# Method of Decomposing Proper Fraction into Partial Fractions

Autor: Andrew James Olete • August 8, 2018 • Coursework • 2,586 Words (11 Pages) • 101 Views

**Page 1 of 11**

Method of Decomposing Proper Fraction into Partial Fractions:

Case I: Factors of the denominator are all linear and no repetition of factors.

Steps: The following steps are itemized by way of an illustrative problem.

Illustrative problem: Decompose into partial fractions[pic 1]

Solution: There are three factors in the denominator. This is to signify that there are three partial fractions in the decomposition. Please note that there will be no variable in the numerators because the denominator factors are all linear.

Step 1. Put a numerator in each denominator factor and put a plus (+) sign in between them to construct the needed equation. The plus (+) sign is only an assumption.

Required Equation: [pic 2]

The two expressions on the left and on the right are equal expressions.

Step 2. To simplify this, multiply both expressions by (x+1) (x+2) (x+3). This is to eliminate all denominators.

[pic 3]

Simplifying this, we have

[pic 4]

[pic 5]

[pic 6]

Please notice that the two expressions are identical.

Step 3. Equate the coefficients of x2, x, and constant to find the values of A, B, and C.

Coefficients:

x2: Left = 1 , Right = A + B + C

x: Left = 10 , Right = 5A + 4B + 3C

k: Left = 13 , Right = 6A + 3B + 2C

Solving for 3 equations and 3 unknown, we have

A + B + C = 1 ` eq. (1)

5A + 4B + 3C = 10 eq. (2)

6A + 3B + 2C = 13 eq. (3)

(2) 5A + 4B + 3C = 10

- x3 3A + 3B + 3C = 3
- -3(1) 2A + B = 7 eq. 4

- 6A + 3B + 2C = 13

(1)x2 2A + 2B + 2C = 2

(3)-2(1) 4A + B = 11 eq. 5

2A + B = 7 eq. 4

(5)-(4) 2A = 4

A = 4 / 2

A = 2 Answer

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