Cubic equations, etc. This is not a piece of homework! This is a question for Mathematics lovers?

Question

Please, read the whole question before asking. This is a serious question.

Thank you.

Some examples:

1) x^2 - 3x^2 + 4x -12 = 0 can be solved so:
(x-3)(x^2+4) = 0

2) x^4 + ax^3+ bx^2 + cx + m^2 = 0, m= c/a can be solved so x+m/x = z

3) ax^5 + bx^4 + cx^3 + ax^2 + bx + c = 0 can be solved so: (x^3 + 1)(ax^2 + bx +c) = 0

Do you know other particular cases?

Regards

Ana, Uruguay

Sorry, there is a typo error in the first example, it should be x^3.

The question is about CUBIC EQUATIONS, QUARTIC EQUATIONS, ETC. not about quadratic equations. Thank you.

you can turn an 8th grade equation into a quartic

You can even do this:

x^8 + bx^7 + cx^6 + dx^5 + ex^4 + dmx^3 + cm^2 x^2 + cm^3 x + m = 0

can be solved using x+mx = z

Later

I know that there are the Tartaglia-Cardana way to solve quartic and cubic equation. But these are very complicated ways to solve equations. I am asking about particular cases who can easily be solved, the kind of issue that you would teach to a highschool students.

PLEASE, IF YOU HAVENT ENABLED THE OPTION SO THAT I CAN CONTACT YOU, WRITE TO MY ACCOUNT AND TELL ME HOW I CAN CONTACT YOU IF IM INTERESTED IN YOUR ANSWER
THANK YOU
ANA

Answer 1

These should be pretty easy to make. After all, to make a factorable cubic or quartic, all you have to do is multiply a quadratic by the appropriate number of linear factors.

Answer 2

There are formulas for solving ANY cubic or quartic equations. General quintic (fifth degree) equations cannot be solved all the time, although of course, specific examples can be solved. Specifically, any 5th degree equation that has a known root can be factored into a linear times a quartic, and the ensuing quartic equation can be solved.

I actually derived the cubic formula myself when i was bored one day. I got very lucky. But you can find it if you look. It is much more involved that then quadratic formula, and often produces very messy answers (square roots inside of cube roots, etc).

I think it was Galois who first proved a general fifth degree (or higher) cannot be solved. I think. It's been so long.

Answer 3

chappa is correct. It was Galois who proved that, in general, equations of degree 5 and higher cannot be solved by means of *finitely* many algebraic terms involving powers and roots.

Tartaglia-Cardan (for cubics) and Ferrari (for quartics) will always work even if the answer is a complex conjugate pair (or a couple of them, in the case of a quartic).

If you just want to generate a polynomial with real roots, start with (x-a0)*(x-a1)*(x-a2) ..... Multiply it out and the roots will be exactly the a0, a1, a2..... terms. It's messy, but higher-order polynomials have a way of being like that


Doug

Answer 4

I actually have an appreciation for what actual math is. The further you delve, the more effective you start up to get exhilaration from what can in uncomplicated words be defined because the straightforward complexities for mathematics. inspite of the undeniable fact that, in extreme college, the point maximum folk study maths to, it isn't about the nature of maths. it truly is about the complicated equations and manipulation of formulation. it is actual that what you spot in college isn't something like professional mathematics. inspite of the undeniable fact that the stuff in college is all maximum see.

Answer 5

What is your answer for real? are you looking a way to solve equation with differents exponential? If that is your question, all I can do is recommend you to look for Pascal's Triangle, it's used to solve polinomial equations with high exponentials.

For more info you can visit:
http://en.wikipedia.org/wiki/Pascal%27s_triangle

You better explain your question beeing clear as possible
However, your question is not quite clear

Answer 6

Reporting a typographical error in the 1st equation: x^2 should be x^3. ?

Answer 7

Basically quadratic equations are of only two degree equations.According to theory of number system it may have only two solutions either real or imaginary.There are particularly three cases in quadratic polynomials.
1.when discriminant =0 ----two real and repeated roots
2.when discriminate >0----two real and distinct roots
3.when discriminant <0-----no real roots
There are various methods to solve them.I have a lot of collection on this topic as i am also a student of mathematics.But unable to write so much on computer.

Answer 8

i don't know any particular case but ilove mathematics....

Answer 9

I know it...

the answer is....ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh


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