# Is there formula for zeres of a cubic polynomial

1. Oct 18, 2012

### 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 ?

2. Oct 18, 2012

### 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.

3. Oct 18, 2012

### arildno

4. Oct 18, 2012

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

Last edited by a moderator: Oct 18, 2012
5. Oct 26, 2012

### vrmuth

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