923 9.4 Partial Fractions Distinct Linear and Quadratic Factors EXAMPLE 3 Finding a Partial Fraction Decomposition Find the partial fraction decomposition of x2 + 3x - 1 1x + 121x2 + 22 . SOLUTION The denominator 1x + 121x2 + 22 has distinct linear and quadratic factors, where neither is repeated. Because x2 + 2 cannot be factored, it is irreducible. The partial fraction decomposition is of the following form. x2 + 3x - 1 1x + 121x2 + 22 = A x + 1 + Bx + C x2 + 2 Multiply each side by 1x + 121x2 + 22. x2 + 3x - 1 = A1x2 + 22 + 1Bx + C21x + 12 (1) First, substitute -1 for x. 1-122 + 31-12 - 1 = A31-122 + 24 + 3B1-12 + C41-1 + 12 -3 = 3A A = -1 Replace A with -1 in equation (1) and substitute any value for x. Let x = 0. 02 + 3102 - 1 = -1102 + 22 + 1B # 0 + C210 + 12 -1 = -2 + C C = 1 Now, letting A = -1 and C = 1, substitute again in equation (1), using another value for x. Let x = 1. 12 + 3112 - 1 = -1112 + 22 + 3B112 + 1411 + 12 3 = -3 + 1B + 12122 6 = 2B + 2 B = 2 Use A = -1, B = 2, and C = 1 to find the partial fraction decomposition. x2 + 3x - 1 1x + 121x2 + 22 = -1 x + 1 + 2x + 1 x2 + 2 Check by combining the terms on the right. S Now Try Exercise 25. Use parentheses around substituted values to avoid errors. For fractions with denominators that have quadratic factors, an alternative method is often more convenient. A system of equations is formed by equating coefficients of like terms on each side of the partial fraction decomposition. For instance, in Example 3, equation (1) was x2 + 3x - 1 = A1x2 + 22 + 1Bx + C21x + 12. (1) Multiply on the right and collect like terms. x2 + 3x - 1 = Ax2 + 2A + Bx2 + Bx + Cx + C 1x2 + 3x - 1 = 1A + B2x2 + 1B + C2x + 1C + 2A2
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