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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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