We can compute the discriminant of any power of a polynomial. For example, the quadratic discriminant is given by . But it gets more complicated for higher-degree polynomials.
The discriminant of a cubic polynomial is given by
If , then the equation has three distinct real roots.
If , then the equation has a repeated root and all its roots are real.
If , then the equation has one real root and two non-real complex conjugate roots.
Alternatively, we can compute the value of the cubic determinant if we know the roots to the polynomial. Let and denote the roots of a certain cubic polynomial, then its discriminant is equal to
Here's a good example that requires us to use cubic discriminant:
For a polynomial having roots (counting multiplicity), its discriminant is
Notice some things:
- When there's a repeated root, the discriminant vanishes.
- If the coefficients of are all real, the discriminant is always a real number.
- If all roots are real and distinct, the discriminant is always positive.
For the case when the polynomial is cubic, i.e. we get .
Since the discriminant is symmetric in the roots of the polynomial, we can express it as elementary symmetric polynomials, i.e. the coefficients of . Although there is a general method to derive the discriminant of any polynomial, this is an elementary and algebraic approach.
By Vieta's formula we have
Let's expand the discriminant:
Let and , then . We can't directly compute and , but since they are cyclic polynomials, it turns out that the elementary symmetric polynomials of and are symmetric in and and hence expressable in terms of the coefficients of . So, let's find them:
The case for requires more tedious expansion:
Finally, use the identity