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Partial Fractions

Department of MathematicsUniversity of Leicester

Content

Quadratic factors

Linear factors

Repeated factors

Introduction

Introduction

Fractions whose algebraic sum is a given fraction are called partial fractions.

E.g.

and are partial fractions of since

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Repeated factors

Quadratic factors

Linear factors

Introduction

Linear factors𝑥+3

𝑥2+3 𝑥+2≡

𝑥+3(𝑥+2 ) (𝑥+1 )

The decomposition of a given fraction into partial fractions is achieved by first factorising the denominator

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Repeated factors

Quadratic factors

Linear factors

Introduction

If the factors are linear then we will have partial fractions of this form

Linear factors𝑥+3

𝑥2+3 𝑥+2≡

𝑥+3(𝑥+2 ) (𝑥+1 )

𝐴𝑥+2

+𝐵𝑥+1

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Repeated factors

Quadratic factors

Linear factors

Introduction

Linear factors𝑥+3

𝑥2+3 𝑥+2≡

𝑥+3(𝑥+2 ) (𝑥+1 )

𝐴𝑥+2

+𝐵𝑥+1

𝐴 (𝑥+1 )+𝐵(𝑥+2)𝑥2+3 𝑥+2

and are found by putting the expression in this form…

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Repeated factors

Quadratic factors

Linear factors

Introduction

Linear factors𝑥+3

𝑥2+3 𝑥+2≡

𝑥+3(𝑥+2 ) (𝑥+1 )

𝐴𝑥+2

+𝐵𝑥+1

𝐴 (𝑥+1 )+𝐵(𝑥+2)𝑥2+3 𝑥+2

𝐴 (𝑥+1 )+𝐵 (𝑥+2 )≡𝑥+3

𝑥+3𝑥2+3 𝑥+2

≡−1𝑥+2

+2

𝑥+1

… and solving this equivalence

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Repeated factors

Quadratic factors

Linear factors

Introduction

Cover-up method

𝑓 (𝑥 )= 𝑥(𝑥−2 ) (𝑥−3 )

≡𝐴

𝑥−2+

𝐵𝑥−3

𝐴= 𝑓 (2 )= 2(𝑥−2 ) (2−3 )

=−2

𝐵= 𝑓 (3 )= 3(3−2 ) (𝑥−3 )

=3

𝑥(𝑥−2 ) (𝑥−3 )

≡−2𝑥−2

+3

𝑥−3

If we “cover-up” the factor associated with the value we want to fin and then evaluate at it’s zero we will achieve the value we were looking for.

You can check this is correct with the previous method.

This only works with linear factors!

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Repeated factors

Quadratic factors

Linear factors

Introduction

Higher order factors𝑥2+1

(𝑥2+2)(𝑥−1)

𝐴𝑥+𝐵𝑥2+2

+𝐶𝑥−1

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Repeated factors

Quadratic factors

Linear factors

Introduction

If there is a quadratic factor we have partial fractions of this form.

Higher order factors𝑥2+1

(𝑥2+2)(𝑥−1)

𝐴𝑥+𝐵𝑥2+2

+𝐶𝑥−1

(𝐴𝑥+𝐵) (𝑥−1 )+𝐶 (𝑥2+2)(𝑥2+2)(𝑥−1)

(𝐴𝑥+𝐵) (𝑥−1 )+𝐶 (𝑥2+2)≡𝑥2+1Next

Repeated factors

Quadratic factors

Linear factors

Introduction

Rearrange to gain this equivalence.

Higher order factors𝑥2+1

(𝑥2+2)(𝑥−1)

𝐴𝑥+𝐵𝑥2+2

+𝐶𝑥−1

(𝐴𝑥+𝐵) (𝑥−1 )+𝐶 (𝑥2+2)(𝑥2+2)(𝑥−1)

(𝐴𝑥+𝐵) (𝑥−1 )+𝐶 (𝑥2+2)≡𝑥2+1

𝐴=13,𝐵=

13,𝐶=

23

𝑥2+1(𝑥2+2)(𝑥−1)

≡𝑥+1

3 (𝑥2+2 )+ 23 (𝑥−1 )Next

Repeated factors

Quadratic factors

Linear factors

Introduction

We can no compare coefficients to find our missing values of and .

Repeated factors𝑥−1

(𝑥+1) (𝑥−2 )2

−29

𝑥+1+

𝐵𝑥−2

+𝐶

(𝑥−2 )2

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Repeated factors

Quadratic factors

Linear factors

Introduction

If there is a repeated factor we have partial fractions of this form. The numerator of the linear factor was found using the cover-up method.

Repeated factors𝑥−1

(𝑥+1) (𝑥−2 )2

−29

𝑥+1+

𝐵𝑥−2

+𝐶

(𝑥−2 )2

𝑥−1≡(− 29 ) (𝑥−2 )2+𝐵 (𝑥+1 ) (𝑥−2 )+𝐶 (𝑥+1 )

𝑥=2  ⟹   𝐶=13 Next

Repeated factors

Quadratic factors

Linear factors

Introduction

Set to eliminate

Repeated factors𝑥−1

(𝑥+1) (𝑥−2 )2

−29

𝑥+1+

𝐵𝑥−2

+𝐶

(𝑥−2 )2

𝑥−1≡(− 29 ) (𝑥−2 )2+𝐵 (𝑥+1 ) (𝑥−2 )+𝐶 (𝑥+1 )

𝑥=2  ⟹   𝐶=13

0=−29+𝐵⇒𝐵=

29

𝑥−1(𝑥+1) (𝑥−2 )2

≡−2

9(𝑥+1)+

29 (𝑥−2 )

+1

3 (𝑥−2 )2

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Repeated factors

Quadratic factors

Linear factors

Introduction

Comparing the coefficients of we can find .

Summary

A linear factor gives a partial fraction of the form:

A quadratic factor gives a partial fraction of the form:

A repeated factor gives a partial fraction of the form:

𝐴𝑎𝑥+𝑏

𝐴𝑥+𝐵𝑎𝑥2+𝑏𝑥+𝑐

𝐴𝑎𝑥+𝑏

+𝐵

(𝑎𝑥+𝑏)2

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Repeated factors

Quadratic factors

Linear factors

Introduction