922 CHAPTER 9 Systems and Matrices The remainder rational expression can be written as the following sum of partial fractions. Use A = 1 2 , B = - 3 2 , and C = 1 in equation 112. 5x - 2 x3 - 4x = 1 2x + -3 21x + 22 + 1 x - 2 1 2 x = 1 2x and - 3 2 x + 2 = -3 21x + 22 The given rational expression can now be written as follows. 2x4 - 8x2 + 5x - 2 x3 - 4x = 2x + 1 2x + -3 21x + 22 + 1 x - 2 Check this result by combining the terms on the right. S Now Try Exercise 29. Repeated Linear Factors EXAMPLE 2 Finding a Partial Fraction Decomposition Find the partial fraction decomposition of 2x 1x - 123 . SOLUTION This is a proper fraction. The denominator is already factored with repeated linear factors. Write the decomposition as shown using Step 3(b). 2x 1x - 123 = A x - 1 + B 1x - 122 + C 1x - 123 Clear the denominators by multiplying each side of this equation by 1x - 123. 2x = A1x - 122 + B1x - 12 + C Substituting 1 for x leads to C = 2. 2x = A1x - 122 + B1x - 12 + 2 (1) The only root has been substituted, and values for A and B still need to be found. However, any number can be substituted for x. For example, when we choose x = -1 (because it is easy to substitute), equation (1) becomes the following. 21-12 = A1-1 - 122 + B1-1 - 12 + 2 Let x = -1 in (1). -2 = 4A - 2B + 2 Simplify each term. -4 = 4A - 2B Subtract 2. -2 = 2A - B Divide by 2. (2) Substituting 0 for x in equation (1) gives another equation in A and B. 2102 = A10 - 122 + B10 - 12 + 2 Let x = 0 in (1). 0 = A - B + 2 Simplify each term. 2 = -A + B Subtract 2. Multiply by -1. (3) Now, solve the system of equations (2) and (3) as shown in the margin to find A = 0 and B = 2. Substitute these values for A and B and 2 for C. 2x 1x - 123 = 2 1x - 122 + 2 1x - 123 Partial fraction decomposition We needed three substitutions because there were three constants to evaluate: A, B, and C. To check this result, we could combine the terms on the right. S Now Try Exercise 17. -2 = 2A - B (2) 2 = -A + B (3) 0 = A Add. If A = 0, then using equation (3), 2 = 0 + B 2 = B.
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