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**Unformatted text preview: **+ D
A (E 3) (E 4)
B =
0
=
1
= −1
= −6 From
From
From
From equating
equating
equating
equating coeﬃcients of x3
coeﬃcients of x2
coeﬃcients of x
the constant terms To solve this system of equations, we could use any of the methods presented in Sections 8.1 through
8.5, but none of these methods are as eﬃcient as the good old-fashioned substitution you learned
in Intermediate Algebra. From E 3, we have A = −1 and we substitute this into E 1 to get C = 1.
Similarly, since E 4 gives us B = −6, we have from E 2 that D = 7. Our ﬁnal answer is
x2 − x − 6
1
6
x+7
x2 − x − 6
=22
=− − 2 + 2
4 + x2
x
x (x + 1)
xx
x +1
which matches the formula given in (2). As we have seen in this opening example, resolving a
rational function into partial fractions takes two steps: ﬁrst, we need to determine the form of
the decomposition, and then we need to determine the unknown coeﬃcients which appear in said
form. Theorem 3.16 guarantees that any polynomial with real coeﬃcients can be factored over
the real numbers as a product of linear fac...

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