# General form of the solutions of polynomials

1. Jun 1, 2014

### Jhenrique

I was studying a article that solves the cube and quartic equation in the inverse sense:

$x = \sqrt[3]{A} + \sqrt[3]{B}$

$x = \sqrt[4]{A} + \sqrt[4]{B} + \sqrt[4]{C}$

https://www.physicsforums.com/attachment.php?attachmentid=70239&stc=1&d=1401676309

I found this relationship too:
https://www.physicsforums.com/attachment.php?attachmentid=70240&stc=1&d=1401676615

http://en.wikipedia.org/wiki/Scipione_del_Ferro#The_Solution_of_the_Cubic_Equation

So arise a question in my mind: if a polynomial can be a generalized format like:
$ax^2+bx+c$
$ax^3+bx^2+cx+d$
$ax^4+bx^3+cx^2+dx+e$

so can the solution have a generalized format too?

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2. Jun 1, 2014

### Matterwave

The general formula for a quadratic equation is taught in high school. Given an equation of the form $ax^2+bx+c=0$ we have two roots:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

The general formula for the cubic equation is quite a bit more annoying. You can find it here: http://en.wikipedia.org/wiki/Cubic_equation#General_formula_for_roots

The quartic equation has an even more annoying general formula:
http://en.wikipedia.org/wiki/Quartic_equation#General_formula_for_roots

There are no general formulas for quintic or higher degree polynomial equations.

3. Jun 1, 2014

### Jhenrique

General formulas in this sense:

for quadratic: $x = A + \sqrt[2]{B}$

for cubic: $x = A + \sqrt[2]{B} + \sqrt[3]{C}$

for quartic: $x = A + \sqrt[2]{B} + \sqrt[3]{C} + \sqrt[4]{D}$

or maybe:

$x = \sqrt[2]{A}$

$x = \sqrt[3]{A} + \sqrt[3]{B}$

$x = \sqrt[4]{A} + \sqrt[4]{B} + \sqrt[4]{C}$

Exist a valid, practice and correct pattern for the solutions?

4. Jun 1, 2014

### Matterwave

Can you define your variables? What's wrong with the general formulas I posted?

5. Jun 2, 2014

### Hertz

I think this is what he means:
$0=c_1+c_2x+c_3x^{1/2}+c_4x^{1/3}+c_5x^{1/4}+...$

Initially I was going to say: For a finite amount of terms (all the way up to $c_{n+1}x^{1/n}$, you can always multiply by $x^n$ to turn it into a normal polynomial.

I realized this is not the case though as you will still have fractional powers of x.

-edit
Actually, I am realizing that this is not what he means. Oh well, may as well leave my post here anyways.
As Matterwave said, there is no general solution to polynomials of degree 5 or greater. The pattern that you are noticing is most likely untrue as you get to higher order equations.

6. Jun 2, 2014

### Jhenrique

First, do you understand what I ask?

7. Jun 2, 2014

### Matterwave

No, I don't know what you're asking. In the OP you asked:
But I gave you the general formulas, and that's apparently not what you wanted.

8. Jun 2, 2014

### Jhenrique

Studying elementar algebra arise equations like: $3x^5 - 2x^2 = x$, $\frac{3x+2}{x^2-1}=0$, $(x^{15}+3)(x^3+x^2+1)$, etc...

So someone thought: "those equations can be written more generically in the form:
$ax^2+bx+c$
$ax^3+bx^2+cx+d$
$ax^4+bx^3+cx^2+dx+e$
and the those equations will be called of polynomials."

Now studying the solutions of polynomials, arises a new format of equations involving radicals:
So, my question is: which would be the general format for equations with radicals?

9. Jun 3, 2014

### Matterwave

Wasn't this what Hertz posted? Or are you asking for the solutions to such equations?

10. Jun 3, 2014

### Jhenrique

Yes!

11. Jun 4, 2014

### HallsofIvy

Matterwave, in his first response, gave links to general solutions to cubic and quartic polynomial equations.

You have also been told that there cannot be a general solution, in terms of radicals, to polynomial equations of degree 5 or higher (because there exist solutions to such polynomials that cannot be written in terms of radicals).

What more do you want?

12. Jun 4, 2014

### Matterwave

I think what he wants is a general solution to the equations:

$$0=a+bx+cx^{1/2}$$
$$0=a+bx+cx^{1/2}+dx^{1/3}$$
$$0=a+bx+cx^{1/2}+dx^{1/3}+ex^{1/5}$$

etc.

My take: You can multiply by the lowest common denominator to obtain polynomial equations of degree LCD.

The first one you can turn into a cubic equation by multiplying by x^2. The second one you can turn into a polynomial of degree six by multiplying by x^6, and the last one you can turn into a polynomial of degree 30...

But I'm not sure if I'm adding or destroying roots or something by doing this procedure since the fractional powers give rise to branch cuts in the complex plane.

13. Jun 4, 2014

### Jhenrique

I want to find the polynomial equation $p(x)$ for a given $x = \sqrt[4]{A} + \sqrt[4]{B} + \sqrt[4]{C}$ or $x = \sqrt[4]{A + \sqrt[3]{B + \sqrt[2]{C}}}$

In other words, given a "radical equation", I want to know which is the polynomial equation of kind $ax^4+bx^3+cx^2+\cdots=0$ for this "radical equation".

14. Jun 4, 2014

### micromass

Ah, so you want this http://en.wikipedia.org/wiki/Minimal_polynomial_(field_theory [Broken])

Last edited by a moderator: May 6, 2017
15. Jun 7, 2014

### Jhenrique

Yeah... how find the minimal polynomial for:
$x = \sqrt[4]{A} + \sqrt[4]{B} + \sqrt[4]{C}$

and for:
$x = \sqrt[4]{A + \sqrt[3]{B + \sqrt[2]{C}}}$

?

Last edited by a moderator: May 6, 2017