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:
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Do any of these ideas appear to work when trying to solve the quintic polynomial?
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How are quintic polynomials generally attempted?
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Does Calculus play a role, as I have yet to see that occur.
Best Answer
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.