5. TEXAS ESSENTIAL KNOWLEDGE AND SKILLS 2A.7.D 2A.7.E Factoring Polnomials Essential Question How can ou factor a polnomial? Factoring Polnomials Work with a partner. Match each polnomial equation with the graph of its related polnomial function. Use the -intercepts of the graph to write each polnomial in factored form. Eplain our reasoning. a. 2 + 5 + = 0 b. 3 2 2 + 2 = 0 c. 3 + 2 2 = 0 d. 3 = 0 e. 5 2 + = 0 f. 2 3 2 + 2 = 0 A. B. C. D. E. F. USING PROBLEM-SOLVING STRATEGIES To be proficient in math, ou need to check our answers to problems and continuall ask ourself, Does this make sense? Factoring Polnomials Work with a partner. Use the -intercepts of the graph of the polnomial function to write each polnomial in factored form. Eplain our reasoning. Check our answers b multipling. a. f() = 2 2 b. f() = 3 2 2 c. f() = 3 2 2 3 d. f() = 3 3 2 + 3 e. f() = + 2 3 2 2 f. f() = 10 2 + 9 Communicate Your Answer 3. How can ou factor a polnomial?. What information can ou obtain about the graph of a polnomial function written in factored form? Section 5. Factoring Polnomials 231
5. Lesson What You Will Learn Core Vocabular factored completel, p. 232 factor b grouping, p. 233 quadratic form, p. 233 Previous zero of a function snthetic division Factor polnomials. Use the Factor Theorem. Factoring Polnomials Previousl, ou factored quadratic polnomials. You can also factor polnomials with degree greater than 2. Some of these polnomials can be factored completel using techniques ou have previousl learned. A factorable polnomial with integer coefficients is factored completel when it is written as a product of unfactorable polnomials with integer coefficients. Finding a Common Monomial Factor Factor each polnomial completel. a. 3 2 5 b. 3 5 8 3 c. 5z + 30z 3 + 5z 2 a. 3 2 5 = ( 2 5) Factor common monomial. = ( 5)( + 1) Factor trinomial. b. 3 5 8 3 = 3 3 ( 2 1) Factor common monomial. = 3 3 ( )( + ) Difference of Two Squares Pattern c. 5z + 30z 3 + 5z 2 = 5z 2 (z 2 + z + 9) Factor common monomial. = 5z 2 (z + 3) 2 Perfect Square Trinomial Pattern Monitoring Progress Help in English and Spanish at BigIdeasMath.com Factor the polnomial completel. 1. 3 7 2 + 10 2. 3n 7 75n 5 3. 8m 5 1m + 8m 3 In part (b) of Eample 1, the special factoring pattern for the difference of two squares was used to factor the epression completel. There are also factoring patterns that ou can use to factor the sum or difference of two cubes. Core Concept Special Factoring Patterns Sum of Two Cubes Eample a 3 + b 3 = (a + b)(a 2 ab + b 2 ) 3 + 1 = () 3 + 1 3 = ( + 1)(1 2 + 1) Difference of Two Cubes Eample a 3 b 3 = (a b)(a 2 + ab + b 2 ) 27 3 8 = (3) 3 2 3 = (3 2)(9 2 + + ) 232 Chapter 5 Polnomial Functions
Factoring the Sum or Difference of Two Cubes Factor (a) 3 125 and (b) 1s 5 + 5s 2 completel. a. 3 125 = 3 5 3 Write as a 3 b 3. = ( 5)( 2 + 5 + 25) Difference of Two Cubes Pattern b. 1s 5 + 5s 2 = 2s 2 (8s 3 + 27) Factor common monomial. = 2s 2 [(2s) 3 + 3 3 ] Write 8s 3 + 27 as a 3 + b 3. = 2s 2 (2s + 3)(s 2 s + 9) Sum of Two Cubes Pattern For some polnomials, ou can factor b grouping pairs of terms that have a common monomial factor. The pattern for factoring b grouping is shown below. ra + rb + sa + sb = r(a + b) + s(a + b) = (r + s)(a + b) Factoring b Grouping Factor (a) 3 + 3 2 7 21 and (b) z 2z 3 27z + 5 completel. a. 3 + 3 2 7 21 = 2 ( + 3) 7( + 3) Factor b grouping. = ( 2 7)( + 3) Distributive Propert b. z 2z 3 27z + 5 = z 3 (z 2) 27(z 2) Factor b grouping. = (z 3 27)(z 2) Distributive Propert = (z 2)(z 3)(z 2 + 3z + 9) Difference of Two Cubes Pattern ANALYZING MATHEMATICAL RELATIONSHIPS The epression 1 81 is in quadratic form because it can be written as u 2 81 where u = 2. An epression of the form au 2 + bu + c, where u is an algebraic epression, is said to be in quadratic form. The factoring techniques ou have studied can sometimes be used to factor such epressions. Factor 1 81 completel. Factoring Polnomials in Quadratic Form 1 81 = ( 2 ) 2 9 2 Write as a 2 b 2. = ( 2 + 9)( 2 9) Difference of Two Squares Pattern = ( 2 + 9)(2 3)(2 + 3) Difference of Two Squares Pattern Monitoring Progress Help in English and Spanish at BigIdeasMath.com Factor the polnomial completel.. a 3 + 27 5. z 5 750z 2. 3 + 2 7. 3 3 + 2 + 9 + 3 8. 1n + 25 9. 5w 25w + 30w 2 Section 5. Factoring Polnomials 233
The Factor Theorem When dividing polnomials in the previous section, the eamples had nonzero remainders. Suppose the remainder is 0 when a polnomial f() is divided b k. Then, f() k = q() + 0 k = q() where q() is the quotient polnomial. Therefore, f() = ( k) q(), so that k is a factor of f(). This result is summarized b the Factor Theorem, which is a special case of the Remainder Theorem. READING In other words, k is a factor of f () if and onl if k is a zero of f. Core Concept The Factor Theorem A polnomial f() has a factor k if and onl if f(k) = 0. STUDY TIP In part (b), notice that direct substitution would have resulted in more difficult computations than snthetic division. Determining Whether a Linear Binomial Is a Factor Determine whether (a) 2 is a factor of f() = 2 + 2 and (b) + 5 is a factor of f() = 3 + 15 3 2 + 25. a. Find f(2) b direct substitution. b. Find f( 5) b snthetic division. f(2) = 2 2 + 2(2) 5 3 15 1 0 25 = + 15 0 5 5 = 3 0 1 5 0 Because f(2) 0, the binomial Because f( 5) = 0, the binomial 2 is not a factor of + 5 is a factor of f() = 2 + 2. f() = 3 + 15 3 2 + 25. Factoring a Polnomial Show that + 3 is a factor of f() = + 3 3 3. Then factor f() completel. ANOTHER WAY Notice that ou can factor f () b grouping. f () = 3 ( + 3) 1( + 3) = ( 3 1)( + 3) = ( + 3)( 1) ( 2 + + 1) Show that f( 3) = 0 b snthetic division. 3 1 3 0 1 3 3 0 0 3 1 0 0 1 0 Because f( 3) = 0, ou can conclude that + 3 is a factor of f() b the Factor Theorem. Use the result to write f() as a product of two factors and then factor completel. f() = + 3 3 3 Write original polnomial. = ( + 3)( 3 1) Write as a product of two factors. = ( + 3)( 1)( 2 + + 1) Difference of Two Cubes Pattern 23 Chapter 5 Polnomial Functions
Because the -intercepts of the graph of a function are the zeros of the function, ou can use the graph to approimate the zeros. You can check the approimations using the Factor Theorem. STUDY TIP You could also check that 2 is a zero using the original function, but using the quotient polnomial helps ou find the remaining factor. Real-Life Application During the first 5 seconds of a roller coaster ride, the function h(t) = t 3 21t 2 + 9t + 3 represents the height h (in feet) of the roller coaster after t seconds. How long is the roller coaster at or below ground level in the first 5 seconds? 1. Understand the Problem You are given a function rule that represents the height of a roller coaster. You are asked to determine how long the roller coaster is at or below ground during the first 5 seconds of the ride. 2. Make a Plan Use a graph to estimate the zeros of the function and check using the Factor Theorem. Then use the zeros to describe where the graph lies below the t-ais. 3. Solve the Problem From the graph, two of the zeros appear to be 1 and 2. The third zero is between and 5. Step 1 Determine whether 1 is a zero using snthetic division. 1 1 9 3 25 3 h( 1) = 0, so 1 is a zero of h 5 3 0 and t + 1 is a factor of h(t). Step 2 Determine whether 2 is a zero. If 2 is also a zero, then t 2 is a factor of the resulting quotient polnomial. Check using snthetic division. 2 5 3 8 3 17 0 So, h(t) = (t + 1)(t 2)(t 17). The factor t 17 indicates that the zero between and 5 is 17, or.25. The zeros are 1, 2, and.25. Onl t = 2 and t =.25 occur in the first 5 seconds. The graph shows that the roller coaster is at or below ground level for.25 2 = 2.25 seconds.. Look Back Use a table of values to verif the positive zeros and heights between the zeros. zero zero 80 0 The remainder is 0, so t 2 is a factor of h(t) and 2 is a zero of h. X.5 1.25 2 2.75 3.5.25 5 X=2 h(t) = t 3 21t 2 + 9t + 3 Y1 33.75 20.25 0-1.88-20.25 0 5 h 1 5 t negative Monitoring Progress Help in English and Spanish at BigIdeasMath.com 10. Determine whether is a factor of f() = 2 2 + 5 12. 11. Show that is a factor of f() = 3 5 2. Then factor f() completel. 12. In Eample 7, does our answer change when ou first determine whether 2 is a zero and then whether 1 is a zero? Justif our answer. Section 5. Factoring Polnomials 235
5. Eercises Dnamic Solutions available at BigIdeasMath.com Vocabular and Core Concept Check 1. COMPLETE THE SENTENCE The epression 9 9 is in form because it can be written as u 2 9 where u =. 2. VOCABULARY Eplain when ou should tr factoring a polnomial b grouping. 3. WRITING How do ou know when a polnomial is factored completel?. WRITING Eplain the Factor Theorem and wh it is useful. Monitoring Progress and Modeling with Mathematics In Eercises 5 12, factor the polnomial completel. (See Eample 1.) 5. 3 2 2 2. k 5 100k 3 7. 3p 5 192p 3 8. 2m 2m 5 + m 9. 2q + 9q 3 18q 2 10. 3r 11r 5 20r 11. 10w 10 19w 9 + w 8 12. 18v 9 + 33v 8 + 1v 7 In Eercises 13 20, factor the polnomial completel. (See Eample 2.) 13. 3 + 1. 3 + 512 15. g 3 33 1. c 3 27 17. 3h 9 192h 18. 9n 51n 3 19. 1t 7 + 250t 20. 135z 11 1080z 8 ERROR ANALYSIS In Eercises 21 and 22, describe and correct the error in factoring the polnomial. 21. 3 3 + 27 = 3( 2 + 9) = 3( + 3)( 3) In Eercises 23 30, factor the polnomial completel. (See Eample 3.) 23. 3 5 2 + 30 2. m 3 m 2 + 7m 7 25. 3a 3 + 18a 2 + 8a + 8 2. 2k 3 20k 2 + 5k 50 27. 3 8 2 + 32 28. z 3 5z 2 9z + 5 29. q + 5q 3 + 8q + 0 30. n 192n 3 + 27n 81 In Eercises 31 38, factor the polnomial completel. (See Eample.) 31. 9k 9 32. m 25 33. c + 9c 2 + 20 3. 3 2 28 35. 1z 81 3. 81a 25 37. 3r 8 + 3r 5 0r 2 38. n 12 32n 7 + 8n 2 In Eercises 39, determine whether the binomial is a factor of the polnomial. (See Eample 5.) 39. f() = 2 3 + 5 2 37 0; 0. g() = 3 3 28 2 + 29 + 10; + 7 22. 9 + 8 3 = ( 3 ) 3 + (2) 3 = ( 3 + 2)[( 3 ) 2 ( 3 )(2) + (2) 2 ] = ( 3 + 2)( 2 + 2 ) 1. h() = 5 15 9 3 ; + 3 2. g() = 8 5 58 + 0 3 + 10; 3. h() = 3 8 2 + 1; +. t() = 8 + 3 3 138 2 3; + 2 23 Chapter 5 Polnomial Functions
In Eercises 5 50, show that the binomial is a factor of the polnomial. Then factor the polnomial completel. (See Eample.) 5. g() = 3 2 20; +. t() = 3 5 2 9 + 5; 5 7. f() = 3 8 + 8; 8. s() = + 3 25; + 9. r() = 3 37 + 8; + 7 50. h() = 3 2 2 3; + 2 ANALYZING RELATIONSHIPS In Eercises 51 5, match the function with the correct graph. Eplain our reasoning. 51. f() = ( 2)( 3)( + 1) 52. g() = ( + 2)( + 1)( 2) 53. h() = ( + 2)( + 3)( 1) 5. k() = ( 2)( 1)( + 2) A. C. 55. MODELING WITH MATHEMATICS The volume (in cubic inches) of a shipping bo is modeled b V = 2 3 19 2 + 39, where is the length (in inches). Determine the values of for which the model makes sense. Eplain our reasoning. (See Eample 7.) 0 20 V B. D. 2 8 5. MODELING WITH MATHEMATICS The volume (in cubic inches) of a rectangular birdcage can be modeled b V = 3 3 17 2 + 29 15, where is the length (in inches). Determine the values of for which the model makes sense. Eplain our reasoning. 2 USING STRUCTURE In Eercises 57, use the method of our choice to factor the polnomial completel. Eplain our reasoning. 57. a + a 5 30a 58. 8m 3 33 59. z 3 7z 2 9z + 3 0. 2p 8 12p 5 + 1p 2 1. r 3 + 729 2. 5 5 10 0 3 3. 1n 1. 9k 3 2k 2 + 3k 8 5. REASONING Determine whether each polnomial is factored completel. If not, factor completel. a. 7z (2z 2 z ) b. (2 n)(n 2 + n)(3n 11) c. 3( 5)(9 2 ). PROBLEM SOLVING The profit P (in millions of dollars) for a T-shirt manufacturer can be modeled b P = 3 + 2 +, where is the number (in millions) of T-shirts produced. Currentl the compan produces million T-shirts and makes a profit of $ million. What lesser number of T-shirts could the compan produce and still make the same profit? V 7. PROBLEM SOLVING The profit P (in millions of dollars) for a shoe manufacturer can be modeled b P = 1 3 +, where is the number (in millions) of shoes produced. The compan now produces 1 million shoes and makes a profit of $25 million, but it would like to cut back production. What lesser number of shoes could the compan produce and still make the same profit? Section 5. Factoring Polnomials 237
8. THOUGHT PROVOKING Fill in the blank of the divisor so that the remainder is 0. Justif our answer. f() = 3 3 2 ; ( + ) 9. COMPARING METHODS You are taking a test where calculators are not permitted. One question asks ou to evaluate g(7) for the function g() = 3 7 2 + 28. You use the Factor Theorem and snthetic division and our friend uses direct substitution. Whose method do ou prefer? Eplain our reasoning. 70. MAKING AN ARGUMENT You divide f() b ( a) and find that the remainder does not equal 0. Your friend concludes that f() cannot be factored. Is our friend correct? Eplain our reasoning. 71. CRITICAL THINKING What is the value of k such that 7 is a factor of h() = 2 3 13 2 k + 105? Justif our answer. 72. HOW DO YOU SEE IT? Use the graph to write an equation of the cubic function in factored form. Eplain our reasoning. 7. REASONING The graph of the function f() = + 3 3 + 2 2 + + 3 is shown. Can ou use the Factor Theorem to factor f()? Eplain. 75. MATHEMATICAL CONNECTIONS The standard equation of a circle with radius r and center (h, k) is ( h) 2 + ( k) 2 = r 2. Rewrite each equation of a circle in standard form. Identif the center and radius of the circle. Then graph the circle. r (h, k) a. 2 + + 9 + 2 = 25 b. 2 + + 2 = 9 (, ) c. 2 8 + 1 + 2 + 2 + 1 = 3 2 2 73. ABSTRACT REASONING Factor each polnomial completel. a. 7ac 2 + bc 2 7ad 2 bd 2 b. 2n 2 n + 1 c. a 5 b 2 a 2 b + 2a b 2ab 3 + a 3 b 2 Maintaining Mathematical Proficienc Solve the quadratic equation b factoring. (Section.1) 77. 2 30 = 0 78. 2 2 10 72 = 0 79. 3 2 11 + 10 = 0 80. 9 2 28 + 3 = 0 Solve the quadratic equation b completing the square. (Section.3) 81. 2 12 + 3 = 1 82. 2 8 11 = 0 83. 3 2 + 30 + 3 = 0 8. 2 + 3 = 0 7. CRITICAL THINKING Use the diagram to complete parts (a) (c). a. Eplain wh a 3 b 3 is equal to the sum of the volumes of the solids I, II, and III. b. Write an algebraic epression for the volume of each of the three solids. Leave II III our epressions in a b factored form. b b c. Use the results from I part (a) and part (b) a to derive the factoring a pattern a 3 b 3. Reviewing what ou learned in previous grades and lessons 238 Chapter 5 Polnomial Functions