Stitz-Zeager_College_Algebra_e-book

# We note here that the reason r 0 is excluded from

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Unformatted text preview: ituted into A + B = 8 gives B = 4. Hence, A = B = 4 and we get 4x3 4 8x 4 √+ √ = 4x + 2 = 4x + 2−2 x x −2 x+ 2 x− 2 5. At ﬁrst glance, the denominator D(x) = x4 + 6x2 + 9 appears irreducible. However, D(x) has three terms, and the exponent on the ﬁrst term is exactly twice that of the second. Rewriting 2 2 D(x) = x2 + 6x2 + 9, we see it is a quadratic in disguise and factor D(x) = x2 + 3 . Since x2 + 3 clearly has no real zeros, it is irreducible and the form of the decomposition is x3 + 5 x − 1 x3 + 5 x − 1 Ax + B Cx + D = =2 + 4 + 6 x2 + 9 2 + 3)2 x x +3 (x (x2 + 3)2 When we clear denominators, we ﬁnd x3 + 5x − 1 = (Ax + B ) x2 + 3 + Cx + D which yields x3 + 5x − 1 = Ax3 + Bx2 + (3A + C )x + 3B + D. Our system is A B 3A + C 3B + D = 1 = 0 = 5 = −1 We have A = 1 and B = 0 from which we get C = 2 and D = −1. Our ﬁnal answer is x3 + 5 x − 1 x 2x − 1 =2 + 4 + 6 x2 + 9 x x + 3 (x2 + 3)2 6. Once again, the diﬃculty in our last example is factoring the denominator. In an attempt to get a quadrat...
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