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Answers to Exercises; Chapter 6 347

2.5. Let the zeros of t3 + at2 + bt + c be 2, y, z. Then

D = (r - Y)~(z - z)~(Y - z)~

= Cx4y2 -I- 2Cx3y2z - 2Cx4yz - 2Xx3$ - 6x2y2z2

= (a2b2 - 2b3 - 3c2 - 2a3c + 4abc) + 2(abc - 3c2)

• 2(a3c - 3abc + 3c2) - 2(b3 - 3abc + 3c2) - 6c2
  = a2b2 - 4a3c + 18abc - 4b3 - 27~~.

2.6. (a) 1, -1, i, -i; -256.

(b) i, -i, 2i, -2i; 5184.

2.7. The discriminant is nonzero _ all the zeros of the polynomial are

distinct, i.e. simple _ the polynomial and its derivative do not have any

zero in common _ the only common divisors of the polynomial and its

derivative are constant.

3.1. First solution. a + b + c = 2 and a2 + b2 + c2 = (e + b + c)” -

2(ab + ac + bc) = 2. Since a3 = 2a2 - a - 5, etc.,

a3 + b3 + c3 = 2(a2 + b2 -I- c”) - (a + b + c) - 15 = -13.

Since o4 = 2a3 - a2 - 5u , etc -,

a4 + b4 + c4 = 2(a3 + b3 + c”) - (u” + b2 + c2) - 5(u + b + c) = -38.

Second sohtion. u4 + b4 + c4 = sf - 4sfs2 + 2s; + 4s3s1 = -38.

3.3. For each i,

tl +U,-ltl-l+ .“+Ulti +Uo = 0.

Multiplying by 2’ and summing over 1 2 i 5 n yields the result.

3.7.

p1 + Cl = (^0) Pl = -cl
p2 +wJ, +2co = 0 p2 = c: - 2co
p3+~1~2+~Clpl=o p3=-c:+3cOcl
p4 +c1p3 +copz = 0 p4 = c: -4c,c:+ 2c;
p5 + qp4 +cop3 = 0 p5 = -c: +5coc:: -5&.
For t2 - 3t + 2, pl = 3, pa = 5, p3 = 9, p4 = 17, p5 = 33. For t2 + t + 1,
p1 = p2 = p4 = p5 = -1, p3 = 2.
3.8. pl = -7; p2 = 61; p3 = -466; p4 = 3621; p5 = -28082.
3.9. Let zr,zz,... , z, be zeros of the polynomial
f(t) = t” + C,-lP--l +. a. f qt + co,
and let pk be the sum of their bth powers. Then, by exercise 6, c,,-k = 6
for k = 1,2,..., n, so that f(t) = t”. Hence, each zi vanishes.