Copyright © Cengage Learning. All rights reserved. Systems of Equations and Inequalities.

Copyright © Cengage Learning. All rights reserved.

Systems of Equations and Inequalities

Copyright © Cengage Learning. All rights reserved.

10.7 Partial Fractions

3

Objectives

► Distinct Linear Factors

► Repeated Linear Factors

► Irreducible Quadratic Factors

► Repeated Irreducible Quadratic Factors

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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.

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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.

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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.

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Distinct Linear Factors

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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.

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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

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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

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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

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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

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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

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Repeated Linear Factors

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Repeated Linear Factors

We now consider the case in which the denominator factors into linear factors, some of which are repeated.

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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

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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

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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

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Irreducible Quadratic Factors

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Irreducible Quadratic Factors

We now consider the case in which the denominator has distinct irreducible quadratic factors.

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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

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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

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Repeated Irreducible Quadratic Factors

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Repeated Irreducible Quadratic Factors

We now consider the case in which the denominator has irreducible quadratic factors, some of which are repeated.

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Example 4 – Repeated Quadratic Factors

Write the form of the partial fraction decomposition of

Solution: