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March 28th, 2014, 10:17 AM   #1
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Tranformation of polynomials

I want to transform a polynomial of kind p(x)=ax+bx+cx+d in another like p(t)=At+B. Is possible?
Is possible to transform a polynomial of kind ax+bx+cx+d in another like t+pt+q.
https://en.wikipedia.org/wiki/Cubic_...epressed_cubic
But, I wish to eliminate the quadratic term and the linear term too...
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March 28th, 2014, 10:44 AM   #2
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Yes. You need some highly technical voodoo tricks to do that, though. See the link in my signature below.

In general, this kinds of forms are called "Principle forms" and you need a quadratic Tschrinhausen instead of linear ones.
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March 29th, 2014, 02:27 AM   #3
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And which are the principal forms of polynomials of 2º, 3º and 4º degree?
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March 29th, 2014, 02:52 AM   #4
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In general, the corresponding principle form of



Is



I'll give a description of the transformation of the cubic for you here, to give a complete answer to your question :

Let the general cubic be , where the second term has been reduced by the affine Tschrinhausen transformation for simplicity. Now let the transformed cubic be



The roots of the latter and the former be related as



All we need is to determine the coefficients, namely, and .

OK, fine. Just sum up the three roots of the former and the latter cubic and use Newton :



Now note that the sum on the left side is 0, as there is no quadratic term in . So and we know the value of .

For the next step, square both sides of and sum it up through the roots again



You can compute the powersums explicitly by using the relations given here, and then solve the quadratic equation in terms of and then plug in the value of found before.

This gives you and , but we still don't have the resulting cubic. Of course one easy way to go would be to sub the transformation in the cubic and plugging in and etc. etc. but that would take lots of algebraic manipulation (in case you don't have a CAS and the tools for computing resultants).

There is a way fortunately, I will show it here if you want. (it's given here, by the way, for a quintic instead)

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March 29th, 2014, 10:32 AM   #5
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This is my try:
Solution for algebraic equations
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March 29th, 2014, 11:36 AM   #6
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This is not correct, no. Also, it is possible to prove that it is impossible to deduce a quintic to such binomial forms.
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