The Partial Fraction Decomposition of a Rational Expression with Distinct Linear Factors in the Denominator. x + 14 (x 4)(x + 2)

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1 Notes: Partial Fraction Decomposition The Partial Fraction Decomposition of a Rational Expression with Distinct Linear Factors in the Denominator 1. Find the partial fraction decomposition of x + 14 (x 4)(x + 2) Solution: We begin by setting up the partial fraction decomposition with unknown constants. So, we write a constant over each of the two linear factors in the denominator. x + 14 (x 4)(x + 2) = A x 4 + B x 2 At this point we will now need to find the constants A and B. Now, we have to find the least common denominator (L.C.D), which in this case is going to be the product of both the denominators. So, the L.C.D is (x 4)(x + 2) x + 14 A(x + 2) + B(x 4) = (x 4)(x + 2) (x 4)(x + 2) x + 14 = A(x + 2) + B(x 4) Since our goal is to find the values of the constants, we can set x = 4, doing this will elimininate the B constant = A(4 + 2) + B(4 4) 18 = 6A 3 = A Now, that we have found A, we next try to find B, to find this we need to eliminate A so we can set x = = A( 2 + 2) + B( 2 4) 12 = 6B 2 = B Now we found both values, that is A = 3 and B = 2 so our partial fraction decomposition has the form: x + 14 (x 4)(x + 2) = 3 x 4 2 x + 2 1

2 Partial Fraction Decomposition with Repeated Linear Factors 2. Find the partial fraction decomposition of x 18 x(x 3) 2 Solution: We begin by setting up the partial fraction decomposition with unknown constants. Now because the linear factor x 3 occurs twice, we must include one fraction with a constant numerator for each power of x 3. So, x 18 x(x 3) 2 = A x + B x 3 + C (x 3) 2 At this point we will now need to find the constants A, B and C. Now, we have to find the least common denominator (L.C.D). So, the L.C.D is x(x 3) 2 x 18 x(x 3) 2 = A(x 3)2 + B(x)(x 3) + C(x) x(x 3) 2 x 18 = A(x 3) 2 + B(x)(x 3) + C(x) Since our goal is to find the values of the constants A, B and C, we begin by setting x = = A(0 3) 2 + B(0)(0 3) + C(0) 18 = A( 3) 2 18 = 9A 2 = A Now we can set x = = A(3 3) 2 + B(3)(3 3) + C(3) 15 = 3C 5 = C At this point we have used x = 0 and x = 3 which allowed us to find the values of A and C but now we need to find the value of B. In order to do this we now we have to choose a value of x along with the values of A and C. So choose x = 1, we have, 2

3 1 18 = A(1 3) 2 + B(1)(1 3) + C(1) 17 = A( 2) 2 + B( 2) + C 17 = 4A 2B + C given, A = 2, C = 5 17 = 4( 2) 2B + ( 5) 17 = 8 2B 5 17 = 13 2B = 2B 4 = 2B 2 = B So, A = 2, B = 2 and C = 5, hence the partial fraction decomposition is x 18 x(x 3) 2 = 2 x + 2 x 3 5 (x 3) 2 The Partial Fraction Decomposition of a Rational Expression with Prime, Nonrepeated Quadratic Factors in the Denominator 3. Find the partial fraction decomposition of 3x x + 14 (x 2)(x 2 + 2x + 4) Solution: We begin by setting up the partial fraction decomposition with unknown constants. Now we observe that the denominator contains both a linear term x 2 and a nonfactorable quadratic term. For the linear term we place a constant A and for the quadratic part we place a linear expression Bx + C. So, 3x x + 4 (x 2)(x 2 + 2x + 4) = A x 2 + Bx + C x 2 + 2x + 4 At this point we will now need to find the constants A, B and C. Now, we have to find the least common denominator (L.C.D). So, the L.C.D is (x 2)(x 2 + 2x + 4) 3x x + 14 (x 2)(x 2 + 2x + 4) = A(x2 + 2x + 4) + (Bx + C)(x 2) (x 2)(x 2 + 2x + 4) 3x x + 14 = A(x 2 + 2x + 4) + (Bx + C)(x 2) Since our goal is to find the values of the constants A, B and C, we begin by setting x = 2 3

4 3(2) (2) + 14 = A((2) 2 + 2(2) + 4) + (Bx + C)(2 2) 3(4) = A( ) 60 = 12A 5 = A Now, since the quadratic term is not factorable, we must substitue two values for x so choose x = 1 and x = 1. So, when x = 1 3(1) (1) + 14 = A( (1) + 4) + (B(1) + C)(1 2) 3(1) = A( ) + (B + C)( 1) 34 = 7A + ( 1)(B + C) given, A = = 7(5) + ( 1)(B + C) 34 = 35 + ( 1)(B + C) = ( 1)(B + C) 1 = ( 1)(B + C) 1 = B + C given, x = 1 3( 1) ( 1) + 4 = A(( 1) 2 + 2( 1) + 4) + (B( 1) + C)( 1 2) 3(1) = A( )( B + C)( 3) = 3A + ( 3)( B + C) 0 = 3A + ( 3)( B + C) given, A = 5. 0 = 3(5) + ( 3)( B + C) 15 = ( 3)( B + C) 5 = B + C So, now we have two equations to solve, to solve we can use the method of substitution or elimination. (1) (2) B + C = 1 B + C = 5 Now, adding both equation we find: 2C = 6 C = 3 Now, substituting C = 3 into equation (1) above we find: B + 3 = 1 B = 2 4

5 So, we have A = 5, B = 2 and C = 3. Hence the partial fraction decompositon is : 3x x + 4 (x 2)(x 2 + 2x + 4) = 5 x 2 + 2x + 3 x 2 + 2x + 4 The Partial Fraction Decomposition of a Rational Expression with a Prime, Repeated Quadratic Factor in the Denominator 4. Find the partial fraction decomposition of Solution: We begin by setting up the partial fraction decomposition with unknown constants. Now we observe that the denominator contains a repeated quadratic factor, so we must include one fraction with a linear numerator for each power of x = Ax + B x Cx + D At this point we will now need to find the constants A, B, C and D. Now, we have to find the least common denominator (L.C.D). So, the L.C.D is = (Ax + B)(x2 + 1) + (Cx + D) = (Ax + B)(x 2 + 1) + (Cx + D) Since our goal is to find the values of the constants A, B,C and D, we begin by setting x = 0 5(0) 3 3(0) 2 + 7(0) 3 = (A(0) + B)((0) 2 + 1) + C(0) + D 3 = B + D Now, since no real choice of x will yield a zero result, we can choose different values of x thereby forming a series of equations that we will have to solve. So choosing x = 1, x = 1 and x = 2. When x = 1 5

6 5( 1) 3 3( 1) 2 + 7( 1) 3 = (A( 1) + B)(( 1) 2 + 1) + C( 1) + D 5( 1) 3(1) 7 3 = ( A + B)(1 + 1) C + D = 2A + 2B C + D 18 = 2A + 2B C + D when x = 1 5(1) 3 3(1) 2 + 7(1) 3 = (A(1) + B)((1) 2 + 1) + C(1) + D = (A + B)(2) + C + D 6 = 2A + 2B + C + D when x = 2 5(2) 3 3(2) 2 + 7(2) 3 = (A(2) + B)((2) 2 + 1) + C(2) + D 5(8) 3(4) = (2A + B)(4 + 1) + 2C + D = 10A + 5B + 2C + D 39 = 10A + 5B + 2C + D So, now we have four (4) equations to solve: (3) (4) (5) (6) B + D = 3 2A + 2B C + D = 18 2A + 2B + C + D = 6 10A + 5B + 2C + D = 39 Adding equations (4) & (5) (7) 4B + 2D = 12 Multiplying equation (3) by -2 and adding the result to equation (7) we find: 2B = 6 B = 3 Now, substituting B = 3 into equation (3) 3 + D = 3 D = 0 Substituting B = 3 and D = 0 into equations (5) 2A + 2( 3) + C + (0) = 6 2A 6 + C = 6 2A + C = 12 6

7 Substituting B = 3 and D = 0 into equations (6) 10A + 5( 3) + 2C + (0) = 39 10A C = 39 10A + 2C = 54 At this point we have two new equations to solve so, (8) (9) 2A + C = 12 10A + 2C = 54 multiplying equation (8) by -2 and adding the result to equation (9) we get 6A = 30 A = 5 Now substituting A = 5 into equation (8) we get 2(5) + C = C = 12 C = 2 So now we see that A = 5, B = 3, C = 2 and D = 0. As a result the partial fraction decomposition is : = 5x 3 x x ALTERNATIVELY Solution: We begin by setting up the partial fraction decomposition with unknown constants. Now we observe that the denominator contains a repeated quadratic factor, so we must include one fraction with a linear numerator for each power of x = Ax + B x Cx + D At this point we will now need to find the constants A, B, C and D. Now, we have to find the least common denominator (L.C.D). So, the L.C.D is = (Ax + B)(x2 + 1) + (Cx + D) = (Ax + B)(x 2 + 1) + (Cx + D) 7

8 At this point we can expand the RHS = Ax 3 + Bx 2 + Ax + B + Cx + D = Ax 3 + Bx 2 + (A + C)x + B + D So, now we can equate the coefficients of powers of x and also equate constant terms (10) (11) (12) (13) A = 5 B = 3 A + C = 7 B + D = 3 Substituting A = 5 into equation (12) we find: 5 + C = 7 C = 2 Substituting B = 3 into equation (13) we find: 3 + D = 3 D = 0 So as a result A = 5, B = 3, C = 2 and D = 0. As a result the partial fraction decomposition is : = 5x 3 x x Practice Problems 1. Find the partial fraction decomposition of the following: a. 5x 1 (x 3)(x + 4) b. x + 2 x(x 1) 2 c. 8x x 20 (x + 3)(x 2 + x + 2) d. 2x 3 + x + 3 Answers a. 2 x x + 4 b. 2 x 2 x (x 1) 2 c. 2 x x 8 x 2 + x + 2 d. 2x x x + 3 8