Polnomial and Rational Functions 5 Figure 1 35-mm film, once the standard for capturing photographic images, has been made largel obsolete b digital photograph. (credit film : modification of work b Horia Varlan; credit memor cards : modification of work b Paul Hudson) CHAPTER OUTLINE 5.1 Quadratic Functions 5. Power Functions and Polnomial Functions 5.3 Graphs of Polnomial Functions 5.4 Dividing Polnomials 5.5 Zeros of Polnomial Functions 5. Rational Functions 5.7 Inverses and Radical Functions 5.8 Modeling Using Variation Introduction Digital photograph has dramaticall changed the nature of photograph. No longer is an image etched in the emulsion on a roll of film. Instead, nearl ever aspect of recording and manipulating images is now governed b mathematics. An image becomes a series of numbers, representing the characteristics of light striking an image sensor. When we open an image file, software on a camera or computer interprets the numbers and converts them to a visual image. Photo editing software uses comple polnomials to transform images, allowing us to manipulate the image in order to crop details, change the color palette, and add special effects. Inverse functions make it possible to convert from one file format to another. In this chapter, we will learn about these concepts and discover how mathematics can be used in such applications. 343
SECTION 5.4 DIVIDING POLYNOMIALS 393 LEARNING OBJECTIVES In this section, ou will: Use long division to divide polnomials. Use snthetic division to divide polnomials. 5.4 DIVIDING POLYNOMIALS Figure 1 Lincoln Memorial, Washington, D.C. (credit: Ron Cogswell, Flickr) The eterior of the Lincoln Memorial in Washington, D.C., is a large rectangular solid with length 1.5 meters ( m ), width 40 m, and height 30 m. [15] We can easil find the volume using elementar geometr. V = l ċ w ċ h = 1.5 ċ 40 ċ 30 = 73,800 So the volume is 73,800 cubic meters (m 3 ). Suppose we knew the volume, length, and width. We could divide to find the height. h V = l ċ w = 73,800 1.5 ċ 40 = 30 As we can confirm from the dimensions above, the height is 30 m. We can use similar methods to find an of the missing dimensions. We can also use the same method if an or all of the measurements contain variable epressions. For eample, suppose the volume of a rectangular solid is given b the polnomial 3 4 3 3 33 + 54. The length of the solid is given b 3; the width is given b. To find the height of the solid, we can use polnomial division, which is the focus of this section. Using Long Division to Divide Polnomials We are familiar with the long division algorithm for ordinar arithmetic. We begin b dividing into the digits of the dividend that have the greatest place value. We divide, multipl, subtract, include the digit in the net place value position, and repeat. For eample, let s divide 178 b 3 using long division. 59 3)178 15 8 7 1 Long Division Step 1: 5 3 = 15 and 17 15 = Step : Bring down the 8 Step 3: 9 3 = 7 and 8 7 = 1 Answer: 59 R 1 or 59 1 3 15. National Park Service. Lincoln Memorial Building Statistics. http://www.nps.gov/linc/historculture/lincoln-memorial-building-statistics.htm. Accessed 4/3/014/
394 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Another wa to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check division in elementar arithmetic. dividend = (divisor ċ quotient) + remainder 178 = (3 ċ 59) + 1 = 177 + 1 = 178 We call this the Division Algorithm and will discuss it more formall after looking at an eample. Division of polnomials that contain more than one term has similarities to long division of whole numbers. We can write a polnomial dividend as the product of the divisor and the quotient added to the remainder. The terms of the polnomial division correspond to the digits (and place values) of the whole number division. This method allows us to divide two polnomials. For eample, if we were to divide 3 3 + 4 + 5 b + using the long division algorithm, it would look like this: We have found or ( 3 + 4 ) 7 + 4 7 ( 3 + 4 ) 7 + 4 ( 7 + 14) 18 + 5 7 + 18 ( 3 + 4 ) 7 + 4 ( 7 + 14) 18 + 5 18 + 3 31 Set up the division problem. 3 divided b is. Multipl + b. Subtract. Bring down the net term. 7 divided b is 7. Multipl + b 7. Subtract. Bring down the net term. 18 divided b is 18. Multipl + b 18. Subtract. 3 3 + 4 + 5 = 7 + 18 31_ + + 3 3 + 4 + 5 + = ( + )( 7 + 18) 31 We can identif the dividend, the divisor, the quotient, and the remainder. 3 3 + 4 + 5 = ( + ) ( 7 + 18) + ( 31) Dividend Divisor Quotient Remainder Writing the result in this manner illustrates the Division Algorithm.
SECTION 5.4 DIVIDING POLYNOMIALS 395 the Division Algorithm The Division Algorithm states that, given a polnomial dividend f () and a non-zero polnomial divisor d() where the degree of d() is less than or equal to the degree of f (), there eist unique polnomials q() and r() such that f () = d()q() + r() q() is the quotient and r() is the remainder. The remainder is either equal to zero or has degree strictl less than d(). If r() = 0, then d() divides evenl into f (). This means that, in this case, both d() and q() are factors of f (). How To Given a polnomial and a binomial, use long division to divide the polnomial b the binomial. 1. Set up the division problem.. Determine the first term of the quotient b dividing the leading term of the dividend b the leading term of the divisor. 3. Multipl the answer b the divisor and write it below the like terms of the dividend. 4. Subtract the bottom binomial from the top binomial. 5. Bring down the net term of the dividend.. Repeat steps 5 until reaching the last term of the dividend. 7. If the remainder is non-zero, epress as a fraction using the divisor as the denominator. Eample 1 Using Long Division to Divide a Second-Degree Polnomial Divide 5 + 3 b + 1. Solution + 1)5 + 3 5 + 1)5 + 3 5 + 1)5 + 3 (5 + 5) 5 + 1)5 + 3 (5 + 5) ( ) 0 Set up division problem. 5 divided b is 5. Multipl + 1 b 5. Subtract. Bring down the net term. divided b is. Multipl + 1 b. Subtract. The quotient is 5. The remainder is 0. We write the result as or 5 + 3 = 5 + 1 5 + 3 = ( + 1)(5 ) Analsis This division problem had a remainder of 0. This tells us that the dividend is divided evenl b the divisor, and that the divisor is a factor of the dividend.
39 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Eample Using Long Division to Divide a Third-Degree Polnomial Divide 3 + 11 31 + 15 b 3. Solution + 5 7 3 ) 3 + 11 31 + 1 ( 3 4 ) 15 31 (15 + 10) 1 + 15 ( 1 + 14) 1 3 divided b 3 is. Multipl 3 b. Subtract. Bring down the net term. 15 divided b 3 is 5. Multipl 3 b 5. Subtract. Bring down the net term. 1 divided b 3 is 7. Multipl 3 b 7. Subtract. The remainder is 1. There is a remainder of 1. We can epress the result as: 3 + 11 31 + 15 = 3 + 5 7 1 + 3 Analsis We can check our work b using the Division Algorithm to rewrite the solution. Then multipl. Notice, as we write our result, the dividend is 3 + 11 31 + 15 the divisor is 3 the quotient is + 5 7 the remainder is 1 (3 )( + 5 7) + 1 = 3 + 11 31 + 15 Tr It #1 Divide 1 3 1 + 0 3 b 4 + 5. Using Snthetic Division to Divide Polnomials As we ve seen, long division of polnomials can involve man steps and be quite cumbersome. Snthetic division is a shorthand method of dividing polnomials for the special case of dividing b a linear factor whose leading coefficient is 1. To illustrate the process, recall the eample at the beginning of the section. Divide 3 3 + 4 + 5 b + using the long division algorithm. The final form of the process looked like this: + + 18 ( 3 + 4 ) 7 + 4 ( 7 14) 18 + 5 (18 + 3) 31 There is a lot of repetition in the table. If we don t write the variables but, instead, line up their coefficients in columns under the division sign and also eliminate the partial products, we alread have a simpler version of the entire problem.
SECTION 5.4 DIVIDING POLYNOMIALS 397 ) 3 4 5 4 7 14 18 3 31 Snthetic division carries this simplification even a few more steps. Collapse the table b moving each of the rows up to fill an vacant spots. Also, instead of dividing b, as we would in division of whole numbers, then multipling and subtracting the middle product, we change the sign of the divisor to, multipl and add. The process starts b bringing down the leading coefficient. 3 4 5 4 14 3 7 18 31 We then multipl it b the divisor and add, repeating this process column b column, until there are no entries left. The bottom row represents the coefficients of the quotient; the last entr of the bottom row is the remainder. In this case, the quotient is 7 + 18 and the remainder is 31. The process will be made more clear in Eample 3. snthetic division Snthetic division is a shortcut that can be used when the divisor is a binomial in the form k where k is a real number. In snthetic division, onl the coefficients are used in the division process. How To Given two polnomials, use snthetic division to divide. 1. Write k for the divisor.. Write the coefficients of the dividend. 3. Bring the lead coefficient down. 4. Multipl the lead coefficient b k. Write the product in the net column. 5. Add the terms of the second column.. Multipl the result b k. Write the product in the net column. 7. Repeat steps 5 and for the remaining columns. 8. Use the bottom numbers to write the quotient. The number in the last column is the remainder and has degree 0, the net number from the right has degree 1, the net number from the right has degree, and so on. Eample 3 Using Snthetic Division to Divide a Second-Degree Polnomial Use snthetic division to divide 5 3 3 b 3. Solution Begin b setting up the snthetic division. Write k and the coefficients. 3 5 3 3 Bring down the lead coefficient. Multipl the lead coefficient b k. 3 5 3 3 15 5 Continue b adding the numbers in the second column. Multipl the resulting number b k. Write the result in the net column. Then add the numbers in the third column. 3 5 3 3 15 3 5 1 0 The result is 5 + 1. The remainder is 0. So 3 is a factor of the original polnomial.
398 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Analsis Just as with long division, we can check our work b multipling the quotient b the divisor and adding the remainder. ( 3)(5 + 1) + 0 = 5 3 3 Eample 4 Using Snthetic Division to Divide a Third-Degree Polnomial Use snthetic division to divide 4 3 + 10 0 b +. Solution The binomial divisor is + so k =. Add each column, multipl the result b, and repeat until the last column is reached. 4 10 0 8 4 0 4 10 0 The result is 4 + 10. The remainder is 0. Thus, + is a factor of 4 3 + 10 0. Analsis The graph of the polnomial function f () = 4 3 + 10 0 in Figure shows a zero at = k =. This confirms that + is a factor of 4 3 + 10 0. 1.8 5 3 14 1 10 8 4 1 8 10 1 14 1 18 0 1 3 4 5 Figure Eample 5 Using Snthetic Division to Divide a Fourth-Degree Polnomial Use snthetic division to divide 9 4 + 10 3 + 7 b 1. Solution Notice there is no -term. We will use a zero as the coefficient for that term. The result is 9 3 + + 8 + 8 + Tr It # _ 1. 1 9 10 7 0 9 1 8 8 9 1 8 8 Use snthetic division to divide 3 4 + 18 3 3 + 40 b + 7. Using Polnomial Division to Solve Application Problems Polnomial division can be used to solve a variet of application problems involving epressions for area and volume. We looked at an application at the beginning of this section. Now we will solve that problem in the following eample.
SECTION 5.4 DIVIDING POLYNOMIALS 399 Eample Using Polnomial Division in an Application Problem The volume of a rectangular solid is given b the polnomial 3 4 3 3 33 + 54. The length of the solid is given b 3 and the width is given b. Find the height, t, of the solid. Solution There are a few was to approach this problem. We need to divide the epression for the volume of the solid b the epressions for the length and width. Let us create a sketch as in Figure 3. Height Length 3 Width Figure 3 We can now write an equation b substituting the known values into the formula for the volume of a rectangular solid. To solve for h, first divide both sides b 3. V = l ċ w ċ h 3 4 3 3 33 + 54 = 3 ċ ( ) ċ h 3 ċ ( ) ċ h 3 = 34 33 33 + 54 3 ( )h = 3 11 + 18 Now solve for h using snthetic division. h = 3 11 + 18 1 1 11 18 18 1 1 9 0 The quotient is + 9 and the remainder is 0. The height of the solid is + 9. Tr It #3 The area of a rectangle is given b 3 3 + 14 3 +. The width of the rectangle is given b +. Find an epression for the length of the rectangle. Access these online resources for additional instruction and practice with polnomial division. Dividing a Trinomial b a Binomial Using Long Division (http://openstacollege.org/l/dividetribild) Dividing a Polnomial b a Binomial Using Long Division (http://openstacollege.org/l/dividepolbild) E : Dividing a Polnomial b a Binomial Using Snthetic Division (http://openstacollege.org/l/dividepolbisd) E 4: Dividing a Polnomial b a Binomial Using Snthetic Division (http://openstacollege.org/l/dividepolbisd4)
400 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 5.4 SECTION EXERCISES VERBAL 1. If division of a polnomial b a binomial results in a remainder of zero, what can be conclude?. If a polnomial of degree n is divided b a binomial of degree 1, what is the degree of the quotient? ALGEBRAIC For the following eercises, use long division to divide. Specif the quotient and the remainder. 3. ( + 5 1) ( 1) 4. ( 9 5) ( 5) 5. (3 + 3 + 14) ( + 7). (4 10 + ) (4 + ) 7. ( 5 5) ( + 5) 8. ( 1) ( + 1) 9. ( 3 + ) ( + ) 10. ( 3 1) ( 5) 11. (3 5 + 4) (3 + 1) 1. ( 3 3 + 5 ) ( ) 13. ( 3 + 3 4 + 15) ( + 3) For the following eercises, use snthetic division to find the quotient. 14. (3 3 + 4) ( + 3) 15. ( 3 7 + ) ( 4) 1. ( 3 10 7 15) ( + 1) 17. (4 3 1 5 1) ( + 1) 18. (9 3 9 + 18 + 5) (3 1) 19. (3 3 + 4) ( + 3) 0. ( 3 + 4) ( 3) 1. ( 3 + 7 13 3) ( 3). (3 3 5 + + 3) ( + ) 3. (4 3 5 + 13) ( + 4) 4. ( 3 3 + ) ( + ) 5. ( 3 1 + 147 343) ( 7). ( 3 15 + 75 15) ( 5) 7. (9 3 + ) (3 1) 8. ( 3 + 5 + ) (3 + 1) 9. ( 4 + 3 3 + 1) ( + 1) 30. ( 4 3 + 1) ( 1) 31. ( 4 + 3 3 + + ) ( + 3) 3. ( 4 10 3 + 37 0 + 3) ( ) 33. ( 4 8 3 + 4 3 + 1) ( ) 34. ( 4 + 5 3 3 13 + 10) ( + 5) 35. ( 4 1 3 + 54 108 + 81) ( 3) 3. (4 4 3 4 + ) ( 1) 37. (4 4 + 3 4 + + ) ( + 1) For the following eercises, use snthetic division to determine whether the first epression is a factor of the second. If it is, indicate the factorization. 38., 4 3 3 8 + 4 39., 3 4 3 5 + 10 40. + 3, 4 3 + 5 + 8 41., 4 4 15 4 4. 1, 4 3 + 1 43. + 1 3, 3 4 + 3 3 + 1 GRAPHICAL For the following eercises, use the graph of the third-degree polnomial and one factor to write the factored form of the polnomial suggested b the graph. The leading coefficient is one. 44. Factor is + 3 45. Factor is + + 4 4. Factor is + + 5 18 18 30 1 1 0 10 4 4 4 10 1 1 0 18 18 30
SECTION 5.4 SECTION EXERCISES 401 47. Factor is + + 1 0 40 0 48. Factor is + + 18 1 4 4 0 0 1 0 18 For the following eercises, use snthetic division to find the quotient and remainder. 49. 43 33 5. 43 1 + 4 TECHNOLOGY 50. 3 + 5 + 3 53. 4 + 51. 33 + 5 1 For the following eercises, use a calculator with CAS to answer the questions. 54. Consider k 1 with k = 1,, 3. What do ou 55. Consider k + 1 for k = 1, 3, 5. What do ou epect 1 + 1 epect the result to be if k = 4? the result to be if k = 7? 5. Consider 4 k 4 for k = 1,, 3. What do ou 57. Consider k with k = 1,, 3. What do ou epect k + 1 epect the result to be if k = 4? k 58. Consider with k = 1,, 3. What do ou epect 1 the result to be if k = 4? EXTENSIONS the result to be if k = 4? For the following eercises, use snthetic division to determine the quotient involving a comple number. 59. + 1 0. + 1 1. + 1 i i + i. + 1 + i REAL-WORLD APPLICATIONS 3. 3 + 1 i For the following eercises, use the given length and area of a rectangle to epress the width algebraicall. 4. Length is + 5, area is + 9 5. 5. Length is + 5, area is 4 3 + 10 + + 15. Length is 3 4, area is 4 8 3 + 9 9 4 For the following eercises, use the given volume of a bo and its length and width to epress the height of the bo algebraicall. 7. Volume is 1 3 + 0 1 3, length is + 3, width is 3 4. 9. Volume is 10 3 + 7 + 4, length is 5 4, width is + 3. 8. Volume is 18 3 1 40 + 48, length is 3 4, width is 3 4. 70. Volume is 10 3 + 30 8 4, length is, width is + 3. For the following eercises, use the given volume and radius of a clinder to epress the height of the clinder algebraicall. 71. Volume is π(5 3 5 9 3), radius is 5 + 1. 7. Volume is π(4 3 + 1 15 50), radius is + 5. 73. Volume is π(3 4 + 4 3 + 4 1 3), radius is + 4.