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Answers to Exercises; Chapter 3 277

2.13. No. 2t2+t-1 = (2t-l)(t+l) is reducible over Z, but irreducible over

Z2. However, if the leading coefficient of the polynomial is not divisible by

m, then irreducibility over Z, implies irreducibility over Z. (Any factor-

ization over Z would yield a factorization over Z, involving polynomials

of the same degree.)

2.14. (a) t2 + t + 1 is irreducible over Zs, hence over Z.

(b) 49t2 + 35t + 11 is irreducible over Zs, hence over Z.

(c) Over Zs, th e p o ly nomial is equal to t3 + t2 + 2t + 1. Since it has no

zero in Zs, it is irreducible over Zs. Since 3tl24, the polynomial must be

irreducible over Z.

2.15. (a) Over Z4, the polynomial is

3t4 + 2t3 + t2 + 2 = 3(t + 1)2(t2 + 2).

Over Z5, it is

3t4 + 3t3 + t2 + 2t = t(t + 3)(3t2 + 4t + 4).

Over Z7, it is

5t3 + 5t2 + 3t + 4 = (t + 5)(5P + t + 5).

Over Zg, it is

7t3 + 2t2 + 7t + 8 = (t + 2)(7t2 + 6t + 4).

(b) The results of (a) suggest a factorization of the form

(at2 + bt + c)(ut2 + vt + w)

with (a, b, c) 3 (7,6,4) and (u, v, w) - (0, 1,2) modulo 9. Since au = 63,
the only possibility is a = 7, u = 9. Since cw = -10, c E 4 (mod 9), w f 2
(mod 9), we must have c = -5, w = 2.
Since the factorization is equivalent modulo 7 to (t + 5)(5t2 + t + 5)
and since 9 E -5 (mod 7), we should have (a, b,c) E (0, -1, -5) and
(21, ?I, w) z (-5, -1, -5) modulo 7. Thus b E 6 (mod 9), b q 6 (mod 7)
leads to b E 6 (mod 63) and v - 1 (mod 9), v - -1 (mod 7) leads to
v E -8 (mod 63).
This leads to the trial (7t2 + 6t - 5)(9t2 - 8t + 2) which works.

2.16. lot5 + 3t4 - 38t3 - 5t2 - 6t + 3 = (5t2 + 9t - 3)(2t3 - 3t2 - t - 1).

3.1. (b) b divides every term but the first of the left side, and so divides
c,a”. Since gcd(u” , b) = 1, b must divide cn.

3.2. This is a consequence of Exercise 1.

3.3. (a) No rational zeros. (b) 4/3. (c) 1, -2/3. (d) -4, 7.