54 2 Algebra
168.LetP(z)andQ(z)be polynomials with complex coefficients of degree greater than
or equal to 1 with the property thatP(z)=0 if and only ifQ(z)=0 andP(z)= 1
if and only ifQ(z)=1. Prove that the polynomials are equal.
169.LetP(x)be a polynomial with all roots real and distinct and such that none of its
zeros is equal to 0. Prove that the polynomialx^2 P′′(x)+ 3 xP′(x)+P(x)also has
all roots real and distinct.
170.LetPn(x)=(xn− 1 )(xn−^1 − 1 )···(x− 1 ),n≥1. Prove that forn≥2,Pn′(x)is
divisible byPn/ 2 (x)in the ring of polynomials with integer coefficients.
171.The zeros of thenth-degree polynomialP(x)are all real and distinct. Prove that the
zeros of the polynomialG(x)=nP (x)P′′(x)−(n− 1 )(P′(x))^2 are all complex.
172.LetP(x)be a polynomial of degreen>3 whose zerosx 1 <x 2 <x 3 <···<
xn− 1 <xnare real. Prove that
P′
(
x 1 +x 2
2
)
·P′
(
xn− 1 +xn
2
)
= 0.
2.2.4 The Location of the Zeros of a Polynomial...................
Since not all polynomial equations can be solved by radicals, methods of approximation
are necessary. Results that allow you to localize the roots in certain regions of the real
axis or complex plane are therefore useful.
The qualitative study of the position of the zeros of a polynomial has far-reaching
applications. For example, the solutions of a homogeneous ordinary linear differential
equation with constant coefficients are stable (under errors of measuring the coefficients)
if and only if the roots of the characteristic equation lie in the open left half-plane (i.e.,
have negative real part). Stability is, in fact, an essential question in control theory, where
one is usually interested in whether the zeros of a particular polynomial lie in the open
left half-plane (Hurwitz stability) or in the open unit disk (Schur stability). Here is a
famous result.
Lucas’ theorem.The zeros of the derivativeP′(z)of a polynomialP(z)lie in the convex
hull of the zeros ofP(z).
Proof.Because any convex domain can be obtained as the intersection of half-planes,
it suffices to show that if the zeros ofP(z)lie in an open half-plane, then the zeros of
P′(z)lie in that half-plane as well. Moreover, by rotating and translating the variablez
we can further reduce the problem to the case in which the zeros ofP(z)lie in the upper
half-plane Imz>0. Here Imzdenotes the imaginary part.
So letz 1 ,z 2 ,...,znbe the (not necessarily distinct) zeros ofP(z), which by hypoth-
esis have positive imaginary part. If Imw≤0, then Imw−^1 zk>0, fork= 1 ,...,n, and
therefore