LESSON 6–4 Partial Fractions

Five-Minute Check (over Lesson 6-3) Then/Now New Vocabulary Example 1: Rational Expression with Distinct Linear Factors Example 2: Improper Rational Expression Example 3: Denominator Example 4: Denominator with Prime Quadratic Factors Concept Summary: Partial Fraction Decomposition of f (x)/d(x) Lesson Menu

D. infinitely many solutions Use an inverse matrix to solve the system of equations, if possible. x + 3y + 4z = –3 x + 2y + 3z = –2 x + 4y + 3z = –6 A. (–1, –2, 1) B. (–5, 2, –1) C. no solution D. infinitely many solutions 5–Minute Check 1

Central High won a track meet with the help of 20 individual-event placers, earning a total of 68 points. A first-place finish earned the team 5 points, a second-place finish earned 3 points, and a third-place finish earned 1 point. If there were as many second-place finishers as first and third-place finishers combined, how many athletes finished in each place? 5–Minute Check 2

A. 10 first-place finishers, 7 second-place finishers, and 3 third-place finishers B. 7 first-place finishers, 3 second-place finishers, and 10 third-place finishers C. 3 first-place finishers, 10 second-place finishers, and 7 third-place finishers D. 7 first-place finishers, 10 second-place finishers, and 3 third-place finishers 5–Minute Check 2

Solve the system of linear equations. 3x – 4y = 4 2x + 2y = 12 B. (4, 2) C. no solution D. infinitely many solutions 5–Minute Check 3

You graphed rational functions. (Lesson 2-4) Write partial fraction decompositions of rational expressions with linear factors in the denominator. Write partial fraction decompositions of rational expressions with prime quadratic factors. Then/Now

partial fraction decomposition Vocabulary

Find the partial fraction decomposition of . Rational Expression with Distinct Linear Factors Find the partial fraction decomposition of . Rewrite the expression as partial fractions with constant numerators, A and B, and denominators that are the linear factors of the original denominators. Form a partial fraction decomposition Multiply each side by the LCD, x 2 – x – 12. Distributive Property Group like terms. Example 1

Rational Expression with Distinct Linear Factors Equate the coefficients on the left and right side of the equation to obtain a system of two equations. To solve the system, you can write it in matrix form CX = D and solve for X. C ● X = D A + B =1 –4A + 3B = –25 You can use a graphing calculator to find X = C –1D. So, A = 4 and B = –3. Use substitution to find the partial fraction decomposition. Example 1

Rational Expression with Distinct Linear Factors Example 1

Rational Expression with Distinct Linear Factors Answer: Example 1

Find the partial fraction decomposition of the rational expression . B. C. D. Example 1

Find the partial fraction decomposition of . Improper Rational Expression Find the partial fraction decomposition of . Because the degree of the numerator is greater than or equal to the degree of the denominator, the rational expression is improper. To rewrite the expression, divide the numerator by the denominator using polynomial division. Example 2

So, the original expression is equal to . Improper Rational Expression ← Subtract and bring down next term. So, the original expression is equal to . Because the remaining rational expression is now proper, you can factor its denominator as x(x – 1) and rewrite the expression using partial fractions. Example 2

Improper Rational Expression Write and solve the system of equations obtained by equating the coefficients. A + B = 7 A = 4 –A = –4 B = 3 Therefore, . Example 2

Partial fraction decomposition Improper Rational Expression Answer: CHECK You can check your answer by simplifying the expression on the right side of the equation. Partial fraction decomposition Rewrite using the LCD, x(x – 1). Example 2

Improper Rational Expression Add. Multiply. Simplify. Example 2

Find the partial fraction decomposition of . B. C. D. Example 2

Find the partial fraction decomposition of Denominator with Repeated Linear Factors Find the partial fraction decomposition of This rational expression is proper, so begin by factoring the denominator as x(x2 – 4x + 4) or x(x – 2)2. Because the factor (x – 2) has multiplicity 2, include partial fractions with denominators of x, (x – 2), and (x – 2)2. Example 3

Denominator with Repeated Linear Factors Form a partial fraction decompostion Multiply each side by the LCD, x(x – 2)2. Distributive Property Group like terms. Once the system of equations obtained by equating coefficients is found, there are two methods that can be used to find the values of A, B, and C. Example 3

3x 2 – 5x + 4 = A (x – 2)2 + Bx(x – 2) + Cx Original equation Denominator with Repeated Linear Factors Method 1 You can write and solve the system of equations using the same method as Example 2. A + B = 3 A = 1 –4A – 2B + C = –5 B = 2 4A = 4 C = 3 Method 2 Another way to solve this system is to let x equal a convenient value to eliminate a variable in the equation found by multiplying each side by the LCD. 3x 2 – 5x + 4 = A (x – 2)2 + Bx(x – 2) + Cx Original equation Example 3

3x 2 – 5x + 4 = A(x – 2)2 + Bx(x – 2) + Cx Original equation Denominator with Repeated Linear Factors 3(0)2 – 5(0) + 4 = A(0 – 2)2 + B(0)(0 – 2) + C(0) Let x = 0 to eliminate B and C. 4 = 4A 1 = A 3x 2 – 5x + 4 = A(x – 2)2 + Bx(x – 2) + Cx Original equation 3(2)2 – 5(2) + 4 = A(2 – 2)2 + B(2)(2 – 2) + C(2) Let x = 2 to eliminate A and B. Example 3

Denominator with Repeated Linear Factors Substitute these values in for A and C and any value for x into the equation to solve for B. Example 3

Therefore, . Answer: Denominator with Repeated Linear Factors Example 3

Find the partial fraction decomposition of . B. C. D. Example 3

Find the partial fraction decomposition of Denominator with Prime Quadratic Factors Find the partial fraction decomposition of This expression is proper. The denominator has one prime quadratic factor of multiplicity 2. Example 4

Denominator with Prime Quadratic Factors Write and solve the system of equations obtained by equating coefficients. A = 1 A = 1 B = 0 B = 0 –3A + C = –1 C = 2 –3B + D = 0 D = 0 Therefore, . Answer: Example 4

Find the partial fraction decomposition of . B. C. D. Example 4

Concept Summary

LESSON 6–4 Partial Fractions