Chapter 7: Matrices and Systems of Equations and Inequalities

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

3. 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.

4. 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

5. 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.

6. 7.8 Finding a Partial Fraction Decomposition Example Find the partial fraction decomposition. Solution Write the fraction as a proper fraction using long division.

7. 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.

8. 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.

9. 7.8 Partial Fraction Decomposition Example with 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:

10. 7.8 Partial Fraction Decomposition Example with 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

11. 7.8 Partial Fraction Decomposition Example with 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

12. 7.8 Partial Fraction Decomposition Example with 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

13. 7.8 Partial Fraction Decomposition Example with 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