921 9.4 Partial Fractions Distinct Linear Factors EXAMPLE 1 Finding a Partial Fraction Decomposition Find the partial fraction decomposition of 2x4 - 8x2 + 5x - 2 x3 - 4x . SOLUTION The given fraction is not a proper fraction—the numerator has greater degree than the denominator. Perform the division (Step 1). 2x x3 - 4x )2x4 - 8x2 + 5x - 2 2x4 - 8x2 5x - 2 The quotient is 2x4 - 8x2 + 5x - 2 x3 - 4x = 2x + 5x - 2 x3 - 4x . Now, work with the remainder fraction. Factor the denominator (Step 2) as x3 - 4x = x1x + 221x - 22. The factors are distinct linear factors (Step 3(a)). Write the decomposition as 5x - 2 x3 - 4x = A x + B x + 2 + C x - 2 , (1) where A, B, and C are constants that need to be found (Step 5). Multiply each side of equation (1) by x1x + 221x - 22 to obtain 5x - 2 = A1x + 221x - 22 + Bx1x - 22 + Cx1x + 22. (2) Equation (1) is an identity because each side represents the same rational expression. Thus, equation (2) is also an identity. Equation (1) holds for all values of x except 0, -2, and 2. However, equation (2) holds for all values of x. We can solve for A by letting x = 0 in equation (2). 5x - 2 = A1x + 221x - 22 + Bx1x - 22 + Cx1x + 22 (2) 5102 - 2 = A10 + 2210 - 22 + B10210 - 22 + C10210 + 22 Let x = 0. -2 = -4A Simplify each term. A = 1 2 Divide by -4. Similarly, letting x = -2 in equation (2) enables us to solve for B. 51-22 - 2 = A1-2 + 221-2 - 22 + B1-221-2 - 22 + C1-221-2 + 22 Let x = -2 in (2). -12 = 8B Simplify each term. B = - 3 2 Divide by 8. Letting x = 2 gives the following for C. 5122 - 2 = A12 + 2212 - 22 + B12212 - 22 + C12212 + 22 Let x = 2 in (2). 8 = 8C Simplify each term. C = 1 Divide by 8.
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