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Partial Fractions

University Mathematics is 100% positioned to meet the demands of undergraduate, postgraduate and professional exams. The book is capable and remarkable, giving students a good grounding in real-world mathematical applications. Inside you will find: • Simplified explanations step by step. • Solved examples of varying difficulties. • Plenty of workouts, fill-in-the-blanks and multiple-choice questions. Every student and every classroom will vastly benefit from University Mathematics.

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42 views3 pages

Partial Fractions

University Mathematics is 100% positioned to meet the demands of undergraduate, postgraduate and professional exams. The book is capable and remarkable, giving students a good grounding in real-world mathematical applications. Inside you will find: • Simplified explanations step by step. • Solved examples of varying difficulties. • Plenty of workouts, fill-in-the-blanks and multiple-choice questions. Every student and every classroom will vastly benefit from University Mathematics.

Uploaded by

Olaniyi Evans
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© © All Rights Reserved
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8

PARTIAL FRACTIONS &


BINOMIAL SERIES

CONTENTS
Partial Fractions 96
Binomial Theorem 100
Applications of Partial Fractions to Series Expansions 103

P A R TI A L F R A C T IO N S
Partial fraction decomposition or partial fraction expansion is the process
of expressing a rational function (where the numerator and the denominator are
both polynomials) as a sum of its initial polynomial fractions. The concept was
discovered by both Johann Bernoulli and Gottfried Leibniz independently in
1702.
Form of Algebraic Fraction Form of Partial Fractions
𝑝𝑥 + 𝑞 𝐴 𝐵
+
(𝑥 + 𝑎)(𝑥 + 𝑏) (𝑥 + 𝑎) (𝑥 + 𝑏)
𝑝𝑥 + 𝑞 𝐴 𝐵
+
(𝑥 + 𝑎)2 (𝑥 + 𝑎) (𝑥 + 𝑏)2
𝑝𝑥 2 + 𝑞𝑥 + 𝑟 𝐴 𝐵 𝐶
+ +
(𝑥 + 𝑎)(𝑥 + 𝑏)(𝑥 + 𝑐) (𝑥 + 𝑎) (𝑥 + 𝑏) (𝑥 + 𝑐)
𝑝𝑥 2 + 𝑞𝑥 + 𝑟 𝐴 𝐵 𝐶
+ 2
+
(𝑥 + 𝑎)2 (𝑥 + 𝑏) (𝑥 + 𝑎) (𝑥 + 𝑎) (𝑥 + 𝑏)
𝑝𝑥 2 + 𝑞𝑥 + 𝑟 𝐴 𝐵𝑥 + 𝐶
+
(𝑥 + 𝑎)(𝑥 2 + 𝑏𝑥 + 𝑐) (𝑥 + 𝑎) (𝑥 2 + 𝑏𝑥 + 𝑐)
where 𝑥 2 + 𝑏𝑥 + 𝑐 cannot be factorised.

For an algebraic fraction to be expressed in partial fractions, the numerator must


be at least one degree less than the denominator.
Chapter 8| Partial Fractions and Binomial Theorems 97

 EXAMPLE 8.1 TWO LINEAR FACTORS


Express in partial fractions
7𝑥 + 16
𝑥² + 2𝑥 − 8
S O L U T I O N tips
Factorise 𝑥 2 + 2𝑥 − 8
7𝑥 + 16 A B
= +
𝑥² + 2𝑥 − 8 𝑥 + 4 𝑥 − 2
Multiply both sides by (𝑥 + 4)(𝑥 − 2)
7𝑥 + 16 = 𝐴(𝑥 − 2) + 𝐵(𝑥 + 4)
Equate the first factor to zero: (𝑥 − 2) = 0 → 𝑥=2
When 𝑥 = 2;
7(2) + 16 = 𝐴(2 − 2) + 𝐵(2 + 4)
14 + 16 = 𝐴(0) + 6𝐵
30 = 6𝐵 → 𝐵=5
Equate the second factor to zero: (𝑥 + 4) = 0 ⇒ 𝑥 = −4
When 𝑥 = −4;
7(−4) + 16 = 𝐴(−4 − 2) + 𝐵(−4 + 4)
−28 + 16 = 𝐴(−6) + 𝐵(0)
−12 = −6𝐴 → 𝐴=2
Therefore
7𝑥 + 16 2 5
= +
𝑥² + 2𝑥 − 8 𝑥 + 4 𝑥 − 2

 EXAMPLE 8.2 REPEATED FACTORS


Express in partial fractions
2𝑥 − 5
(𝑥 − 3)²
S O L U T I O N tips
2𝑥 − 5 𝐴 𝐵
= +
(𝑥 − 3)² 𝑥 − 3 (𝑥 − 3)²
Multiply both sides by the common denominator (𝑥 – 3)2
2𝑥 – 5 = 𝐴 (𝑥 – 3) + 𝐵
(𝑥 − 3) = 0 → 𝑥=3
2(3) – 5 = 𝐴 (3 – 3) + 𝐵
1 = 𝐴 (0) + 𝐵 → 𝐵 = 1
Substituting 𝐵 = 1 in (i)
2𝑥 – 5 = 𝐴 (𝑥 – 3) + 1
98 Olaniyi Evans | University Mathematics

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