**Unformatted text preview: **licity two. We have x3 3
A
3
B
C
=+
=
+
2+x
2
− 2x
x(x − 1)
x
x − 1 (x − 1)2 Clearing denominators, we get 3 = A(x − 1)2 + Bx(x − 1) + Cx, which, after gathering up
the like terms becomes 3 = (A + B )x2 + (−2A − B + C )x + A. Our system is A+B = 0
−2A − B + C = 0 A=3
Substituting A = 3 into A + B = 0 gives B = −3, and substituting both for A and B in
−2A − B + C = 0 gives C = 3. Our ﬁnal answer is x3 3
3
3
3
=−
+
2+x
− 2x
x x − 1 (x − 1)2 3. The denominator factors as x x2 − x + 1 . We see immediately that x = 0 is a zero of
multiplicity one, but the zeros of x2 − x + 1 aren’t as easy to discern. The quadratic doesn’t
factor easily, so we check the discriminant and ﬁnd it to be (−1)2 − 4(1)(1) = −3 < 0. We
ﬁnd its zeros are not real so it is an irreducible quadratic. The form of the partial fraction
decomposition is then x3 3
3
A
Bx + C
=
= +2
2+x
2 − x + 1)
−x
x (x
x
x −x+1 Proceeding as usual, we clear denominators and get 3 = A x2 ...

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