PARTIAL FRACTIONS 5 



In general the best method of dealing with the work on partial 

 fractions is to begin by selecting values of x which will make 

 particular factors disappear ; this can be done for all linear 

 factors : then, if necessary, make x zero and use the results 

 already found. After this, if the solution to the question is still 

 incomplete, a further step must be taken, and this is best done 

 by 'equating coefficients of like powers of x. 



Sometimes we have to deal with an example which can only be 

 done by equating coefficients of like powers of x and solving the 

 resulting simultaneous equations for the quantities A, B, C, &c. 

 x z 



Taking ; 3 as an example of this type. 



x* + x 2 + I 



+ X 2 + I (X 2 + X + 1)(X 2 - X + 1) 



Ax + B Cx + D 



X 2 -\- X + I X 2 X + 1 



Then (Ax + B)(# 2 - x + 1) + (Cx + V)(x 2 + x + 1) =x 2 for all 

 values of x. 



-^ 7 ^7~2 \ 



2(x 2 +0 + 1) 2(x 2 x + 1) 



6. The Binomial Theorem. 

 If (a; + a) n = x n + Aor"- 1 + Ba 2 a?- 2 -f Ca 3 # n - 3 + . . . 



where A, B, C . . . are independent of x and a, 

 Then (x + a) n+1 = x n+l + <x.ax n + ^a 2 ^**- 1 + ja 3 x n ~ 2 + 



where a, (J, y . . . are independent of x and a. 

 For (x + a)"* 1 = (x + a)(x + a) n 

 = (x+ a)(x n + 



= x n + l + Aar n + Ba 2 ^"- 1 + Ca 3 x n ~ 2 . 

 + ax n + AaV- 1 + Ba 3 # n - 2 + 

 = x n+l + (A + l)ar" + (B + A)a-x n ~ l 



+ (C+ 

 - x Ll + o.ax n 



