7-6 Choosing a Factoring Model Extension: Factoring Polynomials with More Than One Variable Essential question: How can you factor polynomials with more than one variable? • What is the connection between x2 - 81 and m2 - n2? They are both the difference of squares. Standards for Mathematical Content 2 2 Factoring by GCF Factoring ax2 + bx + c Factoring special products • In the perfect square trinomial 4x 2 + 12xy + 9y 2, what represents a and what represents b? Math Background Students have factored polynomials using a variety of methods: by grouping, by using special product patterns, by using the factors of the first and third terms. All these polynomials involved a single variable. Now students will factor polynomials that involve more than one variable. They will discover that the methods they learned for single variable polynomials apply equally well to multi-variable polynomials. 2x represents a; 3y represents b. EXTRA EXAMPLE Factor the polynomial completely. Explain each step. -3x2 + 36xy - 108y2 2 -3(x - 12xy + 36y2) Factor the GCF. -3(x - 6y)(x - 6y) Perfect square trinomial Review the methods students have learned to factor polynomials that involve a single variable. Give students an example of each type of polynomial and have them do the factoring. Then give them polynomials with two variables that can be factored using the special product patterns. Lead them to see that the patterns still apply. Ask what they think this means about methods for factoring multivariable polynomials in general. Avoid Common Errors Students may confuse the perfect square trinomial rule with the difference of squares rule and use a2 - b2 = (a - b)(a - b). To help students remember that a2 - b2 = (a - b)(a + b), have them remember the statement “Two terms = Two signs”. T EACH EXPLORE Questioning Strategies • If the second polynomial were 3m2 - 3, how would you factor it? Use the GCF and the difference of squares: 3(m - 1)(m + 1). 397 Lesson 6 © Houghton Mifflin Harcourt Publishing Company Teaching Strategies Have groups of students develop a step-by-step plan to factor polynomials. Ask them to make a list of questions they can ask themselves as they work. Such questions might include Does this polynomial follow the pattern of Difference of Squares? Does it follow the pattern of Perfect Squares? Is there a GCF? What are some factors of the third coefficient? What is the sum of these factors? Allow students to refer to this plan, as necessary, when they factor polynomials. IN T RO DUC E Chapter 7 EXAMPLE Questioning Strategies • How do you know which special product pattern to use when factoring a perfect square trinomial? Look at the first operation sign. If it is +, factor as (a + b)(a + b). If it is -, factor as (a - b)(a - b). Prerequisites 1 2 • How would you factor x2y - r2s ? Use the difference of squares rule: (xy + rs)(xy - rs) A-SSE.1.1b Interpret complicated expressions by viewing one or more of their parts as a single entity.* A-SSE.1.2 Use the structure of an expression to identify ways to rewrite it. Name Class Notes 7-6 Date Choosing a Factoring Method Extension: Factoring Polynomials with More Than One Variable Essential question: How can you factor polynomials with more than one variable? A-SSE.1.2 1 EXPLORE Factoring Polynomials with Two Variables Factor each polynomial completely. Explain each step. 2x2 - 162 2(x2 - 81) Factor the GCF. x(x + 9)(x - 9) Difference of squares 3m2 - 3n2 2 2 3( m - n ) Factor the GCF. 3(m + n)(m - n) Difference of squares REFLECT 1a. Compare the methods of factoring the two polynomials. The methods are the same, except the second polynomial has two variables. 1b. How would the factoring change for the polynomial 3m2 - 12n2? After factoring the GCF, you would factor m2 - 4n2, which is still © Houghton Mifflin Harcourt Publishing Company a difference of squares. To factor a perfect square trinomial with more than variable you can use the same patterns you used with perfect square trinomials in one variable. a2 + 2ab + b2 = (a + b)(a + b) a2 - 2ab + b2 = (a - b)(a - b) A-SSE.1.2 2 EXAMPLE Factoring a Perfect Square Trinomial Factor the trinomial completely. Explain each step. 3g2 + 12gh + 12h2 3 (g2 + 4gh + 4h2) Factor the GCF. 3(g + 2h)(g + 2h) Perfect square trinomial pattern 397 Chapter 7 Lesson 6 REFLECT 2a. Which pattern would you use to factor 16a2 - 48ab + 36b2? Explain how you know and then factor the trinomial. If there is no obvious pattern shown by the trinomial, you can find the factors of the coefficient of the third term and check their sums to find how to factor. A-SSE.1.2 3 EXAMPLE Factoring a Polynomial Factor 3x2 + 21xy + 36y2 completely. Explain your steps. 2 2 3 (x + 7xy + 12y ) Factor the GCF. Find the factors of 12 that add to 7. Factors of 12 Sum of factors 1 and 12 13 2 and 6 8 3 and 4 7 3(x + 3y)(x + 4y) Factor according to the sum. REFLECT 3a. Would finding the factors and their sum be enough to factor the polynomial 2x 2 + 11xy + 12y 2? Why or why not? No, because the x-term of the binomial has a coefficient of 2. This would affect the middle term of the polynomial, so you can’t just find the factors and the sum. 3b. What additional conditions must you consider to factor 2x 2 + 11xy + 12y 2? You have to consider that one factor of 12 has been doubled. You will have to find © Houghton Mifflin Harcourt Publishing Company © Houghton Mifflin Harcourt Publishing Company a2 - 2ab + b2 = (a - b)(a - b); the first operation sign is -; 4(2a - 3b)(2a - 3b) which factor by trying to divide each factor by 2, then finding the sum. Sometimes grouping will allow you to factor a polynomial. Chapter 7 Chapter 7 398 Lesson 6 398 Lesson 6 3 EXTRA EXAMPLE Factor x3 + 2x2y + x + 2y completely. Explain your steps. EXAMPLE Questioning Strategies • What tells you that this polynomial is not a “perfect square” polynomial? 36 is a perfect • Does not have a common GCF. • Group terms with a common factor: (x3 + 2x2y) + (x + 2y) square, but the coefficient of the first term, 3, is not. • Factor out the GCF of each group: x2(x + 2y) + 1(x + 2y) • Describe the relationship between the coefficient of the second term and the coefficient of the third term after the GCF has been factored out. The • Factor out the GCF of the products: (x2 + 1)(x + 2y) coefficient of the second term is the sum of the two factors of the coefficient of the third term. The coefficient of the third term is the product of these two factors. CLOS E Essential Question How can you factor polynomials with more than one variable? EXTRA EXAMPLE Factor m2 - 8md + 15d2 completely. Explain your steps. You can use the same methods you used for polynomials of one variable: difference of squares or perfect square trinomial patterns, sum of the factors method, or factoring by grouping. • Does not have a common GCF. • Does not follow any special pattern. • Multiply the coefficients of the first and third terms: 1 • 15 = 15. The product is positive, so the factors are either both positive or both negative. Summarize Have students write a journal entry describing the aspect of factoring they find most difficult. They should also note any reminders or hints they developed to help overcome this difficulty. • The middle term is negative, so both factors will be negative. • Find the negative factors of 15 and their sums. Factors Sum -1 and -15 -3 and -5 -16 -8 PR ACTICE • Factor according to the sum: (m - 3d)(m - 5d) 4 EXAMPLE Questioning Strategies • Suppose the polynomial were 27x3 - 27xy + 6y - 6x2. What would be the first step of the factoring? Factor out the common GCF of 3. Where skills are practiced 1 EXPLORE EXS. 3, 6, 8 2 EXAMPLE EXS. 1, 11, 15 3 EXAMPLE EXS. 2, 4, 5, 7, 9, 10, 12, 16 4 EXAMPLE EXS. 13, 14 • Do you have to factor 6x2 + 12xy + 3x2 + 4y2 by grouping? Explain. No; you can combine the like terms 6x2 and 3x2 to get 9x2 + 12xy + 4y2 which is a perfect square trinomial. Chapter 7 399 Lesson 6 © Houghton Mifflin Harcourt Publishing Company Where skills are taught • The sum of -3 and -5 is -8, the coefficient of the middle term. Notes A-SSE.1.1b 4 EXAMPLE Factoring a Polynomial Using Grouping Factor 9x3 - 9xy + 2y - 2x2 completely. Explain your steps. There is no GCF for all the terms. Since the polynomial has four terms, factor by grouping. (9x3 - 9xy ) + ( 2y - 2x2 ) Group terms that have a common factor. The common factor of 9x3y and -9xy is 9x . The common factor of 2y and -2x2 is 9x (x 2 -y )+( 2 2 9x ( x - y ) + ( 2 )( 2 y-x ) ( -1 ) ( ) 2 . Factor out the GCF of each group. x2 - y ) The polynomial contains the binomial opposites (x2 - y) and ( ( y - x2 ). ) Write (y - x2) as (-1) x2 - y . 2 9x ( x - y ) - ( 2 ( 9x - 2 )( x2 - y )( x2 - y ) Simplify. ) Factor out ( x2 - y ), the common factor of the products. REFLECT 4a. Describe the steps you would use in factoring 8x2 - 2x + 24xy - 6y. First I would factor out 2, the GCF of all the terms. I would then follow the steps in the Example to complete the factoring, although in this case there are no opposite © Houghton Mifflin Harcourt Publishing Company binomials. PRACTICE Choose a factoring method to factor each polynomial completely. Explain each step. 1. x2 + 6xy + 9x2 (x + 3y)(x + 3y) Perfect square trinomial 2. 4x2 - 4xy - 8y 2 4(x 2 - xy - 2y 2) Factor GCF. 4(x + y)(x - 2y) Factor according to sum. 3. x2 - 4y 2 (x + 2y)(x - 2y) Difference of squares 4. g 2 + 3gh - 10h 2 (g - 2h)(g + 5h) Factor according to sum. 399 Chapter 7 Lesson 6 Factor each polynomial completely. 5. 2m2 + 5mn - 3n2 6. 4x 2 - 9y 2 (2x + 3y)(2x - 3y) 7. g 2 - 7gh + 10h 2 8. 16b2 - 49c2 (g - 5h)(g - 2h) (4b + 7c)(4b - 7c) 9. a3 - 3a2b - 4ab2 10. 6a2 + 3ab - 18b2 a(a 2 - 3ab - 4b 2) 3(2a 2 + ab - 6b 2) a(a + b)(a - 4b) 3(a + 2b)(2a - 3b) 11. 2t 3 + 12t 2w + 18tw2 12. 6c 3 - 27c 2d + 12cd 2 2t (t 2 + 6tw + 9w 2) 3c (2c 2 - 9cd + 4d 2) 2t (t + 3w)(t + 3w) 3c (c - 4d )(2c - d ) 13. x3y - 5x2y + 4x - 20 14. x4y2 + 4x2y2 - 7x2 - 28 x2y(x - 5) + 4(x - 5) x2y2(x2 + 4) - 7(x2 + 4) (x2y + 4)(x - 5) (x2y2 - 7)(x2 + 4) 4 4 15. Factor x - y completely. (Hint: What special form does this polynomial appear to follow?) (x 2 + y 2)(x 2 - y 2) (x 2 + y 2)(x + y)(x - y) 16. Jaime and Sam both factored the polynomial 2x 2 + 10xy + 8y 2. Which student is correct? Explain. Jaime Sam 2x2 + 10xy + 8y2 2 2 2(x + 5xy + 4y ) 2x2 + 10xy + 8y2 (2x + 2y)(x + 4y) 2(x + y)(x + 4y) Jaime is correct. Sam did not completely factor the polynomial. The term (2x + 2y) © Houghton Mifflin Harcourt Publishing Company © Houghton Mifflin Harcourt Publishing Company (m + 3n)(2m - n) can be further factored to 2(x + y). Chapter 7 Chapter 7 400 Lesson 6 400 Lesson 6 ADD I T I O N A L P R AC T I C E AND PRO BL E M S O LV I N G Assign these pages to help your students practice and apply important lesson concepts. For additional exercises, see the Student Edition. Answers Additional Practice 1. yes 2. no; 5m(m + 9) 3. no; 2p(p2 + 3)(p2 − 3) 2 4. yes 5. no; 3jk3(5k + 19) 6. no; 14(7g4 − 2g + 5) 7. 8y(3xy + 5) 8. 5r(r2 − 2s) 9. x2y(3x + y) 10. −3b(a − 2)2 11. (5t + 3)(t − 3s)(t + 3s) 12. 2(y + 4x)(y − 7x) 13. 3a(2a + 3)(a + 5) 14. xy(x − 3y)(x + 3y) 15. 12(n3 − 4) 16. 3c2(c + 4d)2 17. not factorable 2 18. 10w2(w + 4v2)(w − 2v)(w + 2v) © Houghton Mifflin Harcourt Publishing Company Problem Solving 1. 3(5x − 4y)(x + y) 2. 4π(2k − 1)2; 4π m2 3. 4(2x + y)(2x − y) 4. −5(x − 19)(x + 3); 525 attendees 5. D 6. G 7. C 8. J Chapter 7 401 Lesson 6 Name Class Notes 7-6 Date © Houghton Mifflin Harcourt Publishing Company Additional Practice 401 Chapter 7 Lesson 6 Problem Solving © Houghton Mifflin Harcourt Publishing Company Q Q Q Chapter 7 Chapter 7 402 © Houghton Mifflin Harcourt Publishing Company Lesson 6 402 Lesson 6

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