4 PRACTICAL MATHEMATICS 



Then A(a? - 2) (20 +1) + B(2# + 1) + C(a? - 2) 2 = 4a? +3 for all 

 values of x, 



when a? = 2 5B 11 B - ~ 



5 



1 25 ,,4 



whena; = -- T C C = - 



when aj = -2A+B+4C =3 A = - ^ 



M& 



2 11 4 



The partial fractions are : - -r + - + 



25(x - 2) 5(x - 2) 2 



Case IV. When the denominator of the fraction contains a 

 quadratic factor which cannot be resolved into linear factors. 



x 2 __ x*_ _ A Ex + C 



a? - I " (x - \}(x z + a; + 1) ~ x - 1 + x 2 + x + 1 

 Then A(x 2 + a? + 1) + (B# + C)(a* - 1) = x 2 for all values of x, 



when #=13A =1 A = - 



o 



when a? = A-C =0 C=J 



o 



when a? = 2 7A+2B+C =4 B = | 



o 



The partial fractions are - + - - 



3(* 1) 3(# 2 + a? + 1) 



5. If the numerator is of higher degree than the denominator, 

 then the denominator must be divided into the numerator, and 

 the fraction whose numerator is the remainder must be split up 

 into partial fractions. 



x* 8x 



Thus by division ^ _ g = x + ^-^ 



8x ___ 8x ___ A_ Ex + C 



* Hell o > / *-*\ / o .*-__ ~* j \ r _" i~ l 



x 3 - 8 (^ __ 2)(x z + 2x + 4) x - 2 a? 2 + 2x + 4 

 A(x 2 +'2x +4) + (Ex + C)(a; 2) = 8a; for all values of a;, 



4 



when x = 2 12A =16 A = - 



o 



when a:=0 4A-2C =0 C=| 



o 



when a:=l 7A-B-C =8 B = -- 



o 



a; 4 4 8 - 4a; 



a; 3 -8 = X f 3(2? - 2) + 3(x 2 + 2x + 4) 



