174 CHAPTER 3 Inequalities
R(f) = det
a b c d 0
0 a b c d
3 a 2 b c 0 0
0 3a 2 b c 0
0 0 3a 2 b c
Show thatR(f) =−aD(f),whereD=D(f) is the discriminant
of f(x). (This is the generalization of the result of Exercise 6 to
cubic polynomials.)3.7 The Discriminant (Optional Discussion)
In this section I’ll give a couple of equivalent definitions of the discrim-
inant of a polynomial of arbitrary degree. Not all proofs will be given,
but an indication of what’s involved will be outlined. To this end, let
there be given the polynomial
f(x) =anxn+an− 1 xn−^1 +···+a 1 x+a 0 ,wherean 6 = 0 and where all coefficients are real. Denoting byx 1 , x 1 , ..., xn
the zeros off(x) (which may include several complex-conjugate pairs
of imaginary zeros), we know that (by the Factor Theorem)
f(x) =an(x−x 1 )(x−x 2 )···(x−xn).In analogy with the above work, we define thediscriminantoff(x)
by setting
∆ = ∆(f) =a^2 nn−^2 det
1 x 1 x^21 ··· xnn−^1
1 x 2 x^22 ··· xn 2 −^1
... ...1 xn x^2 n ··· xnn−^1
2.
The above involves the determinant of the so-called Vandermonde
matrix,