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Given an arbitrary polynomial:

$$a_0 + a_1x + a_2x^2 ... a_nx^n$$

Does there exist a series of substitutions (or single substitution if you choose to combine them) that leaves this function in the form:

$$p_1w + p_2w^r$$

I am aware there are substitutions (referred to as polynomial depression) that leave the polynomial in the form:

$$p_1 + p_2w + rw^n$$

For example in this article:

http://en.wikipedia.org/wiki/Bring_radical#Bring.E2.80.93Jerrard_normal_form

Thomas Andrews
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Sidharth Ghoshal
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  • No and no. The first binomial form is solvable, and any radical transformation would contradict insolvability of general polynomials. The second is a trinomial form and it is provably true that a sextic cannot be reduced to that which is essentially of the form $z^n + z + t$. See Hilbert's $13$th problem for an essentially similar conjecture for septics. – Balarka Sen Jan 27 '14 at 13:56
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    Actually, it's yes. He didn't specify that it be in radicals, only if there was such a well-defined transformation. And the Tschirnhausen can do it (but one must be careful to point out that the transformation is generally not in radicals). – Tito Piezas III Jan 27 '14 at 17:51
  • Yes, it's yes if nonradical transformations are allowed, as I recall you did from one of your blogpost transforming a Brioshic to a DeMoivre via modular $j$ (This is replaced by my previous arguments. Surely $N = F(X, Y)$ is possible, which slipped from my mind for a bit) – Balarka Sen Jan 27 '14 at 18:36

1 Answers1

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Yes, given the general equation,

$$a_nx^n+a_{n-1}x^{n-1}+\dots+a_0 =0\tag{1}$$

you can use a deg $n-1$ Tschirnhausen transformation,

$$y =b_nx^{n-1}+b_{n-1}x^{n-2}+\dots b_1$$

to reduce (1) to binomial form,

$$y^n+c_0 = 0$$

Unfortunately, in general the unknowns $b_i$ entail solving an equation of at least $(n-1)!$ degree, hence is not in radicals for $n\geq5$. (For $n=4$, the system results in a solvable sextic.)

But you can eliminate, in radicals, the three terms $x^{n-1},x^{n-2},x^{n-3}$ simultaneously. A clear step-by-step description is given here.

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Tito Piezas III
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  • This is quite informative but doesn't answer exactly what I was asking. Can we transform the equation into a linear term + some-power term as opposed to constant term and normal power term as you did? – Sidharth Ghoshal Jan 25 '14 at 05:01
  • But the form $P(w)=p_1w+p_2w^r = w(p_1+p_2w^{r-1})$ is factorable. You do know that the transformation you cited (Bring quintics) is applied on equations, hence the polynomial is necessarily equated to $P(w) = 0$, and one can factor out that $w$. – Tito Piezas III Jan 25 '14 at 05:22
  • Does that mean knowing such a transformation implies you can solve the the equation if the original polynomial set to 0? – Sidharth Ghoshal Jan 26 '14 at 05:35
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    Ah, you are using "polynomials" and "equations" interchangeably. The transformations that you cited are applied to *equations* (in particular, the quintic for the Bring radical) . Kindly edit your post and *put zeros* at the end of your three equations, and you will see that the second one is addressed by my answer. – Tito Piezas III Jan 26 '14 at 06:33
  • I see why as it is – Sidharth Ghoshal Jan 27 '14 at 12:20