Partial Fractions Part1 - Linear Fractions.

INTRODUCTION
 
Partial Fractions  problems has few types.First type is Linear Fractions.

 

 

                                                                              

 

  

 Look at this example
 Resolve into partial fractions
      
                         3X+7       
                    X² - 3X +2
Since X² -3X+ 2= (X-1)(X-2) corresponding to each linear factor we have a fraction with a constant in a numerator.
Now we can write ,

                         3X+7         =        A   +      B                Where A and B are constant.
                    X² - 3X +2            (X-2)      (X-1)     
Clearing off the fractions , we obtain ,
                      3X + 7= A(X-1) +B(X-2)          --------------- (1)
Equating now the coefficients of like power of X and constant on both sides ,we have

                                    A+B=3 and    -A-2B=7, ---------------(2)

                  solution is A= 13  and B = -10
 Now we can write,

                                    3X+7         =      13    -     10         (Answer)      
                               X² - 3X +2            (X-2)      (X-1)

 Example 1,
 Resolve into partial fractions
      
                         3X+2      
                    X² - 5X +6
Since X² - 5X +6= (X-3)(X-2) corresponding to each linear factor we have a fraction with a constant in a numerator.
Now we can write ,

                         3X+2         =        A   +      B                Where A and B are constant.
                    X² - 5X +6           (X-3)      (X-2)     
Clearing off the fractions , we obtain ,
                      3X + 2= A(X-2) +B(X-3)          --------------- (1)
Equating now the coefficients of like power of X and constant on both sides ,we have

                                    A+B=3 and    -3A-2B=2, ---------------(2)

                  solution is A= -8  and B = 11
 Now we can write,

                                    3X+7         =      -8      +   11         (Answer)      
                               X² - 3X +2            (X-2)      (X-1)

 Example 2,
 Resolve into partial fractions
      
                           2X      
                    X² - 8X +12
Since X² - 8X +12= (X-2)(X-6) corresponding to each linear factor we have a fraction with a constant in a numerator.
Now we can write ,

                             2X         =          A   +       B                Where A and B are constant.
                     X² - 8X +12           (X-2)      (X-6)     
Clearing off the fractions , we obtain ,
                      2X= A(X-6) +B(X-2)          --------------- (1)
Equating now the coefficients of like power of X and constant on both sides ,we have

                                    A+B=2 and    -6A-2B=0, ---------------(2)

                  solution is A= -1  and B =3
 Now we can write,

                                     2X            =         -1      +    3         (Answer)      
                                X² - 8X +12           (X-2)      (X-6)


 Example 3,

      X²+X-3  

(X-1)(X-3)(X+1)

Let ,

         X²+X-3        =         A          +  B         +  C               

(X-1)(X-3)(X+1)           (X-1)         (X-3)        (X+1)

Clearing off the fractions ,We obtain,

X²+X-3 = A(X-3)(X+1)+B(X-1)(X+1)+C(X-1)(X-3)------(1)

Setting X=1 yields

                        -2A = -1

Setting X=2 yields

                         3B = 3

Setting X=1 yields

                          6C=-3

So that,

     A = ½  B = 1 and C = - ½

           X²+X-3        =    ½           +  1         -   ½                  (Answer)

(X-1)(X-3)(X+1)           (X-1)         (X-3)        (X+1)