7.4
PARTIAL FRACTIONS
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What You Should Learn
• Recognize partial fraction decompositions of
rational expressions.
• Find partial fraction decompositions of rational
expressions.
2
Introduction
3
Introduction
4
Introduction
5
Partial Fraction Decomposition
6
Example 1 – Distinct Linear Factors
Write the partial fraction decomposition of
.
Solution:
x2 – x – 6 = (x – 3)(x + 2)
Write form of decomposition.
x + 7 = A(x + 2) + B(x – 3)
Basic equation
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Example 1 – Solution
For instance, let x = –2. Then,
–2 + 7 = A(–2 + 2) + B(–2 – 3)
cont’d
Substitute –2 for x.
5 = A(0) + B(–5)
5 = – 5B
–1 = B.
To solve for A, let x = 3 and obtain
3 + 7 = A(3 + 2) + B(3 – 3)
Substitute 3 for x.
10 = A(5) + B(0)
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Example 1 – Solution
cont’d
10 = 5A
2 = A.
So, the partial fraction decomposition is
9
Or…
𝑥 + 7 = 𝐴 𝑥 + 2 + 𝐵(𝑥 − 3)
𝑥 + 7 = 𝐴𝑥 + 2𝐴 + 𝐵𝑥 − 3𝐵
𝑥 + 7 = 𝐴 + 𝐵 𝑥 + (2𝐴 − 3𝐵)
𝐴+𝐵 =1
2𝐴 − 3𝐵 = 7
𝐴=2
𝐵 = −1
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Partial Fraction Decomposition
11
Example
Write the partial fraction decomposition of
8𝑥 3 +13𝑥
𝑥 2 +2 2
Solution:
It is proper fraction.
8𝑥 3 + 13𝑥 𝐴𝑥 + 𝐵
𝐶𝑥 + 𝐷
= 2
+ 2
2
2
𝑥 +2
𝑥 +2
𝑥 +3 2
⇒ 8𝑥 3 + 13𝑥 = 𝐴𝑥 + 𝐵 𝑥 2 + 2 + (𝐶𝑥 + 𝐷)
= 𝐴𝑥 3 + 2𝐴𝑥 + 𝐵𝑥 2 + 2𝐵 + 𝐶𝑥 + 𝐷
= 𝐴𝑥 3 + 𝐵𝑥 2 + 2𝐴 + 𝐶 𝑥 + (2𝐵 + 𝐷)
𝐴=8
𝐴=8
𝐵=0
𝐵=0
⇒
⇒
2𝐴 + 𝐶 = 13
𝐶 = −3
2𝐵 + 𝐷 = 0
𝐶=0
12
Therefore,
8𝑥 3 + 13𝑥 𝐴𝑥 + 𝐵
𝐶𝑥 + 𝐷
8𝑥
−3𝑥
= 2
+ 2
=
+ 2
𝑥2 + 2 2
𝑥 +2
𝑥 + 3 2 𝑥2 + 2
𝑥 +3
2
13
Example
Write the partial fraction decomposition of
3𝑥 2 +4𝑥+4
.
𝑥 3 +4𝑥
Solution:
𝑥 3 + 4𝑥 = 𝑥(𝑥 2 + 4)
It is proper fraction
3𝑥 2 + 4𝑥 + 4 𝐴 𝐵𝑥 + 𝐶
= + 2
3
𝑥 + 4𝑥
𝑥 𝑥 +4
⇒ 3𝑥 2 + 4𝑥 + 4 = 𝐴 𝑥 2 + 4 + 𝐵𝑥 + 𝐶 𝑥
= 𝐴 + 𝐵 𝑥 2 + 𝐶𝑥 + 4𝐴
𝐴=1
𝐴+𝐵 =3
⇒
𝐶=4 ⇒ 𝐵=2
𝐶=4
4𝐴 = 4
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Therefore,
3𝑥 2 + 4𝑥 + 4 𝐴 𝐵𝑥 + 𝐶 1 2𝑥 + 4
= + 2
= + 2
3
𝑥 + 4𝑥
𝑥 𝑥 +4 𝑥 𝑥 +4
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Partial Fraction Decomposition
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