PARTIAL FRACTIONS 3 



Then A(x + 3)(2x - 5) + B(x - 2)(2x - 5) + C(x - 2)(x + 3) 



80; + 2 for all values of x, 



o 



\vhen x = 2 5A s A = 



5 



when # = - 3 + 55B =-7 B=-^ 



55 



5 11., 19 n 38 



when * = 2 T C = 2 - = n 



The partial fractions are -3^) - g^-^ + n( J 8 . g) 



In this type o example, and in the other types when possible, 

 it is much better to choose the values of a; so that they make the 

 factors disappear in turn. 



Case II. When the denominator of the fraction consists of 

 one factor raised to an integral power. 



4# 2 - 2x + 8 

 Taking _ as an example, 



put a = 8x 2, then x = -(a + 2) 



3 



Then 4z 2 - 2x + 8 = ^(a + 2) 2 - |(a + 2) +3 



4 , 10 31 



= -a a -\ 



9 99 



4a 2 - lOa + 31 



The fraction becomes 



9a 3 



4 10 31 



This may be written as -5 -\ = 



9a 9a 2 9a 3 



Replacing a by 8x 2, we get as the partial fractions 

 4 10 31 



- 2) 9(3x - 2) 2 9(3a; - 2) 3 



This result shows us how the partial fractions should be 



arranged when a certain factor in the denominator of a fraction 

 is raised to a power. 



A B C D 



x + I Sx - 2 (3x - 2) 2 (3x - 2) 3 



Case III. When the denominator of the fraction is the pro- 

 duct of a certain number of linear factors, but one of these factors 

 is raised to a power. 



4c + 3 A B C 



T 



(x - 2) 2 (2a? +1) x - 2 (x - 2) 2 2x + 1 



