Is there formula for zeres of a cubic polynomial

[vrmuth]

is there any general formula to find out zeros of a cubic polynomial that will give you all the zeros ? if not please tell me what are the different methods to find out the zeros , guessing and trial and error , numerical etc. i want to see where are each methods useful and is there any subcategory of cubic polynomial for which we have some specific method to find out all the zeros ?

 

[Number Nine]

Yes. It's long and inefficient. A google search for "cubic root formula" should do the trick; or just look up cubic polynomials on wikipedia.

 

[arildno]

It is called Cardano's formula, and its full derivation can be found here:
http://www.proofwiki.org/wiki/Cardano's_Formula

A quartic polynomial has also an explicit formula for its zeroes; it is even worse, and you can find it here:
http://en.wikipedia.org/wiki/Quartic_function

 

[HallsofIvy]

Just because I enjoy doing it, here is a derivation of Cardano's formula:

If a and b are any two real numbers then (a- b)^3= a^3- 3a^2b+ 3ab^2- b^3 and 3ab(a- b)= 3a^2b- 3ab^2 so that (a- b)^3+ 3ab(a- b)= a^3- b^3. If we let x= a- b, m= 3ab, and n= a^3- b^3, we have the x^3- mx= n. That is a "reduced" cubic equation. (Reduced because there is no "x^2" term. Given any cubic equation, x^3+ px^2+ qx+ r= 0, we can always replace x by y- a, then choose a so that the coefficient of y^2 is 0.)

Now, the question is, suppose we know m and n, can we solve for a and b and so find x? The answer is, of course, yes. From m= 3ab, b= m/3a so that n= a^3- b^3= a^3- m^3/3^3a^3 and, multiplying through by a^3, na^3= n(a^3)^2- (m/3)^3= 0. That is a quadratic equation for a^3 that we can solve using the quadratic equation:
a^3= \frac{n\pm\sqrt{n^2- 4(m/3)^3}}{2}= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}

Since a^3- b^3= n,
b^3= a^3- n= -\frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}

 

[vrmuth]

Just because I enjoy doing it, here is a derivation of Cardano's formula:

i also enjoyed it but find little tough to get it ,i need to apply it for a polynomial thanks .