LIAL HORNSBY SCHNEIDER 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 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 + e is 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.

Solution FINDING A PARTIAL FRACTION DECOMPOSITION Example 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 DECOMPOSITION Example 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 DECOMPOSITION Example 1 Multiply both sides of equation (1) by x(x + 2)(x – 2) to obtain (2)

FINDING A PARTIAL FRACTION DECOMPOSITION Example 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 DECOMPOSITION Example 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.

Solution FINDING A PARTIAL FRACTION DECOMPOSITION Example 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 DECOMPOSITION Example 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 DECOMPOSITION Example 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 DECOMPOSITION Example 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.

Find the partial fraction decomposition of FINDING A PARTIAL FRACTION DECOMPOSITION Example 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

Use parentheses around substituted values to avoid errors. FINDING A PARTIAL FRACTION DECOMPOSITION Example 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 DECOMPOSITION Example 3 Replace A with – 1 in equation (1) and substitute any value for x. If x = 0, then

FINDING A PARTIAL FRACTION DECOMPOSITION Example 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 DECOMPOSITION Example 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 DECOMPOSITION Example 3 Multiplying on the right and collecting like terms, we have Now, equating the coefficients of like powers of x gives the three equations

Solution FINDING A PARTIAL FRACTION DECOMPOSITION Example 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 DECOMPOSITION Example 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 DECOMPOSITION Example 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.

FINDING A PARTIAL FRACTION DECOMPOSITION Example 4

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 Method 2 For Quadratic Factors Step 1 Multiply both sides of the resulting rational equation by the common denominator. Step 2 Collect like terms on the right side of the equation. Step 3 Equate the coefficients of like terms to get a system of equations. Step 4 Solve the system to find the constants in the numerators.