# Finding the discriminant and roots of a polynomial

How is the discriminant of a polynomial determined? I know that for a quadratic function, the roots (where $f(x)=0$) are found by $$x=\frac{-b\pm\sqrt{\Delta}}{2a}$$
and here $\Delta$ is the discriminant. The discriminant is what defines the nature of the roots (if they'll be real or complex, depending on whether $\Delta>0$ or $\Delta<0$).

How do you determine the discriminant of a cubic polynomial and higher polynomials?
In regards to quadratic polynomials: $$x=\frac{-b\pm\sqrt{\Delta}}{2a}$$ is found by completing the square, which can be done to all quadratic polynomials, and then this formula is applied to find the roots.
But for cubic polynomials, quatric polynomials, degree $7$ polynomials, etc,..., how do you find the roots? (Also how do you determine the discriminant?)

If you have a cubic polynomial, sometimes you can factor it and make it a quadratic polynomial multiplied by another term: $$9t^3-18t^2+6= 3t(3t^2-6t+2)$$ but there are cases when you can't simplify it like this, right?

So I am just curious about how to find the roots/discriminant in higher degree polynomials.

-
There are explicit methods for polynomials up to degree $4$. After that, all polynomials of higher degree only have formulas for specific forms. The fundamental theorem dealing with polynomials guarantees roots across the complex numbers, but there is no guarantee that they are easy to find. Look up the results due to Galois, and how to look at the roots of a polynomial as a set together. – abiessu Aug 29 '13 at 22:59

## 2 Answers

If $x_1, \ldots, x_n$ are the roots of a polynomial $f$, i.e. $f(x)=(x-x_1)\cdots (x-x_n)$, then the discriminant of $f$ is defined as $\Delta=\prod_{1\le i<j\le n}(x_i-x_j)$. Then $\Delta^2$ is an expression that is symmetric in the $x_i$, hence can be expressed using the elementary symmetric polynomials, which are the coefficients of $f$. While $\Delta$ tells us something about the behavipur of the roots (especially, $\Delta=0$ iff there are multiple roots), it is not the end of the story for higher polynomial degrees. As has been hinted in the comment by abiessu, Galois theory shows that no general method using radicals exists to solve degrees five and above.

-

If a polynomial is given numerically, (coefficients $a_0..a_n$ are given), the resultant method can be used to get at a numerical value of the discriminant. The coefficients of the polynomial and its derivative are put in a (n+2)- squared Sylvester matrix. Then the determinant is the discriminant wanted. Where writing out the discriminant of a matrix containing symbols is prohibitive, the discriminant can be calculated swiftly numerically using existing matrix packages in $O(N^3)$ time.

It is very nicely explained in

http://www.math.uu.se/~svante/papers/sjN5.pdf

This is example 4.7

If $f(x) = ax^4 + bx^3 + cx^2 + dx + e$, then Theorem 3.3 yields

                             | a  b  c  d  e  0  0  |
| 0  a  b  c  d  e  0  |
| 0  0  a  b  c  d  e  |
| 4a 3b 2c d  0  0  0  |
| 0  4a 3b 2c d  0  0  |
| 0  0  4a 3b 2c d  0  |
| 0  0  0  4a 3b 2c d  |


= $b^2c^2d^2 - 4b^2c^3 e - 4b^3d^3 + 18b^3cde - 27b^4e^2 - 4ac^3d^2 + 16ac^4e + 18abcd^3 - 80abc^2de - 6ab^2d^2e + 144ab^2ce^2 - 27a^2d^4 + 144a^2cd^2e - 128a^2c^2e^2 - 192a^2bde^2 + 256a^3e^3$

-