Partial Fraction Decomposition Honors Precalculus Mr. Velazquez Rm. 254

Partial Fraction Decomposition Honors Precalculus Mr. Velazquez Rm. 254

Adding and Subtracting Rational Expressions Recall that we can use multiplication and common denominators to write a sum or difference of two rational expressions (fractions): 5 x 2 4 x + 4 = 5 x + 4 x 2 x + 4 4 x 2 x 2 x + 4 = 5x + 20 4x + 8 (x 2)(x + 4) = x + 28 (x 2)(x + 4) Usually this is done to simplify the expression, or to cancel out certain terms.

Reversing the Process Certain operations in Calculus will require us to reverse this process: x+12 (x 2)(x+4) is expressed as the sum of two fractions. Partial fraction x + 28 (x 2)(x + 4) = 5 x 2 + Partial fraction 4 x + 4 This is the partial fraction decomposition of x+12 (x 2)(x+4)

Partial Fraction Decomposition P x A sum or difference of partial fractions can be written in the form, Q x where functions P and Q have no common factors, and the degree (highest power) of Q is higher than the degree of P. 9x 2 9x + 6 (2x 1)(x + 2)(x 2) P x = 9x 2 9x + 6; degree = 2 Q x = (2x 1)(x + 2)(x 2); degree = 3

Four Cases of Partial Fraction Decomposition For a rational expression of the form P(x), we will consider Q(x) four different cases for the factors of the denominator: 1. Q(x) is a product of distinct linear factors. 2. Q(x) is a product of linear factors, some of which are repeated. 3. Q x has prime quadratic factors, none of which are repeated. 4. Q(x) has a repeated prime quadratic factor.

Case 1: Q(x) is a Product of Distinct Linear Factors This applies when the denominator Q(x) of the rational expression can be factored into linear expressions that are not repeated. Each linear factor in the denominator produces a partial fraction in the form of a constant over a linear factor. 9x 2 9x + 6 (2x 1)(x + 2)(x 2) = A 2x 1 + B x + 2 + C x 2

Case 1: Q(x) is a Product of Distinct Linear Factors Example: Let us find the partial fraction decomposition of the expression shown below: x + 14 (x 4)(x + 2) = A x 4 + B x + 2 Step 1 is to write the expression as a sum of rational expressions, with each of its linear factors serving as the denominators, and constant terms in the numerator. Our goal here is to find A and B.

Case 1: Q(x) is a Product of Distinct Linear Factors Step 2: Multiply both sides by the denominator Q(x). (x 4)(x + 2) x + 14 x 4 x + 2 = (x 4)(x + 2) A x 4 + B x + 2 Using distribution on the right side will result in an equation with no more rational expressions. Simplify. x 4 x + 2 A x 4 x + 2 B x + 14 = + x 4 x + 2 x + 14 = x + 2 A + x 4 B x + 14 = Ax + 2A + Bx 4B

Case 1: Q(x) is a Product of Distinct Linear Factors Step 3: Rearrange the terms to identify the coefficients of the varying powers of x (in this case, there is only one power of x and a constant term). x + 14 = Ax + 2A + Bx 4B x + 14 = Ax + Bx + 2A 4B 1x + 14 = A + B x + (2A 4B) x term coefficient constant term These can be set up into a system of equations involving two variables, A and B.

Case 1: Q(x) is a Product of Distinct Linear Factors Step 4: Set up a system of equations, equating the coefficients and the constant terms. Solving this system will produce the values of A and B. 1x + 14 = A + B x + (2A 4B) A + B = 1 2A 4B = 14 A = B + 1 2 B + 1 4B = 14 2B + 2 4B = 14 6B = 12 B = 2, which means A = 2 + 1 = 3

Case 1: Q(x) is a Product of Distinct Linear Factors Step 5: Plug your values for A and B back into the original expression. x + 14 (x 4)(x + 2) = A x 4 + B x + 2 = 3 x 4 + 2 x + 2 Or, x+14 (x 4)(x+2) = 3 x 4 2 x+2

Try it Out! Use partial fraction decomposition to find A and B and confirm the result obtained earlier, shown below: x + 28 (x 2)(x + 4) = A x 2 + B x + 4

Try it Out! Use partial fraction decomposition to find A and B and C for the rational equation shown below: 4x 2 + 13x 9 x(x 1)(x + 3) = A x + B x 1 + C x + 3

Try it Out! / Extra Credit Use partial fraction decomposition to find A, B and C from the expression shown below: (Place in the bin for extra credit) 9x 2 9x + 6 (2x 1)(x + 2)(x 2) = A 2x 1 + B x + 2 + C x 2

Case 2: Q(x) is a Product of Linear Factors Some of Which are Repeated This occurs when any of the factors in the denominator are raised to any power greater than 1. To decompose such expressions, we must create a series of terms for each power of the repeated factor. As before, Step 1 is to write it as a sum of partial fractions, only you must create a separate term for each power of the repeated factor. x 18 x x 3 2 = A x + B x 3 + C x 3 2 Since x 3 is raised to the 2 nd power, we need two terms to decompose the expression.

Case 2: Q(x) is a Product of Linear Factors Some of Which are Repeated Step 2: Multiply both sides by the denominator Q(x). x x 3 2 x 18 x x 3 2 = x x 3 2 A x + B x 3 + C x 3 2 Using distribution on the right side will result in an equation with no more rational expressions. Simplify. x 18 = x x 3 2 A x + x x 3 2 B x 3 + x x 3 2 C x 3 2 x 18 = A x 3 2 + Bx x 3 + Cx x 18 = Ax 2 6Ax + 9A + Bx 2 3Bx + Cx x 18 = Ax 2 + Bx 2 6Ax 3Bx + Cx + 9A

Case 2: Q(x) is a Product of Linear Factors Some of Which are Repeated Step 3: Group the terms and factor. Identify the coefficients of the various powers of x again, in this case, there are only two terms on the left side, so this example will be easier x 18 = Ax 2 + Bx 2 6Ax 3Bx + Cx + 9A 0x 2 + 1x 18 = A + B x 2 + 6A 3B + C x + (9A) Step 4: Based on this, we can create a new system of equations, and solve. A + B = 0 6A 3B + C = 1 9A = 18 9A = 18 A = 2 A + B = 0 2 + B = 0 B = 2 6A 3B + C = 1 6 2 3 2 + C = 1 12 6 + C = 1 C = 5

Case 2: Q(x) is a Product of Linear Factors Some of Which are Repeated Step 5: Substitute these values into the original equation for A, B and C. x 18 x x 3 2 = A x + B x 3 + C x 3 2 x 18 x x 3 2 = 2 x + 2 x 3 + 5 x 3 2

Try it Out! Find the partial fraction decomposition of the following expressions: x + 2 x x 1 2 x 2 x 1 2 (x + 1)

Case 3: Q(x) is a Product of Distinct Prime Quadratic Factors This occurs when any of the factors in the denominator are a quadratic expression that cannot be factored further. To decompose such expressions, one of the partial fractions we use must have a linear numerator. 3x 2 + 17x + 14 (x 2)(x 2 + 2x + 4) = A x 2 + Bx + C x 2 + 2x + 4 The numerator of this term must contain both B and C, along with an x.

Case 3: Q(x) is a Product of Distinct Prime Quadratic Factors This occurs when any of the factors in the denominator are a quadratic expression that cannot be factored further. To decompose such expressions, one of the partial fractions we use must have a linear numerator. Step 1: Rewrite the expression as the sum of partial fractions, one of which has a linear numerator. 3x 2 + 17x + 14 (x 2)(x 2 + 2x + 4) = A x 2 + Bx + C x 2 + 2x + 4 The numerator of this term must contain both B and C, along with an x.

Case 3: Q(x) is a Product of Distinct Prime Quadratic Factors Step 2: Multiply both sides by the denominator Q x. Simplify the right side. (x 2)(x 2 + 2x + 4) 3x 2 + 17x + 14 = x 2 x 2 + 2x + 4 3x 2 + 17x + 14 x 2 x 2 + 2x + 4 = (x 2)(x2 + 2x + 4) A x 2 + A x 2 + (x 2)(x2 + 2x + 4) 3x 2 + 17x + 14 = A x 2 + 2x + 4 + (Bx + C)(x 2) 3x 2 + 17x + 14 = Ax 2 + 2Ax + 4A + Bx 2 2Bx + Cx 2C 3x 2 + 17x + 14 = Ax 2 + Bx 2 + 2Ax 2Bx + Cx + 4A 2C Step 3: Group terms together based on powers of x. 3x 2 + 17x + 14 = A + B x 2 + 2A 2B + C x + (4A 2C) Bx + C x 2 + 2x + 4 Bx + C x 2 + 2x + 4

Case 3: Q(x) is a Product of Distinct Prime Quadratic Factors Step 4: Create a system of equations for each coefficient and solve. 3x 2 + 17x + 14 = A + B x 2 + 2A 2B + C x + (4A 2C) A + B = 3 2A 2B + C = 17 4A 2C = 14 Multiplying the 2 nd equation by 2 and adding it to the third: 8A 4B = 48 Substituting the 1 st equation: 8A 4(3 A) = 48 8A 12 + 4A = 48 12A = 60 A = 5 A + B = 3 (5) + B = 3 B = 2 4A 2C = 14 4(5) 2C = 14 C = 3

Case 3: Q(x) is a Product of Distinct Prime Quadratic Factors Step 5: Substitute these values back into the original equation. 3x 2 + 17x + 14 (x 2)(x 2 + 2x + 4) = A x 2 + Bx + C x 2 + 2x + 4 3x 2 + 17x + 14 (x 2)(x 2 + 2x + 4) = 5 x 2 + 2x + 3 x 2 + 2x + 4

Case 4: Q(x) is a Product of Prime Quadratic Factors Some of Which are Repeated This occurs when any of the factors in the denominator are quadratic expressions that cannot be factored further, and are raised to a power greater than 1. To decompose such expressions, we must create a partial fraction for every ascending power of the quadratic factor, just as we did for Case 2. Step 1: Rewrite the expression as a sum of partial fractions. Make sure to include separate terms for each ascending power of any repeated quadratics. 5x 3 3x 2 + 7x 3 x 2 + 1 2 = Ax + B x 2 + 1 + Cx + D x 2 + 1 2 Since x 2 + 1 is raised to the 2 nd power, we must separate it into two terms.

Case 4: Q(x) is a Product of Prime Quadratic Factors Some of Which are Repeated Step 2: Multiply both sides by the denominator Q x. Simplify the right side. x 2 + 1 2 5x3 3x 2 + 7x 3 x 2 + 1 2 = x 2 + 1 2 Ax + B x 2 + 1 + Cx + D x 2 + 1 2 5x 3 3x 2 + 7x 3 = x 2 + 1 2 Ax + B x 2 + 1 + x2 + 1 2 Cx + D x 2 + 1 2 5x 3 3x 2 + 7x 3 = x 2 + 1 Ax + B + (Cx + D) 5x 3 3x 2 + 7x 3 = Ax 3 + Bx 2 + Ax + B + Cx + D Step 3: Group terms together based on powers of x. 5x 3 3x 2 + 7x 3 = Ax 3 + Bx 2 + A + C x + (B + D)

Case 4: Q(x) is a Product of Prime Quadratic Factors Some of Which are Repeated Step 4: Create a system of equations for each coefficient, then solve. 5x 3 3x 2 + 7x 3 = Ax 3 + Bx 2 + A + C x + (B + D) A = 5 B = 3 A + C = 7 B + D = 3 We can see that even though this system has 4 equations and 4 variables, it is still quite easy to solve: We already know A = 5, and B = 3. Substituting these values into the other equations gives us C and D. A + C = 7 (5) + C = 7 C = 2 B + D = 3 ( 3) + D = 3 D = 0

Case 4: Q(x) is a Product of Prime Quadratic Factors Some of Which are Repeated Step 5: Substitute these values back into the original equation. 5x 3 3x 2 + 7x 3 x 2 + 1 2 = Ax + B x 2 + 1 + Cx + D x 2 + 1 2 5x 3 3x 2 + 7x 3 x 2 + 1 2 = 5x 3 x 2 + 1 + 2x x 2 + 1 2

Case 4: Q(x) is a Product of Prime Quadratic Factors Some of Which are Repeated Step 5: Substitute these values back into the original equation. 5x 3 3x 2 + 7x 3 x 2 + 1 2 = Ax + B x 2 + 1 + Cx + D x 2 + 1 2 5x 3 3x 2 + 7x 3 x 2 + 1 2 = 5x 3 x 2 + 1 + 2x x 2 + 1 2

Try it Out! Compute the partial fraction decomposition for the following expressions: 8x 2 + 12x 20 (x + 3)(x 2 + x + 2) 2x 3 + x + 3 x 2 + 1 2

EXTRA CREDIT Complete the following two problems and turn them into the bin for extra credit: 1. 7x 4 5x 2 6x + 7 2. x 2 x 12 (x 1)(x 2 + 1) HINT: Factor the denominator!