Partial Fraction – SS3 Further Mathematics Lesson Note (First Term)

Improper Fraction

Equations of the degree of the numerator of the fraction we wish to resolve into partial fractions may be equal to or higher than the degree of the denominator. In this case we first use long division to divide the numerator by the denominator before we proceed to resolve into partial fractions Worked Examples

1. Resolve  2𝑥2+3𝑥+10  into partial fraction 

(𝑥+1)(𝑥+2) 

Solution:

Expanding the denominator

(𝑥 + 1)(𝑥 + 2) = 𝑥2 + 3𝑥 + 2 

                         2

𝑥2 + 3𝑥 + 2  2𝑥2 + 3𝑥 + 10

                        2𝑥2 + 6𝑥 + 4

                              −3𝑥 + 6

2+3𝑥+2     

                   (𝑥+1)(𝑥+2)              𝑥

               2+3𝑥+2       (𝑥+1)(𝑥+2)        

𝑥

                        A               B

∴ −3𝑥 + 6 = 𝐴(𝑥 + 2) + 𝐵(𝑥 + 1) 

𝑝𝑢𝑡 𝑥 = −2 

        −𝐵 = 12                

 𝐵 = −12 

𝑝𝑢𝑡 𝑥 − 1 

       𝐴 = 9 

 𝑥2+3𝑥+2  = 𝑥+1 − 𝑥−2  Hence:

   2𝑥2+3𝑥+10 = 2 + 9 − 12

             (𝑥+1)(𝑥+2)              𝑥+1   𝑥+2

1. Resolve into partial fractions 3𝑥3+22𝑥2+38𝑥+7 into partial fraction

(𝑥+1)(𝑥+3)

Solution:

  3𝑥3(+𝑥22+1𝑥)2(𝑥++383)𝑥+7 = 3𝑥3+𝑥222+𝑥22𝑥+−383𝑥+7

                   3𝑥 + 6 

𝑥2 + 2𝑥 − 3 3𝑥3 + 22𝑥2 + 38𝑥 + 7

             3𝑥3 + 6𝑥2 + 9𝑥

                   16𝑥2 + 47𝑥 + 7

                   16𝑥2 + 32𝑥 − 48

                         15𝑥 + 55

 15𝑥 ∓ 55 = 𝐴(𝑥 + 3) + 𝐵(𝑥 − 1) 

𝑝𝑢𝑡𝑡𝑖𝑛𝑔 𝑥 = −3 

        10 = −4𝐵

𝑝𝑢𝑡𝑡𝑖𝑛𝑔 𝑥 = 1 

       70 = 4𝐴

ASSIGNMENT: 

• Resolve into partial fraction

𝑥2

1)   

(𝑥−1)(𝑥+3)

2)  𝑥𝑥32−−14

2𝑥2+3𝑥+1

3)   

𝑥(𝑥+1)

REFERENCE:

1. Multipurpose Mathematics for SSCE, NECO and GCE 2nd Edition By J. Olowofero.
2. New mathematics project 3 by M.R Tuttuh, Adegun et al