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7.8 Partial Fractions

7.8 Partial Fractions. Partial Fraction Decomposition of Step 1 If is not a proper fraction (a fraction with the numerator of lower degree than the denominator), divide f ( x ) by g ( x ). For example, Then apply the following steps to the remainder, which is a proper fraction.

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7.8 Partial Fractions

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  1. 7.8 Partial Fractions Partial Fraction Decomposition of Step 1 If is not a proper fraction (a fraction with the numerator of lower degree than the denominator), divide f(x) by g(x). For example, Then apply the following steps to the remainder, which is a proper fraction. Step 2 Factor g(x) completely into factors of the form (ax + b)m or (cx2 + dx + e)n, where cx2 + dx + e is irreducible and m and n are integers.

  2. 7.8 Partial Fractions • Partial Fraction Decomposition of (continued) • Step 3 (a) For each distinct linear factor (ax + b), the decomposition must include the term • (b) For each repeated linear factor (ax + b)m, the decomposition must include the terms

  3. 7.8 Partial Fractions Partial Fraction Decomposition of (continued) Step 4 (a) For each distinct quadratic factor (cx2 + dx + e), the decomposition must include the term (b) For each repeated factor (cx2 + dx + e)n, the decomposition must include the terms Step 5 Use algebraic techniques to solve for the constants in the numerators of the decomposition.

  4. 7.8 Finding a Partial Fraction Decomposition Example Find the partial fraction decomposition. Solution Write the fraction as a proper fraction using long division.

  5. 7.8 Finding a Partial Fraction Decomposition Now work with the remainder fraction. Solve for the constants A, B, and C by multiplying both sides of the equation by x(x + 2)(x – 2), getting Substituting 0 in for x gives –2 = –4A, so Similarly, choosing x = –2 gives –12 = 8B, so Choosing x = 2 gives 8 = 8C, so C = 1.

  6. 7.8 Finding a Partial Fraction Decomposition The remainder rational expression can be written as the following sum of partial fractions: The given rational expression can be written as Check the work by combining the terms on the right.

  7. 7.8 Repeated Linear Factors Example Find the partial fraction decomposition. Solution This is a proper fraction and the denominator is already factored. We write the decomposition as follows:

  8. 7.8 Repeated Linear Factors Multiplying both sides of the equation by (x – 1)3: Substitute 1 for x leads to C = 2, so Since any number can be substituted for x, choose x = –1, and the equation becomes

  9. 7.8 Repeated Linear Factors Substituting 0 in for x in gives Now solve the two equations with the unknowns A and B to get A = 0 and B = 2. The partial fraction decomposition is

  10. 7.8 Distinct Linear and Quadratic Factors Example Find the partial fraction decomposition. Solution The partial fraction decomposition is Multiply both sides by (x + 1)(x2 + 2) to get

  11. 7.8 Distinct Linear and Quadratic Factors First, substitute –1 in for x to get Replace A with –1 and substitute any value for x, say x = 0, in to get Solving now for B, we get B = 2, and our result is

  12. 7.8 Techniques for Decomposition into Partial Fractions Method 1 For Linear Factors Step 1 Multiply each side of the resulting rational equation by the common denominator. Step 2 Substitute the zero of each factor into the resulting equation. For repeated linear factors, substitute as many other numbers as is necessary to find all the constants in the numerators. The number of substitutions required will equal the number of constants A, B, . . .

  13. 7.8 Techniques for Decomposition into Partial Fractions Method 2 For Quadratic Factors Step 1 Multiply each side of the resulting rational equation by the common denominator. Step 2 Collect terms on the right side of the equation. Step 3 Equate the coefficients of like terms to get a system of equations. Step 4 Solve the system to find the constants in the numerators.

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