Cubic and Quartics

[prog_tom]

Anyone know how to solve any one of them? Given in the form of

ax^3 + bx^2 + cx + d = 0

ax^4 + bx^3 + cx^2 + dx + e = 0

I have read some guides, but i still don't know how to solve them after 10 minutes!!!

[Peter1]

I think you have to find F'(X) and F''(X). Or summuts like that.

Pete

[Spooner]

Here's a starting point at Mathworld and you can follow some of the links.

[bugzpodder]

you can use softwares (Mathematica) to guess the answers, use factor theorem (if f(a)=0 then x-a is a factor of f(x)) or use cubic/quartic formula which can be found www.sosmath.com

[sanstylo]

There are two ways to solve these sort of equations...

1.) Generally if these kinda problems comein exams ...99.9% of the time ...one of the roots is (1,2,3 or -1,-2,-3, 0)...try all of them out on the equation(resultant value shud be '0') and then suppose u found one root to be 'A', divide the whole equation by (x-A)... u will end up with a quadratic if the original euation was cubic... there u go... im sure u will be able to solve the quadratic..

2.) this method is a bit more tougher than the above solution....u need a lot of practice...take bout five cubic or above equations wghich u allready have worked out the answers foir...and try to rearrange the eqautions in two parts ...where u can take summin and try and work out some come multiplicator..........if u get somewhere rest of the path shud be easy enough....cant really explain this method properly without a couple of practical examples...not possible on the message board...

neways...let meknow..if u were able to work out the roots by and of the above methods..

Cheers

Sandip
(Stop Existing Start Living)

[Guv]

A search of this forum should turn up something. Similar questions have been asked here before.

In modern times, successive approximations methods are used to find polynomial roots. These start with a guess at a root, followed by the use of an algorithm which provides a more accurate root. Repeated use of the algorithm usually results in an extremely accurate root.

There are special situations which cause trouble, requiring extra analysis.

For an odd polynomial, there is always at least one real root. For even polynomials, all the roots might be complex.

Complex arithmetic is usually used so that a root can almost always be found. If a complex root is found, there is a simple rule for determining another root, and a quadratic factor. Dividing by the quadratic factor results in a lower order polynomial. If a real root is found, the polynomial is divided by a linear factor to reduce it.

Try searching the web for Tartaglia or Cardano. this should result in finding a now obsolete method for solving a cubic. There is a method for 4th order polynomials, which is a nightmare (it requires solving a cubic first).

The Newton method of successive approximations is easy to understand and use. It is not too tuff to write a VB application to find roots of almost any polynomial. I wrote such an application just for fun.

If you want, I can explain the Newton method in more detail. I can also explain the Tartaglia analytical method (not recommended). I have long since forgotten the analytical method for 4th order polynomials. The analytic methods have theoretical and historical significance. I doubt that any one has used them for 50 or more years, except for a student in some math course.

BTW: It was proven over 100 years ago that there can be no general analytic solution for 5th order of higher polynomials, although there are many special cases that can be solved.

[sql_lall]

I found a nice way to solve a cubic of form x^3 + (3b^2)x = 2c^3, by Thomas Harriot, using the substitution
x = (e²-b²)/e

See here->
http://www-groups.dcs.st-and.ac.uk/h...ations.html#69
(abot 5/6th down the page)