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COLLEGE ALGEBRA. LIAL HORNSBY SCHNEIDER. 5.4. Partial Fractions. Decomposition of Rational Expressions Distinct Linear Factors Repeated Linear Factors Distinct Linear and Quadratic Factors Repeated Quadratic Factors. Decomposition of Rational Expressions.

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COLLEGE ALGEBRA

LIAL

HORNSBY

SCHNEIDER

5.4

Partial Fractions

Decomposition of Rational Expressions

Distinct Linear Factors

Repeated Linear Factors

Decomposition of Rational Expressions

The sums of rational expressions are found by combining two or more rational expressions into one rational expression.

Now consider the reverse process: Given one rational expression, express it as the sum of two or more rational expressions. A special type of sum of rational expressions is called the partial fraction decomposition; each term in the sum is called a partial fraction.

Partial Fraction Decomposition of

Step 1 If is not a proper fraction (a fraction with the numerator of lesser degree than the denominator), divide (x) by g(x). For example,

Then apply the following steps to the remainder, which is a proper fraction.

Partial Fraction Decomposition of

Step 2 Factor the denominator g(x) completely into factors of the form(ax + b)m or (cx2 + dx + e)n, where cx2 + dx + eis irreducible and m and n are positive integers.

Partial Fraction Decomposition of

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

Partial Fraction Decomposition of

Step 4 (a)For each distinct quadratic factor

(cx2 + dx + e), the decomposition must include the term

(b) For each repeated quadratic factor

(cx2 + dx + e)n, the decomposition must include the terms

Partial Fraction Decomposition of

Step 5 Use algebraic techniques to solve for the constants in the numerators of the decomposition.

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 1

Find the partial fraction decomposition of

Solution

The given fraction is not a proper fraction; the numerator has greater degree than the denominator. Perform the division.

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 1

The quotient is

Now, work with the remainder fraction. Factor the denominator as x3 – 4x = x(x + 2)(x – 2). Since

the factors are distinct linear factors, use Step 3(a) to write the decomposition as

(1)

where A, B, and C are constants that need to be found.

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 1

Multiply both sides of equation (1) by

x(x + 2)(x – 2) to obtain

(2)

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 1

Equation (1) is an identity since both sides represent the same rational expression.

Thus, equation (2) is also an identity. Equation (1) holds for all values of x except 0, –2, and 2. However, equation (2) holds for all values of x. In particular, substituting 0 for x in equation (2) gives –2 = –4A, so A = ½ . Similarly, choosing

x = 2 gives –12 = 8B, so B = –3/2. Finally, choosing x = 2 gives 8 = 8C, so C = 1.

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 1

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.

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 2

Find the partial fraction decomposition of

Solution

This is a proper fraction. The denominator is already factored with repeated linear factors. Write the decomposition as shown, by using Step 3(b).

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 2

Clear the denominators by multiplying both sides of this equation by (x – 1)3.

Substituting 1 for x leads to C = 2, so

(1)

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 2

The only root has been substituted, and values for A and B still need to be found. However, any number can be substituted for x. For example, when we choose x = –1 (because it is easy to substitute), equation (1) becomes

(2)

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 2

Substituting 0 for x in equation (1) gives

(3)

Now, solve the system of equations (2) and (3) to get A = 0 and B = 2. The partial fraction decomposition is

We needed three substitutions because there were three constants to evaluate, A, B, and C. To check this result, we could combine the terms on the right.

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 3

Find the partial fraction decomposition of

Solution

The denominator (x +1)(x2 + 2) has distinct linear and quadratic factors, where neither is repeated. Since x2 + 2 cannot be factored, it is irreducible.

The partial fraction decomposition is

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 3

Multiply both sides by (x +1)(x2 + 2) to get

First, substitute –1 for x to get

Use parentheses around substituted values to avoid errors.

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 3

Replace A with –1 in equation (1) and substitute any value for x. If x = 0, then

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 3

Now, letting A = –1 and C = 1, substitute again in equation (1), using another value for x. If x = 1, then

Using A = –1, B = 2, and C = 1, the partial fraction decomposition is

Again, this work can be checked by combining terms on the right.

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 3

For fractions with denominators that have quadratic factors, another method is often more convenient. The system of equations is formed by equating coefficients of like terms on both sides of the partial fraction decomposition. For

instance, in Example 3, after both sides were multiplied by the common denominator, the equation was

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 3

Multiplying on the right and collecting like terms, we have

Now, equating the coefficients of like powers of x gives the three equations

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 4

Find the partial fraction decomposition of

Solution

This expression has both a linear factor and a repeated quadratic factor. By Steps 3(a) and 4(b) from the beginning of this section,

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 4

Multiplying both sides by (x2 +1)2 (x –1) leads to

(1)

If x = 1, then equation (1) reduces to 2 = 4E, or E = ½ . Substituting ½ for E in equation (1) and expanding and combining terms on the right gives

(2)

FINDING A PARTIAL FRACTION DECOMPOSITIONExample 4

To get additional equations involving the unknowns, equate the coefficients of like powers of x on the two sides of equation (2). Setting corresponding coefficients of x4 equal, 0 = A + ½ or A = –½ . From the corresponding coefficients of x3, 0 = –A + B. Since A = –½, B = –½. Using the coefficients of x2, 0 = A – B + C + 1. Since A = –½ and B = –½, C = –1. Finally, from the coefficients of x, 2 = –A + B + D – C. Substituting for A, B, and C gives D = 1. With A = –½, B = –½, and C = –1, D = 1, and E = ½ , the given fraction has a partial fraction decomposition.

Techniques for Decomposition Into Partial Fractions

Method 1 For Linear Factors

Step 1 Multiply both sides of the resulting rational equation by the common denominator.

Step 2 Substitute the zero of each factor in the resulting equation. For repeated linear factors, substitute as many other numbers as necessary to find all the constants in the numerators. The number of substitutions required will equal the number of constants

Techniques for Decomposition Into Partial Fractions