My Method for solving Polynomial where Order = 3

[NotLKH]

How's My Math?


{of course, once you know your Y's, you can find a, b, and then the other two roots of X }

-Lou

[nishantp]

Lol I just learned quadratics at school...so i dont understand any of that

[Guv]

NotLKH: Assuming that your formula is correct, it still seems like a lot of work to find a root of a special case cubic.

It might be fun to develop such a formula, but it does not seem very practical. Newton-Raphson seems to be easy to program for any arbitrary cubic. Once you get one root of a cubic, it is fairly easy to determine the quadratic factor and find the other two roots.

I would not like to deal with anything higher than a quadratic without some programable device.

My HP calculator has a built in polynomial root finder that seems to work for order ten or more. I do not know what the limit is.

I developed a VB application which will find all roots (real & complex) of a polynomial. I have tried it for various polynomials up to order 21, and believe that it will do any order for which I am willing to enter the coefficients. I am not sure that I can accurately enter more than 30 or so coefficients.

BTW: As roots are found, my application uses synthetic division to reduce the order of the polynomial.

While all but the first one or two roots are found using reduced polynomials, Newton-Raphson versus the original polynomial is used to refine the precision of roots. I have discovered that roundoff errors in the coefficients of the reduced polynomials can result in significant loss of precision in the last 2-6 roots found for a high order polynomial.

[bugzpodder]

Here is a brief of the cubic and quartic formula -- read on only if you are interested:

(note, if there is a co-efficient in front of the highest degree of x, divide the whole equation by that)

cardino's cubic formula:

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

sub in (y-b/3), and expand, you get:

y^3+ex=f (the coefficient infront of x^2 is 0)

if you sub in y=s-t,e=3st,f=s^3-t^3, you'll see that it is an identity (always true) -- where e,f are constants
so,

e=3st, s=e/3t
sub s=e/3t in f=s^3-t^3
e^3/9t^3-t^3=f
multiply both sides by t^3

e^3/9-t^6=f(t^3)
t^6+f(t^3)-e^3/9=0
this is a quadratic in terms of t^3. solve for t^3 and take the cube root of it.
similarily, solve for s.

since y=s-t is true, and you solved for s,t. you know what y is
since x=y-b/3, you solved for one of the roots of x
use synetic division to reduce the original equation and you get a quadratic in which you can use the quadratic formula.

Quartic formula

x^4+bx^3+cx^2+dx+e=0
complete the square for the first 3 terms and move leftovers to the right hand side
so now you have

(....)^2=fx^2+gx+e

now add in an extra variable y and make sure you DON'T change the equation by adding equilvalent stuff to the RHS

(...+y)^2=some mumble jumble quadratic in terms of x
since the LHS is a perfect square, that means the RHS has a discriminant of 0 (not going to go into theries). set the discriminant to 0 and solve for y for a numerical value

take the square-root of BS and you have a quadratic in terms of x. solve for x and reduce the qartic to cubic using synthetic division and use the cubic formula

[NotLKH]

Originally posted by bugzpodder
Here is a brief of the cubic and quartic formula -- read on only if you are interested:

(note, if there is a co-efficient in front of the highest degree of x, divide the whole equation by that)

cardino's cubic formula:

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

sub in (y-b/3), and expand, you get:

y^3+ex=f (the coefficient infront of x^2 is 0)

if you sub in y=s-t,e=3st,f=s^3-t^3, you'll see that it is an identity (always true) -- where e,f are constants
so,

e=3st, s=e/3t
sub s=e/3t in f=s^3-t^3
e^3/9t^3-t^3=f
multiply both sides by t^3

e^3/9-t^6=f(t^3)
t^6+f(t^3)-e^3/9=0
this is a quadratic in terms of t^3. solve for t^3 and take the cube root of it.
similarily, solve for s.

since y=s-t is true, and you solved for s,t. you know what y is
since x=y-b/3, you solved for one of the roots of x
use synetic division to reduce the original equation and you get a quadratic in which you can use the quadratic formula.

Quartic formula

x^4+bx^3+cx^2+dx+e=0
complete the square for the first 3 terms and move leftovers to the right hand side
so now you have

(....)^2=fx^2+gx+e

now add in an extra variable y and make sure you DON'T change the equation by adding equilvalent stuff to the RHS

(...+y)^2=some mumble jumble quadratic in terms of x
since the LHS is a perfect square, that means the RHS has a discriminant of 0 (not going to go into theries). set the discriminant to 0 and solve for y for a numerical value

take the square-root of BS and you have a quadratic in terms of x. solve for x and reduce the qartic to cubic using synthetic division and use the cubic formula

Sorry, too many words... whats this mean???

[bugzpodder]

solving the roots of a cubic and quartic equation (3rd/4th degree polynomial)

[Guv]

The posted methods for algebraic solutions of cubics and quartics seem strange or perhaps erroneous to me.

You can always force the high order coefficient of a cubic to be one by dividing by the high order coefficient if it is not one. Therefore the problem is to solve the following.

x³ + bx² + cx + d = 0

If you make the substitution x = (y - b/3), you get a cubic with no second order term.

y³ + py + q, where (p, q) are simple expressions in (b, c, d).

Up to here, the posted solutions seem correct, but then get unnecessarily complicated.

At this point, you can make the substitution y = (z - p/3z), resulting in the following.

Z³ + q + p³/27z³ = 0

Now multiply by Z³ and you get a quadratic in z³, which can readily be solved. Then make the appropriate substitutions to get a root of the original cubic. Then factor out the known root and solve the remaining quadratic.

The introduction of s & t seem to be an unnecessary complication.

I do not remember the details of the solution of the fourth order polynomial. I have a vague memory of some algebraic manipulations followed by solving a cubic This resulted in a quartic which was the square of a quadratic, an easily solvable form.

BTW: Over 100 years ago, it was proven that no such solutions are possible for polynomials of order greater than four.

[bugzpodder]

just a clearafication,

Over 100 years ago, it was proven that no such solutions are possible for polynomials of order greater than four.

It is not possible to develop a general formula to solve for the roots for an equation of a polynomial with a power of 5 or greater. special cases are still possible, such as: solve for x if x^5=0. of course, the only solution is when x=0 the equation holds.

[DavidHooper]

is not possible to develop a general formula to solve for the roots for an equation of a polynomial with a power of 5 or greater. special cases are still possible, such as: solve for x if x^5=0. of course, the only solution is when x=0 the equation holds.

Don't you think we already know this?

[bugzpodder]

to ppl whos as smart as u, yep. to other ppl, it might be somewhat misleading.

[[Digital-X-Treme]]

http://mathworld.wolfram.com/CubicEquation.html

[Gordon Bennett]

It is entirely possible to develop solutions to 5th and above order polynomials when they are put equal to zero, only such general solutions would not be of any real use since that order polynomial rarely comes up in applied science.

[bugzpodder]

No. there is no *general* solutions for equations of polynomials of order 5 and greater! it is proven to be mathematically impossible. however in special cases such as
(x-1)(x-2)...(x-100)=0

FYI the equation doesn't need to be equal to 0 when you solve it.

You can just do something to both sides (such as subtract 5 on both sides) to make it equal to 0 and then solve.