16 at 23:44
  • $\begingroup$ I think the inverse transformation would be most useful, if we can get from a polynomial with complicated root to the one with a simple root $\endgroup$ – Yuriy S Jul 4 '16 at 16:19
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Regarding your three specific questions:

Do any of these ideas appear to work when trying to solve the quintic polynomial?

In general, no, but, as you discovered, there are techniques that work for certain special cases.

How are quintic polynomials generally attempted?

By using numerical methods. Even for cubic and quartic equations, where formulae exist, they are numerically unstable and difficult to implement reliably. See the answers to this question and this question.

Does Calculus play a role

In numerical methods, yes, sometimes. Some of them (like the Newton-Raphson method) use derivatives.

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For brevity, I'll only answer your main question, that is, "Applying trigonometry in solving quintic polynomials?"

The answer is Yes, the general quintic can be solved by trigonometric methods. But it is much more complicated than the cubic version and uses special functions like elliptic integrals, etc.

If you look at the trigonometric solution of the cubic, it depends on an identity,

$$4x^3-3x+b=0\tag1$$

where $x = \cos(\theta)$ and $b$ is a parameter. It is possible, using a Tschirnhausen transformation, to reduce the cubic to a simpler form and match it coefficient-by-coefficient to $(1)$.

For the quintic, the relevant trigonometric relation is,

$$x^5+5x^3+5x+b = 0\tag2$$

Fortunately, the quintic can also be reduced to a simpler one-parameter form. And it turns out this parameter can be matched with $b$.

The explanation is quite detailed, but these posts should be helpful.

  1. Reducing the quintic to Brioschi form.
  2. Solving the Brioschi using trigonometric functions.
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