5.2. Tests for Real Zeros 171
- Descartes’ rule of signs. Let p(t) be a polynomial over R. Prove that
p(t) cannot have more positive zeros counting multiplicity than there
are sign changes in its coefficients, and cannot have more negative
zeros counting multiplicity than there are sign changes in the coeffi-
cients of p(-t). - Use Descartes’ Rule of Signs to verify that tl’ + t” - 3t5 + t4 + t3 -
2t2 + t - 2 has at most 5 positive and 2 negative zeros. Deduce that
it has at least 4 nonreal zeros. - Verify Descartes’ Rule of Signs for each of the following quadratics
by actually determining its zeros:
(a) 8t2 - 8t + 1
(b) 4t2 - 4t + 1
(c) t2 - t + 1.
- Suppose in the real polynomial ant” +... + alt + ao, u,,ue # 0 and
ak = ak+l = 0 for some k with 1 5 k 5 n - 2. Prove that the
polynomial cannot have all its zeros real. - Prove that the real polynomial u,t”+a,-~t”-l+~~ .+u3t3+t2+t+1
cannot have all its zeros real. - There are various ways of obtaining estimates of an upper bound for
the real zeros of a real polynomial. For example, suppose that
f(t) = a&” + a,-lP-l +... + aptP +... + alt + a0
is a polynomial over R such that
(1) Q* >o
(2) there is at least one negative coefficient
(3) p is the largest value of i for which ai < 0.
Let M = maX{ Iail : ai < 0).
(a) Verify that, if r > 1, then
f(r) > a,r” - M(P+’ - l)(r - 1)-l.
(b) Prove that, if r > 1 + (M/cI,)“(~-~), then
a,(r - l)“-p > A4
and deduce that anrn(r - 1) > M+‘+‘.
(c) Deduce from (a) and (b) that if r > 1 + (M/a,)‘I(“-P), then
f(r) > 0.