# Polynomial 3 degrees

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1. Jan 26, 2016

### Stephanus

Dear PF Forum,
As we know in polynomial 2 degrees AX2 + BX + C = 0, there's a formula for solving it.
What about 3 degrees for example: AX3 + BX2 + CX + D = 0, there's is really no formula for solving it?
The only way to solve it is by hand?
I have several methods in my head, at least programming method, but none having formulas in it.

So, there's really no formula?

Thanks.

2. Jan 26, 2016

### Staff: Mentor

I don't know what your difference between 'formula' and 'by hand' is. But here are the formulas:
https://en.wikipedia.org/wiki/Cubic_function#General_formula_for_roots

There is also a solution for polynomials of 4th degree on Wikipedia but unfortunately not in English.
For 5th degree and higher there do not exist formulas, i.e. it can be proven that they cannot exist in general.

EDIT: Sorry. Here are the roots of quartic polynomials:
https://en.wikipedia.org/wiki/Quartic_function#General_formula_for_roots

Last edited: Jan 26, 2016
3. Jan 30, 2016

### bigfooted

It can be proven that you cannot write the roots of a fifth degree polynomial in terms of radicals (basically a formula containing only a finite number of the operations +,-,x,/ and powers). If you are willing to accept more complicated functions like the elliptic modulus or hypergeometric functions you can still find closed form solutions, but the results are not easy to interpret anymore (and some might even go as far as to say these results are quite useless):
http://mathworld.wolfram.com/QuinticEquation.html

4. Jan 30, 2016

### WWGD

I think the solvability depends on the associated Galois group of the polynomial. Then, e.g., $A_5$ is not solvable.

5. Jan 30, 2016

### Staff: Mentor

Yes, but only in the classical sense of algebraic field extensions. The link lead us to the dark side of mathematics ...

6. Jan 30, 2016

### WWGD

Sorry, my algebra is kind of rusty, would you expand?

7. Jan 30, 2016

### Staff: Mentor

One cannot explicitly solve polynomials with degree 5 or higher. As you've mentioned the Galois groups $A_n$ aren't solvable anymore.
I have to be cautious because otherwise I mess it up with one of the 3 classics: squaring the circle, doubling the cube or divide angles into three by means of compass and ruler.
Were the Galois groups solvable, then one could get a chain of algebraic extensions of the prime field ℚ in which the roots can be expressed, i.e. in terms of algebraic formulas. The automorphism groups of the (algebraic, finite) field extensions correspond to the symmetries of the polynomial's roots.
The formulas in the link for "solving" polynomials of degree 5 are found in ℂ and not fun to read. Therefore the dark side of math.

8. Jan 30, 2016

### Staff: Mentor

There are some 5th and higher degree polynomial equations that you can solve, such as $x^5 - x^4 - 13x^3 + 13x^2 + 36x - 36 = 0.$ This one can be factored into $(x - 1)(x^4 - 13x^2 + 36) = 0$, yielding roots of $x = 1, \pm 2, \pm 3$.

But as Galois proved, there does not exist a formula for solving an arbitrary 5th or higher degree polynomial equation.

9. Jan 30, 2016

### Staff: Mentor

I once have had his originally published version in hand. I couldn't find anything known to me in it ...

10. Feb 1, 2016

### micromass

Staff Emeritus
Actually, it was proven by Abel and Ruffini. https://en.wikipedia.org/wiki/Abel–Ruffini_theorem Galois offered another solution only many years after Abel did. Galois' solution is the best one though.

11. Feb 1, 2016

### WWGD

I think a solution by radicals exists when the Galois group is solvable.

12. Feb 1, 2016

### Staff: Mentor

That might be so. All I was doing was providing a counterexample to what fresh_42 said,
There are some 5th and higher degree polynomial equations that can be solved, but there is no general formula for solving these higher degree polynomials.

13. Feb 1, 2016

### WWGD

Right, and you can guarantee that there will exist some that cannot, like (assuming I am correct) those whose associated group is not solvable. Sorry to keep at it, just want to make it clear, in part for my own sake.

14. Feb 1, 2016

### Staff: Mentor

That has been an abbreviation to what I've already said before: "i.e. it can be proven that they cannot exist in general."

In the given example there are no field extensions needed and therefore the Galois group is trivial.
If The Galois group is solvable, then there is a chain of normal subgroups $\{1\} < G_1 < ... < G_n = G$. To each quotient there is an algebraic filed extension (starting with ℚ) in which more and more roots can be expressed by radicals. The correspondence between algebraic field extensions and Galois groups as their automorphism groups is 1:1.
Since the groups $A_n$ are all simple for n > 4 Abel-Ruffini follows.