178 CHAPTER 3 Inequalities
Next, writef(x) =an(xn+a′n− 1 xn−^1 +···+a′ 1 x+a′ 0 ), a′i=ai/an, i= 0, 1 ,...,n−1;
similarly, write
g(x) =bm(xm+b′m− 1 xm−^1 +···+b′ 1 x+b′ 0 ), b′j=bj/bm, j= 0, 1 ,...,n−1;
It follows easily thatR(f,g) =amnbnmR(f/an,g/bm), which reduces com-
putations to resultants ofmonicpolynomials.
Theorem 4.. Letf(x), g(x), andh(x)be polynomials with real coef-
ficients. Then
R(fg,h) =R(f,h)R(g,h).
Proof. From the above, it suffices to assume that all polynomials are
monic (have leading coefficient 1). Furthermore, by theFundamen-
tal Theorem of Algebraf(x) splits into linear factors, and so it is
sufficient to prove that the above result is true whenf(x) =x+ais
linear. Here, we let
g(x) =xn+an− 1 xn−^1 +···+a 1 x+a 0 , andh(x) =xm+bm− 1 xm−^2 +···+b 1 x+b 0.
First of all, one obviously has thatR(x+a,h) = det
S(x+a,h(x)) Z(^0) n,m+1 In
whereZ is the (m+ 1)×nmatrix
Z=
0 ··· 0 0
... ... ... ...
0 ··· 0 0
−an− 1 ··· −a 1 −a 0
.
Next, it is equally clear that