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284 Answers to Exercises and Solutions to Problems

Solutions to Problems

Chapter 3

7.1. (a) (2x2 + 1)2 - 4x2 = (2x2 +2x + 1)(2x2 - 2x + 1).
(b) (x2 - 2)2 - 16x2 = (x2 +4x - 2)(z2 - 4x - 2).
(c) (xy - x - y + l)(xy - 1) = (Z - l)(y - l)(zy - 1).
(d) (xz - l)(yz - l)(xy - 1).
(e) 3(a - b)(b - c)(c - a).
(f) (a + c)(b + c)(a + b - c).
(d (a f b)(b + c>(c + a).
(h) (a + b)(b + c)(c - a).
(i) (x2 + x + f)(xd - x7 + x5 - x4 + x3 - x + 1).
i”k: :‘z;;z,x - 2zz + z2 + 3Y2).
z - xyt - 2)(X + y + z - xyz + 2).
(1) (f- Y)(Y - z>(z - x>(xy + yz + zx).
(m) 12abc(a + b + c).
(n) 3ubc(a + b)(b + c)(c + u).
(0) 2ubc(a + b + c).
(P) (z + Y + s)“[(z + Y + z)3 - 5(x + y + z)(xy + yz + ZX) + 15xyz] =
(x + y + z)“[x” + y3 + z3 - 2x2y - 2xy2 - 2y2z - 2yz2 - 2z2x - 22x2 + 6xyz].
(d (x + Y + z + w>(x + Y - z - W)(X - y + w - z)(x - y - z + w).
(r) (xy - z2)(yz - x2)(.2x - y2).
(s) (x2 + y2 - s2)(y2 + 22 - X2)(%” + x2 - y2),
(t> (x + Y>(Y + z)(z + 3).
(u) [(a” + b2) + (2ab - ~“)][(a~ + b2) - (2ab - c2)] = [(a + b)2 - c”]
[(a - b)2 + c”] = (a $ b + c)(u + b - c)(a” + b2 + c2 - 2ab).
7.2. (x4 - x3 + x2 + 2x - 6) = (x2 - x + 3)(x2 - 2).
(x4+x3+3x2+4x+6)=(x2-x+3)(x2+2x+2).
7.3. Since [P(x)]~ - x should vanish when x = 1, 2, 3, we require that
p(1) = 1, p(2) = 21j5, p(3) = 31i5. A possible polynomial p is the quadratic
ax2 + bx f c, where

2~ = 31i5 - 26/5 + 1, b = 21i5 - 1 - 3a, c = 1 - a - b.

7.4. Since i should be a zero of the polynomial, we have that (-a + b- 1) +
(a + b - 5)i = 0. H ence a = 2, b = 3, and

(2x + 3)(x5 + 1) - (5x + 1) = 2(x6 + 1) + 3x(x4 - 1)

= (x2 + 1)[2(x4 - x2 + 1) +3x(x2 - l)].

7.5. (a) The zeros of x2 + px + 1 are reciprocals, say r and l/r. These are
also zeros of ax3 + bx + c. Since the sum of the zeros of the cubic is 0, its