How to Factor a Sextic Equation into Two Cubics?

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Discussion Overview

The discussion revolves around the factorization of a sextic equation into two cubic polynomials. Participants explore methods and patterns related to this algebraic manipulation, with a focus on recognizing underlying structures within the equation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant inquires about effective methods to factor the given sextic equation without extensive manipulation of the resulting cubic equations.
  • Another participant suggests that such problems are designed to recognize patterns from simpler examples and encourages breaking down the equation into manageable parts, particularly noting the presence of a factor related to the number 13.
  • A third participant points out that, despite the equation appearing arbitrary, its roots are specifically related to sine functions of rational multiples of pi.
  • A later reply details a successful factorization of the sextic equation into two cubic polynomials, providing the specific forms of these cubics and the relationships between their coefficients.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to factor the sextic equation, and multiple perspectives on the problem remain present throughout the discussion.

Contextual Notes

Some participants express uncertainty about the transcription of the original equation and its implications for the factorization process. The discussion includes references to specific roots and relationships among coefficients that may require further clarification or exploration.

Who May Find This Useful

This discussion may be of interest to those studying algebraic factorization, particularly in the context of higher-degree polynomials, as well as individuals looking for insights into recognizing patterns in mathematical problems.

elfboy
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Anyone know of a good way to factor

x^6+x^4(13^.5/2-13/2)+x^2(13-2*13^.5)+3*13^.5/2-13/2 into 2 cubics?

without having to manipulate the messy cubic equation?
 
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These things are intended as exercises in recognising patterns of which you have already dealt with simpler examples, woven into something more elaborate.

The rules are you are supposed to show an effort. Try and deal with different bits of it. For instance can you not do something with the part that has as factor a straight 13? The rest of it looks more difficult. I would ask you check whether you have really transcribed it exactly right. These things (I cannot imagine it is anything but an artificially constructed problem) are meant to work out to something that rhymes with sense.
 
Last edited:
The equation looks arbitrary but its roots are plus minus sin(pi/13), sin(3pi/13), and sin(4pi/13)
 
After a month of working on this problem on and off I finally factored it in the way desired

(x^3+bx^2+c*x+(-3*13^.5/2+13/2)^.5)*(x^3-bx^2+c*x-(-3*13^.5/2+13/2)^.5)

13^.5/2-13/2=-b^2+2c
13-2*13^.5=c^2-b(26-6*13^.5)^.5solve for b and c and plug into the above expression to obtain the two cubics:

(x^3+(13/2+3*13^.5/2)^.5*x^2+13^.5*x+(-3*13^.5/2+13/2)^.5)*
(x^3-(13/2+3*13^.5/2)^.5*x^2+13^.5*x-(-3*13^.5/2+13/2)^.5)=0

i didn't think this was possible to get a nice expression but I've done it
 

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