MODERATE EXERCISES

(3) Find the constants A and B so that the given equation is true on its domain. 


  5x2 - 9x + 19      A      Bx + C
 --------------- = ----- + --------
 (x - 4)(x2 + 5)   x - 4    x2 + 5
SOLUTION: 

Since   5x2 - 9x + 19      A      Bx + C
       --------------- = ----- + --------
       (x - 4)(x2 + 5)   x - 4    x2 + 5

when we find the common denominator on the left hand side we get:

         5x2 - 9x + 19      A(x2+5) + (Bx+C)(x-4)
        ---------------  =  ---------------------
        (x - 4)(x2 + 5)         (x-4)(x2 + 5)

Therefore 5x2 - 9x + 19 = Ax2 + 5A + Bx2 -4Bx + Cx - 4C
                         = (A + B)x2 + (-4B + c) + (5A - 4C)

    A + B = 5   ==>  A = 5 - B
  -4B + C = -9  ==>  C = 4B - 9
  5A - 4C = 19  ==>  5(5 - B) - 4(4B - 9) = 19  ==>  25 - 5B - 16B + 36 = 19 
                                                ==>  -21B =-42  ==>  B = 2

    A = 5 - 2 = 3
    C = 4(2) - 9 = -1

Hence the solution is:

        5x2 - 9x + 19      3      2x - 1
       --------------- = ----- + --------
       (x - 4)(x2 + 5)   x - 4    x2 + 5

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