Introduction to Abstract Algebra, Solutions Manual, 4th Edition

4.4 Partial Fractions

1. Suppose r₀+ r₁p + = s₀ + s₁p + where p is monic in R[x] and each r_i and s_i is either zero or has degree less than p. Then if t_i = r_i − s_i we have t₀ + t₁p + t₂p² + = 0, so t₀ + (t₁ + t₂p + )p = 0. The uniqueness in the division algorithm (Theorem 4 §4.1) shows t₀ = 0 and t₁ + t₂p + = 0. Then do it again to get t₁ = 0 and t₂ + t₃p + = 0. This continues to show t_i = 0 for all i.

2.

a. x² − x + 1x(x² + x + 1) = ax + bx + cx² + x + 1, so

Evaluating at 0, 1 and −1 gives 1 = a, 1 = 3a + b + c, 3 = a + b − c. Hence a = 1, b = 0, c = − 2.

c. x + 1x(x² + 1)² = ax + bx + cx² − 1 + dx + e(x² + 1)², so

Evaluating at 0, gives 1 = a; the coefficients x⁴, x³, x², x give ...