SECTION 3.6 Cubic Discriminant 169
Case (i): ∆>0. That is to say, (x 2 −x 1 )^2 >0 and so certainly the
zerosx 1 andx 2 are distinct. Furthermore, if they were not real,
then they would have to be complex conjugates of one another and
this would force (think about it!) (x 2 −x 1 )^2 <0 (asx 2 −x 1 would
be purely imaginary). Therefore∆>0 =⇒ Qhas two distinct real zeros.Case (ii): ∆ = 0. This is clear in that one immediately has that
x 1 =x 2. That is to say∆ = 0 =⇒ Qhas a double zero.Case (iii): ∆<0. Since (x 2 −x 1 )^2 <0 we certainly cannot have both
x 1 andx 2 real. Therefore, they’re both complex (non-real) as they
are complex conjugate. Therefore∆<0 =⇒ Qhas two complex (non-real) zeros.That is to say,Dand ∆ satisfy the same trichotomy law!Whereas the definition of D does not suggest a generalization to
higher-degree polynomials, the definition of ∆ can be easily generalized.
We consider first a natural generalization to the cubic polynomial
P(x) =ax^3 +bx^2 +cx+d, a, b, c, d∈R, a 6 = 0.By theFundamental Theorem of Algebra, we know that (counting
multiplicities),P(x) has three zeros; we shall denote them by x 1 , x 2 ,
andx 3. They may be real or complex, but we do know that one of
these zeros must be real.
We set∆ =a^4 det
1 x 1 x^21
1 x 2 x^22
1 x 3 x^23
2.