Copyright © 2007 Pearson Education, Inc. Slide 7-1

Copyright © 2007 Pearson Education, Inc. Slide 7-2 Chapter 7: Matrices and Systems of Equations and Inequalities 7.1Systems of Equations 7.2Solution of Linear Systems in Three Variables 7.3Solution of Linear Systems by Row Transformations 7.4Matrix Properties and Operations 7.5Determinants and Cramer’s Rule 7.6Solution of Linear Systems by Matrix Inverses 7.7Systems of Inequalities and Linear Programming 7.8Partial Fractions

Copyright © 2007 Pearson Education, Inc. Slide Partial Fractions Partial Fraction Decomposition of Step 1If 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 2Factor g(x) completely into factors of the form (ax + b) m or (cx 2 + dx + e) n, where cx 2 + dx + e is irreducible and m and n are integers.

Copyright © 2007 Pearson Education, Inc. Slide 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

Copyright © 2007 Pearson Education, Inc. Slide 7-5 Partial Fraction Decomposition of (continued) Step 4(a)For each distinct quadratic factor (cx 2 + dx + e), the decomposition must include the term (b)For each repeated factor (cx 2 + dx + e) n, the decomposition must include the terms Step 5Use algebraic techniques to solve for the constants in the numerators of the decomposition. 7.8 Partial Fractions

Copyright © 2007 Pearson Education, Inc. Slide 7-6 ExampleFind the partial fraction decomposition. SolutionWrite the fraction as a proper fraction using long division. 7.8 Finding a Partial Fraction Decomposition

Copyright © 2007 Pearson Education, Inc. Slide 7-7 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 = Finding a Partial Fraction Decomposition

Copyright © 2007 Pearson Education, Inc. Slide 7-8 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. 7.8 Finding a Partial Fraction Decomposition

Copyright © 2007 Pearson Education, Inc. Slide 7-9 ExampleFind the partial fraction decomposition. SolutionThis is a proper fraction and the denominator is already factored. We write the decomposition as follows: 7.8 Partial Fraction Decomposition Example with Repeated Linear Factors

Copyright © 2007 Pearson Education, Inc. Slide 7-10 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 7.8 Partial Fraction Decomposition Example with Repeated Linear Factors

Copyright © 2007 Pearson Education, Inc. Slide 7-11 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 7.8 Partial Fraction Decomposition Example with Repeated Linear Factors

Copyright © 2007 Pearson Education, Inc. Slide 7-12 ExampleFind the partial fraction decomposition. SolutionThe partial fraction decomposition is Multiply both sides by (x + 1)(x 2 + 2) to get 7.8 Partial Fraction Decomposition Example with Distinct Linear and Quadratic Factors

Copyright © 2007 Pearson Education, Inc. Slide 7-13 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 7.8 Partial Fraction Decomposition Example with Distinct Linear and Quadratic Factors