For the general definition, suppose
1 a_{n1} a_{n2} . . . a_{0} 0 . . . 0 0 1 a_{n1} a_{n2} . . . a_{0} 0 . . 0 0 0 1 a_{n1} a_{n2} . . . a_{0} 0 . 0 . . . . . . . . . . . . . . 0 0 0 0 0 1 a_{n1} a_{n2} . . . a_{0} n (n1)a_{n1} (n2)a_{n2} . . 1a_{1} 0 0 . . . 0 0 n (n1)a_{n1} (n2)a_{n2} . . 1a_{1} 0 0 . . 0 0 0 n (n1)a_{n1} (n2)a_{n2} . . 1a_{1} 0 0 . 0 . . . . . . . . . . . . . . 0 0 0 0 0 n (n1)a_{n1} a_{n2} . . 1a_{1} 0 0 0 0 0 0 0 n (n1)a_{n1} a_{n2} . . 1a_{1}
In the case n=4, this matrix looks like this:
1 a_{3} a_{2} a_{1} a_{0} 0 0 0 1 a_{3} a_{2} a_{1} a_{0} 0 0 0 1 a_{3} a_{2} a_{1} a_{0} 4 3a_{3} 2a_{2} 1a_{1} 0 0 0 0 4 3a_{3} 2a_{2} 1a_{1} 0 0 0 0 4 3a_{3} 2a_{2} 1a_{1} 0 0 0 0 4 3a_{3} 2a_{2} 1a_{1}
The discriminant of p(x) is thus equal to the resultant[?] of p(x) and p'(x).
One can show that, up to sign, the discriminant is equal to
In order to compute discriminants, one does not evaluate the above determinant each time for different coefficient, but instead one evaluates it only once for general coefficients to get an easytouse formula. For instance, the discriminant of a polynomial of third degree is a_{1}^{2}a_{2}^{2}  4a_{0}a_{2}^{3} 4a_{1}^{3} + 18 a_{0}a_{1}a_{2}  27a_{0}^{2}.
The discriminant can be defined for polynomials over arbitrary fields, in exactly the same fashion as above. The product formula involving the roots r_{i} remains valid; the roots have to be taken in some splitting field of the polynomial.
Search Encyclopedia

Featured Article
