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

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The discussion focuses on solving the higher order polynomial equation ax + x^3 - x^5 = 0, where a is a positive integer. The participants identify that x = 0 is one solution and suggest that the remaining solutions can be found by solving the quadratic equation a + x^2 - x^4 = 0. A substitution of t = x^2 simplifies the problem, leading to a solvable quadratic equation. The conversation highlights the complexity of finding roots for polynomials of degree 5 or higher, noting that general solutions do not exist for such cases. Overall, the thread emphasizes the challenges and methods for tackling higher order polynomial equations.
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 n \ge 5 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 n \ge 5 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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