264 Answers to Exercises and Solutions to Problems
9.20. Since (Z + Y)~ = 8xy and (z - Y)~ = 4xy, the answer is a.
Answers to Exercises
Chapter 2
1.5. -1062.6855.1.6. 7t5 - 2t3 - 3t2 + t + 2; 6; 53900.
1.7. 3t4 + 5t3 + t2 - 2t + 6; 2; 94.
1.10. (c) 325301 + 372391(t - 6) + 182521(t - 6)2 + 49656(t - 6)3+
8099(t - 6)4 + 792(t - 6)5 + 43(t - 6)6 + (t - 6)7.
1.11. -357 + 1157(y + 2) - 1585(y + 2)2 + 1184(y + 2)3 - 518(y + 2)4 +
132(y + 2)5 - lS(y + 2)6 + (y + 2)7,
1.13. If f(t) is the given polynomial, then the polynomial sought is f(t -3).
Expand f(t) in P owers of t +^3 to obtain f(t) = (t + 3)4 - 15(t + 3)3 +
83(t + 3)2 - 196(t + 3) + 163. The required polynomial is t4 - 15t3 + 83t2 -
196t + 163.
2.2. By Exercise 1, p(t) = (t - c)q(t) +p(c), from which the result follows.
2.3. If q(s) = 0, then p(s) = 0. Suppose 0 = p(s) = (s - r)q(s). Since an
integral domain has no zero divisors and s - r is nonzero, q(s) = 0.
2.5. Since u(b) = 0, we have that u(t) = (t - b)w(t) so that f(t) =
(t - a)(t - b)w(t) and v(t) = (t - a)w(t). Hence u(t) - v(t) = (a - b)w(t)
and the result follows.
2.6. t4-5t-6 = (t+l)(t3-t2+t-6) = (t-2)(t3+2t2+4t+3). The remaining
zeros of this quartic are those of (t3 + 2t2 + 4t + 3) - (t3 - t2 + t - 6) =
3(t2 + t + 3). (Ob serve that t4 - 5t - 6 = (t + l)(t - 2)(t2 + t + 3).)
2.7. (d) 4t5 - 3t4 - 7t2 + 6 = (t3 + 7t2 + 3t - 2)(4t2 - 31t + 205)+
(-1341t2 - 677t + 416).
2.9. Consider the set S = {f - gh : h E F[t]}. It contains a polyno-
mial r = f - gq of lowest degree. Suppose, if possible, deg r 2 deg g. Let
g(t) = a&“’ + ... and r(t) = bm+ktm+6 + .... The polynomial r(t) -
bm+ka;‘tkg(t) = f - s(q + b,+uG^1 t k ) belongs to S, but has degree less
than deg r(t), contradicting the choice of r. Hence deg r < deg g.
2.11. f(t) = q(t)(t - a)(t - b) + At + B, for some polynomial q(t) and
constants A and B. Substituting t = a and t = b yields
A = fta) - fcb),
a-bB = ‘ftb) - bf(a)
a-b ’2.13. (e) f = sls$ - 2s:s3 - S2S3.