ADVANCED EXERCISES

Decompose each fraction into partial fractions:

(3)   x2 + x + 1
     ---------------
      2x4 + 3x2 + 1

Let   x2 + x + 1        Ax + B      Cx + D
    -------------  =  ---------- + --------
    2x4 + 3x2 + 1      (2x2 + 1)    (x2 + 1)

Then we get:

    x2 + x + 1 = (Ax + B)(x2 + 1) + (Cx + D)(2x2 + 1)
			  = Ax3 + Ax + Bx2 + B + 2Cx3 + Cx + 2Dx 2 + D
			  = (A + 2C)x3 + (B + 2D)x2 + (A + C)x + (B + D)

Therefore:

    A + 2C = 0  ==>  A = -2C
    B + 2D = 1  ==>  B = 1 - 2D
    A + C  = 1  ==>  -2C + C = 1  ==>  -C = 1  ==>  C = -1
    B + D  = 1  ==>  B = 1 - D

    1 - 2D = 1 - D  ==> -D = 0  ==>  D = 0

    A = -2(-1) = 2
    B = 1 - 0 = 1

Hence the solution is:

      x2 + x + 1        2x + 1        -1x 
    -------------  =  ---------- + --------
    2x4 + 3x2 + 1      (2x2 + 1)    (x2 + 1)