Solving Higher Order Polynomial: ax + x^3 - x^5 = 0

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    Higher order Polynomial
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Discussion Overview

The discussion revolves around solving the higher order polynomial equation ax + x^3 - x^5 = 0, where a is a positive integer. Participants explore methods for finding algebraic solutions, particularly focusing on the roots of the polynomial and the implications of its degree.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant notes that x = 0 is a solution and seeks to find the other two solutions.
  • Another participant suggests substituting t = x^2 to simplify the remaining polynomial a + x^2 - x^4 = 0, indicating that this leads to a quadratic equation.
  • A different participant points out that the polynomial is of degree 5, implying there should be five solutions in total.
  • Some participants express uncertainty about finding roots for higher order polynomials, particularly for degrees n ≥ 3, and mention that general solutions do not exist for n ≥ 5.
  • One participant reflects on their lack of familiarity with formulas for cubic and quartic equations, expressing relief that they are not alone in struggling with higher order polynomials.

Areas of Agreement / Disagreement

Participants generally agree that x = 0 is a solution and that the remaining polynomial can be simplified. However, there is no consensus on the methods for finding all solutions, particularly for the degree 5 nature of the polynomial, and uncertainty remains regarding the existence of general solutions for higher order polynomials.

Contextual Notes

Participants acknowledge limitations in their understanding of higher order polynomial solutions, particularly for degrees greater than 4, and the discussion reflects varying levels of familiarity with relevant mathematical concepts.

Peregrine
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I haven't done this in ages, and I'm having trouble recalling how to factor a higher order polynomial. I almost always do this graphically, but for this case I'm interested in an algebraic solution. Specifically, I'm looking at ax + x^3 - x^5 = 0 (with a = an integer >0.)

Clearly 0 is one solution since x(a + x^2 - x^4) =0. Is there a way to solve for the 2 others? Thanks.
 
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Peregrine said:
I haven't done this in ages, and I'm having trouble recalling how to factor a higher order polynomial. I almost always do this graphically, but for this case I'm interested in an algebraic solution. Specifically, I'm looking at ax + x^3 - x^5 = 0 (with a = an integer >0.)

Clearly 0 is one solution since x(a + x^2 - x^4) =0. Is there a way to solve for the 2 others? Thanks.

Generally i do not know how to find the roots of higher order polynomials myself, of an order n>=3. but in yor case it looks easy,because it is not of a complete form.
like you said one trivial answer is x=0, the others will be the solutions to the equation
a + x^2 - x^4=0, so u may want to take a substituton like this t=x^2, so you will end up with sth like this

a+t-4t^2=0, now you now how to solve this right?
and after that just go back and find solutions for x.
 
Clearly 0 is one solution since x(a + x^2 - x^4) =0. Is there a way to solve for the 2 others? Thanks.
The remaining polynomial is quadratic in x2. Therefore you can get x2=(1+-(1+4a)1/2)/2. I'm sure you can finish.
 
Thanks, I hadn't thought to substitute. I got it.
 
Note that it's a degree 5 polynomial, so you should have found 5 solutions.

sutupidmath said:
Generally i do not know how to find the roots of higher order polynomials myself, of an order n>=3.
There are formulas for cubic and quartic equations, but I don't think anyone knows them by heart (ok, maybe a few, but certainly very few people). For [itex]n \ge 5[/itex] general solutions do not even exist (I believe this can be proven using ring theory).
 
CompuChip said:
Note that it's a degree 5 polynomial, so you should have found 5 solutions.


There are formulas for cubic and quartic equations, but I don't think anyone knows them by heart (ok, maybe a few, but certainly very few people). For [itex]n \ge 5[/itex] general solutions do not even exist (I believe this can be proven using ring theory).

That's a bit of relief, i started feeling stupid for not being able to find general solutions for these higher order polynomials. For those of a degree of 3 i have seen some formulas, but haven't actually tried to look closely what is going on. I am going to have to have a look upon these things, because they come up sometimes in calculus one as well! And thnx, for enlightening me for polynomials of a degree of 5 or higher!
 

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