326 Answers to Exercises and Solutions to Problems2.21. (a) The real zeros are -a, 4, approx. 3.087.
(b) Zeros are -7, 2/3, -5/8, (l/2)(1 f ifi).
(c) 2t4 + 5t3 + t2 + 5t + 2 = (2P - t + 2)(P + 3t + 1).
(d) Polynomial is (4t - 7)(4t5 + 7t4 + 13t3 + 22t2 + 3t + 3). Real zeros
are 714 and approx. -1.6896.
(e) There are three real and four nonreal zeros. The polynomial factors
as (4t + 3)(t2 + t + 2)(4t4 - 7t3 - llt + 6). The nonrational zeros are
approximately 0.4912 and 2.1829.
3.1. (b) If [WI 5 1, then the inequality is satisfied. Otherwise 1~1~~ < 1
and (a) can be applied.
3.2. (b)
I%ol = I(%0 + 21 + *.. + 2,) - (Zl + *.. + &)I
< Izo+... +z*I+Itl+z2+*~~+%,~.Applying (a) and transferring terms yields the result. Strict equality holds
iff all the zi have the same argument.
3.3. (b) If ]w] < 1, the inequality holds. Otherwise, 1~l-l < 1 and, from
(4,
0 2 ]a,ur”][l - K(]w]-’ + ]w]-~ + +.. + ]w]-“1
> lanw”I(IwI - l)-l(]w] - 1- K).3.5. (a) By Descartes’ Rule of Signs, there is at most one positive zero.
Since the polynomial is negative at t = 0 and positive for large t, there is
at least one positive zero.
(b) Since the polynomial in (a) is positive for t > r, it follows that for
I4 > r,Iwn + an-lwnsl +... + alw + a01^2 lwnl - la,-lwn-l I - ... - laoI > 0.[A generalization of Exercise 2 and this Exercise appears in Emeric Deutsch,
Bounds for the zeros of polynomial, Amer. Math. Monthly 88 (1981), 205-
206.1
3.6. (1 - w)g(w) = b. + (b, - bo)w +... + (b, - b,&w” - b,,w”+‘. For
/WI 5 1, w # 1, we have from Exercise 2(b) thatI(1 - w)s(w)l > bo - lb1 - bol - lb2 - bll - .e. - lb, - bn-11 - lbnl= b. + (bl - bo) + (b2 - bl) +... + (b, - b,-l) - b, = 0.
Since, also, 1 is not a zero of g(t), the required result follows.
(b) w is a zero of g(t) iff w-l is a zero of hot” + bit”-’ +.. .+ b,-It + b,,
whence the required result follows from (a).
(c) It is straightforward to see that 0 < a,un 5 a,-lu”-’ 5 ... 5 alu 5
ao. If bi = aiui, then w is a zero of f(t) iff w/u is a zero of g(t). By (a),