What is the general formula for the discriminant of a polynomial of degree n?

1 Answer
Jun 14, 2017

Answer:

See explanation...

Explanation:

The discriminant of a polynomial #f(x)# of degree #n# can be described in terms of the determinant of the Sylvester matrix of #f(x)# and #f'(x)# as follows:

Given:

#f(x) = a_nx^n+a_(n-1)x^(n-1)+...+a_1x+a_0#

We have:

#f'(x) = na_(n-1)x^(n-1)+(n-1)a_(n-1)x^(n-2)+...+a_1#

The Sylvester matrix of #f(x)# and #f'(x)# is a #(2n-1)xx(2n-1)# matrix formed using their coefficients, similar to the following example for #n=4# ...

#((a_4, a_3, a_2, a_1, a_0, 0, 0),(0, a_4, a_3, a_2, a_1, a_0, 0),(0, 0, a_4, a_3, a_2, a_1, a_0),(4a_4, 3a_3, 2a_2, a_1, 0, 0, 0),(0,4a_4,3a_3,2a_2,a_1,0,0),(0, 0, 4a_4, 3a_3, 2a_2, a_1, 0),(0, 0, 0, 4a_4,3a_3,2a_2,a_1))#

Then the discriminant #Delta# is given in terms of the determinant of the Sylvester matrix by the formula:

#Delta = (-1)^(1/2n(n-1))/a_nabs(S_n)#

For #n=2# we have:

#Delta = (-1)/a_2abs((a_2,a_1,a_0),(2a_2,a_1,0),(0,2a_2,a_1))=a_1^2-4a_2a_0#

(which you might find more recognisable in the form #Delta = b^2-4ac#)

For #n=3# we have:

#Delta = (-1)/a_3abs((a_3, a_2, a_1, a_0, 0),(0, a_3, a_2, a_1, a_0),(3a_3, 2a_2, a_1, 0, 0),(0, 3a_3, 2a_2, a_1, 0), (0, 0, 3a_3, 2a_2, a_1))#

#color(white)(Delta) = a_2^2a_1^2-4a_3a_1^3-4a_2^3a_0-27a_3^2a_0^2+18a_3a_2a_1a_0#

The discriminants for quadratics (#n=2#) and cubics (#n=3#) are the most useful in that they tell you exactly how many real, repeated or non-real complex zeros a polynomial has.

The interpretation of the discriminant for higher order polynomials is more limited, but always has the property that the polynomial has repeated zeros if and only if the discriminant is zero.

#color(white)()#
Further reading

See http://www2.math.uu.se/~svante/papers/sjN5.pdf