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- 1. 2014/08/07 1 Copyright © Cengage Learning. All rights reserved. Systems of Equations and Inequalities 2 Information: Thursday 7 August 2014 1. Homework Task 4 on par 10.7 & 10.8 is due on Wednesday 13 August. 2. Find all memos on uLink. 3. Consultation times for Ms Durandt on Thursdays and Fridays at 10:30. Copyright © Cengage Learning. All rights reserved. 10.7 Partial Fractions 4 Objectives ► Distinct Linear Factors ► Repeated Linear Factors ► Irreducible Quadratic Factors ► Repeated Irreducible Quadratic Factors 5 Partial Fractions To write a sum or difference of fractional expressions as a single fraction, we bring them to a common denominator. For example, But for some applications of algebra to calculus we must reverse this process—that is, we must express a fraction such as 3x/(2x2 – x – 1) as the sum of the simpler fractions 1/(x – 1) and 1/(2x + 1). These simpler fractions are called partial fractions; we learn how to find them in this section. 6 Partial Fractions Let r be the rational function r(x) = where the degree of P is less than the degree of Q. By the Linear and Quadratic Factors Theorem, every polynomial with real coefficients can be factored completely into linear and irreducible quadratic factors, that is, factors of the form ax + b and ax2 + bx + c, where a, b, and c are real numbers.
- 2. 2014/08/07 2 7 Partial Fractions For instance, x4 – 1 = (x2 – 1) (x2 + 1) = (x – 1)(x + 1)(x2 + 1) After we have completely factored the denominator Q of r, we can express r(x) as a sum of partial fractions of the form and This sum is called the partial fraction decomposition of r. Let’s examine the details of the four possible cases. 8 Distinct Linear Factors 9 Distinct Linear Factors We first consider the case in which the denominator factors into distinct linear factors. The constants A1, A2, . . . , An are determined as in the next example. 10 Example 1 – Distinct Linear Factors Find the partial fraction decomposition of . Solution: The denominator factors as follows. x3 + 2x2 – x – 2 = x2(x + 2) – (x + 2) = (x2 – 1) (x + 2) = (x – 1) (x + 1) (x + 2) This gives us the partial fraction decomposition 11 Example 1 – Solution Multiplying each side by the common denominator, (x – 1)(x + 1)(x + 2), we get 5x + 7 = A(x + 1)(x + 2) + B(x – 1)(x + 2) + C(x – 1)(x + 1) = A(x2 + 3x + 2) + B(x2 + x – 2) + C(x2 – 1) = (A + B + C)x2 + (3A + B)x + (2A – 2B – C) If two polynomials are equal, then their coefficients are equal. Thus since 5x + 7 has no x2-term, we have A + B + C = 0. cont’d Expand Combine like terms 12 Example 1 – Solution Similarly, by comparing the coefficients of x, we see that 3A + B = 5, and by comparing constant terms, we get 2A – 2B – C = 7. This leads to the following system of linear equations for A, B, and C. A + B + C = 0 3A + B = 5 2A – 2B – C = 7 cont’d Equation 1: Coefficients of x2 Equation 2: Coefficients of x Equation 3: Constant coefficients
- 3. 2014/08/07 3 13 Example 1 – Solution We use Gaussian elimination to solve this system. A + B + C = 0 – 2B – 3C = 5 – 4B – 3C = 7 A + B + C = 0 – 2B – 3C = 5 3C = –3 cont’d Equation 2 + (–3) Equation 1 Equation 3 + (–2) Equation 1 Equation 3 + (–2) Equation 2 14 Example 1 – Solution From the third equation we get C = –1. Back-substituting, we find that B = –1 and A = 2. So the partial fraction decomposition is cont’d 15 Repeated Linear Factors 16 Repeated Linear Factors We now consider the case in which the denominator factors into linear factors, some of which are repeated. 17 Example 2 – Repeated Linear Factors Find the partial fraction decomposition of . Solution: Because the factor x – 1 is repeated three times in the denominator, the partial fraction decomposition has the form Multiplying each side by the common denominator, x(x – 1)3, gives x2 + 1 = A(x – 1)3 + Bx(x – 1)2 + Cx(x – 1) + Dx 18 Example 2 – Solution = A(x3 – 3x2 + 3x – 1) + B(x3 – 2x2 + x) + C(x2 – x) + Dx = (A + B)x3 + (–3A – 2B + C)x2 + (3A + B – C + D)x – A Equating coefficients, we get the following equations. A + B = 0 –3A – 2B + C = 5 3A + B – C + D = 7 –A = –1 cont’d Expand Combine like terms Coefficients of x3 Coefficients of x2 Coefficients of x Constant coefficients
- 4. 2014/08/07 4 19 Example 2 – Solution If we rearrange these equations by putting the last one in the first position, we can easily see (using substitution) that the solution to the system is A = –1, B = 1, C = 0, D = 2, so the partial fraction decomposition is cont’d 20 Irreducible Quadratic Factors 21 Irreducible Quadratic Factors We now consider the case in which the denominator has distinct irreducible quadratic factors. 22 Example 3 – Distinct Quadratic Factors Find the partial fraction decomposition of . Solution: Since x3 + 4x = x(x2 + 4), which can’t be factored further, we write Multiplying by x(x2 + 4), we get 2x2 – x + 4 = A(x2 + 4) + (Bx + C)x = (A + B)x2 + Cx + 4A 23 Example 3 – Solution Equating coefficients gives us the equations A + B = 2 C = –1 4A = 4 so A = 1, B = 1, and C = –1. The required partial fraction decomposition is cont’d Coefficients of x2 Coefficients of x Constant coefficients 24 Repeated Irreducible Quadratic Factors
- 5. 2014/08/07 5 25 Repeated Irreducible Quadratic Factors We now consider the case in which the denominator has irreducible quadratic factors, some of which are repeated. 26 Example 4 – Repeated Quadratic Factors Write the form of the partial fraction decomposition of Solution:

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