Advanced High-School Mathematics

180 CHAPTER 3 Inequalities

zerosα 1 , ..., αn. Then

R(f,g) =amn

∏n

i=1

g(αi).

Proof. Letg(x) have zerosβ 1 ,...,βm. Since

g(x) =bm(x−β 1 )···(x−βm),

we see that g(αi) = bm(αi−β 1 )···(αi−βm). From this, the result
follows instantly.

3.7.2 The discriminant as a resultant

As already given above, the discriminant of the polynomial f(x) =
anxn+ lower-degree terms and having zerosx 1 , x 2 , ..., xnis given by

∆(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

=a^2 nn−^2

∏

i<j

(xj−xi)^2.

We relate the above to the resultant as follows. Let f(x) = anxn

• lower-degree terms, where an 6 = 0, and let α 1 ,...,αn be the zeros
  off(x). Then, using Corollary 3 above, and lettingf′=f′(x) be the
  derivative off, we have that

R(f,f′) =ann−^1

∏n

i=1

f′(αi).

Next, we havef(x) =an(x−α)···(x−α); applying the product rule
for differentiation leads quickly to

f′(x) =an

∑n

i=1

(x−α 1 )···(¤x−αi)···(x−αn),