Math

[NathanS]

in a equation like this: ax^2+bx+c, I can do some math and get the quadratic formula. But how do I isolate the x in this equation:
ax^3+bx^2+cx+d

 

[f95toli]

The equation az^3+bz^2+cz+d can by the substitution z=x-b/3a be reduced to

x^3+px+q=0

this equation has a solution given by Cardano's formula (with one, two or three roots).
Another way to solve it is by rewritting it as a system of three equations.

I suggest you look this up in a book, algebra written in ASCII has a tendency to get messy. Or you can try the following link
to Mathworld

 

[andyman7]

you could also do synthetic division if you know how to do that

 

[TDSLB]

I prefer organic....

 

[Mday]

the cubic formula is not one you want to memorize... nor is the one for 4th order...

 

[simonsky]

OK, hum, sorry about that but, how is this highly technical?

 

[blahblah99]

If you're looking for the roots of any order polynomial, there are numerical methods. To find roots of 2nd, 3rd, 4th order polynomials, there are formulas.

 

[oboeguy]

This IS a highly technical question. Try to give me a formula for a sixth degree polynomial... have fun. Seriously, did you even know if there is a formula for third degree polynomials, or stop to think if we have formulae for nth degree? :Q

For numerical methods, look-up stuff for zero-finding like "Newton's Method" or "Bisection".

 

[Shalmanese]

Isnt it provable that for n >= 5, there are no rational roots? Or s it only for 5 and 6?

 

[Special K]

Isnt it provable that for n >= 5, there are no rational roots? Or s it only for 5 and 6?

I think its for 5 and higher; here is a link I found:

link

Also I think what you meant (correct me if I'm wrong) to ask is that there is no explicit formula for finding the roots of a polynomial of degree 5 or greater. You can make a polynomial of degree 5 or higher with rational roots.

The polynomial x^6 - 5*x^5 - 51*x^4 + 161*x^3 + 818*x^2 - 564*x - 2520

has roots 7,6,2,-2,-3,and -5, for example

 

[f95toli]

Originally posted by: Shalmanese
Isnt it provable that for n >= 5, there are no rational roots? Or s it only for 5 and 6?

No. An example:
The equation

(x-1)(x-2)(x-.3)(x-4)(x-5)=0

is a 5th order equation and has the solution x=1,2,3,4,5 as you can easily verify..

 

[Shalmanese]

Sorry, I meant there exists non-rational roots for polynomials of degrees >=5. Got the wording mixed up.