Divide (12x² – 8x – 32) by (x - 2), factor is +2

(X – 2)/₂ 12x² - 8x -32

+24x + 32

--------------------------

12x + 16 R = 0

Divide 7x² + 5x + 3 by x – 1 Q = 7x + 12 R = 15

Divide 7x² + 5x + 3 by x + 1 Q = 7x –2 R = 5

Divide x³ + 7x² + 6x + 5 by x – 2 Q = x² + 9x + 24 R = 53

Divide x4 – x3 + x2 + 3x +5 by x² – x – 1 Factors are x + 1

Q = x² + 0x + 2 R = 5x + 7

Divide 6x⁴ + 13 x³ + 39 x² + 37x + 45 by X² – 2x – 9

Factors 2x + 9 Q = 6x2 + 25x + 143 R = 548x + 1332

Divide x⁴ + x² + 1 by x² – x +1 (add 0x³ & 0x)

X⁴ + 0x³ + x² + 0x + 1 Factors are: x - 1

Q = x² + x + 1 R = 0

Divide 1 2 3 4 by 1 1 2 Factors -1 –2 Q = 11 R = 2

Divide 1 3 4 5 6 by 1 1 2 3 Q = 12 R = - 20

The Reminder cannot be negative. Hence Q = 11 R = 1 1 0 3

Divide 1 3 9 0 5 by 1 1 3 Q = 1 24 R = -107

Q = 123 R = 6

The following examples will help you to understand how to apply the above sutra to get partial equations in a simple and short manner.

1.       x² + 5x + 3             ....................Eq(1)

(x + 2)(x - 7)(x - 4)

This can be written in the form

            A     +   B    +    C       ............. Eq(2)   
            (x + 2)   (x - 7)   (x - 4)                      here A ,  B , C are constants to be determined.  

Now equate each term of the denominator to zero to get three different values of x. Now leaving out the term containing the particular value of x, substitute for x in all the other terms of the given equation to obtain respective constant for the term in the second equation.

i.e.    Equating  (x + 2) to zero we obtain   x + 2 = 0    ==>    x = -2

Now substituting for x in the first equation by (-2) we obtain the constant A.

 
      Therefore   A  =   (-2)2 + 5(-2) + 3  =  -11
                                           (-2 - 7)(-2 - 4)      54

      Similarly   B  =   (7)2 + 5(7) + 3    =   52
                                      (7 + 2)(7 - 4)            27

      Similarly   C  =   (4)2 + 5(4) + 3    =   31
                                      (4 - 7)(4 + 2)           -18

The above steps can be put in the form of a few simple formulae as follows : 

If the equation is of the form :     Lx2 + Mx + N    
                                                       (x - a)(x - b)(x - c)
 

Then the partial fractions can be written as    

               A    +    B    +    C  
                (x - a)   (x - b)   (x - c)                               

 where the constants A, B, C are given as

   A =  La2 + Ma + N     B =  Lb2 + Mb + N      C =  Lc2 + Mc + N
               (a - b)(a - c)            (b - a)(b - c)                   (c - a)(c - b)                 

2.            8x2 + 9x + 11     
          (x - 12)(x + 13)(x + 14)  

 This can be written in the form     A     +     B     +     C

                                                     (x - 12)    (x + 13)    (x + 14)

    Using the above explained formulas we get

    A = 1271/650   B = -1246/25    C = 1453/26