924 CHAPTER 9 Systems and Matrices Now, equate the coefficients of like powers of x to obtain three equations. 1 = A + B 3 = B + C -1 = C + 2A Solving this system for A, B, and C gives the partial fraction decomposition. Repeated Quadratic Factors EXAMPLE 4 Finding a Partial Fraction Decomposition Find the partial fraction decomposition of 2x 1x2 + 1221x - 12 . SOLUTION This expression has both a linear factor and a repeated quadratic factor. Use Steps 3(a) and 4(b) from the box at the beginning of this section. 2x 1x2 + 1221x - 12 = Ax + B x2 + 1 + Cx + D 1x2 + 122 + E x - 1 Multiply each side by 1x2 + 1221x - 12. 2x = 1Ax + B21x2 + 121x - 12 + 1Cx + D21x - 12 + E1x2 + 122 (1) If x = 1, then equation (1) reduces to 2 = 4E, or E = 1 2 . Substitute 1 2 for E in equation (1), and expand and combine like terms on the right. 2x = Ax4 - Ax3 + Ax2 - Ax + Bx3 - Bx2 + Bx - B + Cx2 - Cx + Dx - D+ 1 2 x4 + x2 + 1 2 2x = aA + 1 2bx4 + 1-A + B2x3 + 1A - B + C + 12x2 + 1-A + B - C + D2x + a-B - D+ 1 2b (2) To obtain additional equations involving the unknowns, equate the coefficients of like powers of x on the two sides of equation (2). Setting corresponding coefficients of x4 equal, 0 = A + 1 2 , or A = - 1 2 . From the corresponding coefficients of x3, 0 = -A + B. Because A = - 1 2 , it follows that B = - 1 2 . Using the coefficients of x2, 0 =A-B+C+1. Since A= - 1 2 and B= - 1 2 , it follows that C = -1. From the coefficients of x, 2 = -A + B - C + D. Substituting for A, B, and C gives D= 1. With A = - 1 2 , B = - 1 2 , C = -1, D= 1, and E = 1 2 , the given fraction has partial fraction decomposition as follows. 2x 1x2 + 1221x - 12 = - 1 2 x - 1 2 x2 + 1 + -1x + 1 1x2 + 122 + 1 2 x - 1 Substitute for A, B, C, D, and E. 2x 1x2 + 1221x - 12 = -1x + 12 21x2 + 12 + -x + 1 1x2 + 122 + 1 21x - 12 Simplify complex fractions. S Now Try Exercise 31.
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