PARTIAL FRACTIONS

 

WEEK 3 1^ST TERM

SUBJECT: Further Mathematics

CLASS: SS 3

TOPIC: PARTIAL FRACTIONS

CONTENT:

1. Basic definitions
2. Proper rational functions with linear factors (distinct and repeated)

Subtopic 1: Basic Definitions

You know what a fraction is. ¼ is a fraction. When we have 3/5 + 2/x, , we say that 3/5 and 2/x are the partial fractions of

Let p(x) and q(x) be polynomials in a variable x. is called a rational fraaction in q(x) ≠ 0.

When the degree of p(x) is less than the degree of q(x), then the rational function is called a proper rational function.

Any rational function can be expressed as the sum of a polynomail function and a proper rational function.

Note that the sum and difference of two rational functions is a rational function.

Example

1. Simplify
2. Express as partial fraction

= = =

If we reverse these examples then, this is called the resolution of proper rational function into partial fractions of the form = +

The degree of numerator must be less than the degree of the denominator. We also take note of the factors of the denominator.

Example 2:

can be resolved into partial fraction

= ; 3x+2 = A(x-5) + B(x+2)

Let x = 5, then B =

Let x = -2, then A =

=

A(x-5)+B(x+2)

Let x=5, then B = ;

We can equally resolve the using equation of coefficency.

=

3x+2 = A(x-5) + B(x+2)

3x +2 = x(A+B) + 2B-5A

…….. (1)

-5A + 2B =2 ……..(2)

-5A + 2B = 2

7B = 17 ; B =

A + = 3 ; A = 3 -2 =

+

EVALUATION

1. Make these fractions single fraction
2. + –
3. Resolve into partial fraction

Sub Topic 2: Proper rational functions with linear factors (distinct and repeated).

Partial fractions are of many types depending on the nature of the factors of the denominator we are asked to resolve.

There are four different categories.

1. Non-repeated linear factors at the denominator
2. Non-linear factors that are not repeated
3. Repeated factors linear or non-linear
4. Improper fractions
5. Non-repeated linear factors at the denominator

Those polynomial whose degree is less than they degree of the product of the polynomial in the denominator. This is of the form

For each factor

Example 3: Resolve into partial fractions

Solution:

=

4x + 3 = A(x+4) + B(x+2)

4x+3 = x(A+B) + (4A+2B)

A+B = 4

4A + 2B = 3 ……. (1)

2A +2B = 8 ……. (2)

Subracting equation (2) from (1) we have

2A = -5 ; A = -5/2 and B = 13/2

Then = -= –

Example 4:

Resolve into partial fraction

=

= + +

= + B

= A((+C

2x² + 3x – 1 = A (x²-4) + B(x²+2x) + C (x²-2x)

Let x = -2

1 = 8c ,

(A+B+C)x = 2 ; + = 2 ; A = 2- =

= + +

1. Non-repeated factors at the denominators follow the examples to understand this. When the denominator is not factorisable then we asign a power less by 1 i.e. if the power is n then we assign to the numerator variable of power n-1.

e.g. =

= +

Example 5: Resolve into partial fraction.

Solution :

=

2x²-3x+1 = (A+B)x²+x(5A+B+C) + A+C

A+B= 2………………………….i

A = 2-B………………………..ii from (ii) B = 2-A

5A+B+C =-3

A+C = 1⇒A=1-CC=1-A

5A+2-A+1-A = 3-3 = -6⇒3A = -6 ; A=-2, B=4, C = 3

=

1. Repeated factors at the denominator when the denominator has a repeated linear factor of the form (ax+b)ⁿ. We assign partial fractions of the form:

• …+

e.g. = +

Example 6

Resolve into partial fractions

Solution: = +

2x²+3x+3=

If x = 0 ; 3 = 9D ; D =

Letx = 1

7 = 16A + 16B + 4C +D

7 = 16A +

21 = 16A + 4C +

48A + 12C = 63-17

48A + 12C = 46

-12A + 6C = -2

48A +12C = 46

948A+24C= -8

36C = 38

C = =

-12A +6( = -2

-12A = -2-6 = -8; A = -x =

= ++

(d) improper fractions

We have improper fractions when the degree of the numerator is greater thanthe denominator. To handle this, we use long division method before resolving.

Example 7.

Resolve

Solution

16x²+47x+7

16x²+32x-48

15x+55

15x + 55= A(x+3) + B(x-1)=15x + 55 = (A+B)x + 3A – B

A+B = 55[mediator_tech]

A = ; B = 15 – 17 = -2

EVALUATION

Resolve the following to partial fractions

GENERAL EVALUATION

1. Express in partial fractions