7-2 Factoring by GCF Going Deeper Essential question: How can you factor polynomials completely by grouping? 2 Standards for Mathematical Content Questioning Strategies • Are (x + 5) and (5 + x) binomial opposites? Explain. No; (x + 5) and (5 + x) are equal by A-SSE.1.1 Interpret expressions that represent a quantity in terms of its context.* 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. the commutative property. To be opposites, the binomials would have to be (x - 5) and (5 - x); that is, they would have to have the same variables and numbers in the reverse order with a subtraction sign between them. EXTRA EXAMPLE Use grouping and binomial opposites to factor 2x3 + 6x2 - 3 - x. (2x2 - 1)(x + 3) Math Background Factoring by grouping is a factoring method that can be used with some polynomials that have four terms. This method can be used when the terms of the polynomial can be grouped so that each group has the same greatest common factor (GCF). CLOS E Essential Question How can you factor polynomials completely by grouping? IN T RO DUC E Summarize Have students write a journal entry in which they describe how they factored one of the polynomials in the Practice exercises. They should include how analyzing the terms indicated that grouping was an appropriate method to use and a justification for each step that they took in their work. T EACH EXAMPLE Questioning Strategies • A student factored 2x4 + 6x3 - 8x2 - 24 as 2x(x2 - 4)(x + 3). Is this correct? Explain. No; (x2 - 4) can be factored further as (x - 2)(x + 2). The correct factorization is 2x(x - 2)(x + 2)(x + 3). PR ACTICE Where skills are taught EXTRA EXAMPLE Use grouping to factor -3x4 - 6x3 + 3x2 + 6x. -3x(x + 2)(x - 1)(x + 1) Chapter 7 377 Where skills are practiced 1 EXAMPLE EXS. 1–2 2 EXAMPLE EXS. 3–4 Lesson 2 © Houghton Mifflin Harcourt Publishing Company First, determine whether all the terms have a common GCF and, if so, factor that out. Then, group the terms that have a common factor and factor out the GCF of each group. Use that GCF as a common factor to rewrite the polynomial in factored form. As a final step, check that none of the binomial factors can be factored further. Review how to use the GCF to factor an expression with a common binomial factor, such as 3(x + 1) - 2x(x + 1). Once students recognize that the common factor in both terms is the binomial (x + 1), remind them that a binomial can be factored out of a term just as a monomial can. Tell students that they can factor some polynomials with four terms by grouping the terms in pairs and factoring out common monomials and binomials. 1 EXAMPLE Name Class Notes 7-2 Date Factoring by GCF Going Deeper Essential question: How can you factor polynomials completely by grouping? When a polynomial has four terms, you may be able to make two groups of terms and factor out the GCF from each group. Before doing this, however, you should check if there is a GCF for all the terms that can be factored out first. A-SSE.1.2 1 EXAMPLE Factoring out a GCF and Grouping Factor -2x4 - 2x3 - 6x2 - 6x. -2x ( x3 + x2 + 3x + 3 ) Factor out -2x , the GCF of all the terms. ( -2x )[ ( x3 + x2 ) + ( 3x + 3 ) ] Group terms that have a common factor. The common factor of x3 and x2 is x2 of 3x and 3 is ( -2x ) [ ( x + 1 )+ 2 x 3 ( x + 1 )] ( -2x )( x + 1 ) ( x2 + 3 ) . The common factor 3 . Factor out the GCF of each group. Factor out ( x + 1 ) , the common factor of the products inside the brackets. © Houghton Mifflin Harcourt Publishing Company REFLECT 1a. Why did the operation signs change when -2x was factored out of each term? Subtracting terms with positive coefficients is the same as adding terms with negative coefficients. The polynomial could be written as -2x4 + (-2x3) + (-6x2) + (-6x). When -2x is factored out, the coefficients become positive again, but the operation signs remain as +. 1b. Does performing grouping before factoring out -2x change the factorization? Explain. No; if grouping is done first, the steps would be: (-2x4 - 2x3) + (-6x2 - 6x) -2x3(x + 1) + (-6x)(x + 1) -2x3(x + 1) - 6x(x + 1) (-2x3 - 6x)(x + 1) -2x(x2 + 3)(x + 1) 377 Chapter 7 Lesson 2 Binomials have opposites. For example, (3 - x) and -(3 - x) are opposites. The expression -(3 - x) can be rewritten as (-3 + x) or (x - 3). So, (3 - x) and (x - 3) are opposites. 2 A-SSE.1.1b EXAMPLE Factoring with Binomial Opposites ( 4x3 - 8x2 ) + ( 6 - 3x ) 4x2 ( x - 2 )+ 3 ( 2-x ) Group terms in the order in which they appear. The GCF of the terms in the first group is 4x2 . The GCF of the terms in the second 3 . Factor out the GCF of each group. group is 4x2 ( x - 2 )+ 3 ( x - 2 )- 3 ( -1 )( x - 2 ) The polynomial contains the binomial ( x-2 ) Simplify. opposites ( x - 2 ) and ( 2 - x ). Write (2 - x) as (-1)(x - 2 ). 4x2 ( x - 2 )( 4x2 - 3 ) Factor out ( x - 2 ) , the common factor of the products. REFLECT 2a. How could you have rearranged the terms of the polynomial so that you would not have encountered binomial opposites when factoring? Show how you would have factored the polynomial if you had first rearranged the terms. Rearrange the terms so that subtracting 3x comes before adding 6. 4x3 - 8x2 - 3x + 6 © Houghton Mifflin Harcourt Publishing Company © Houghton Mifflin Harcourt Publishing Company Factor 4x3 - 8x2 + 6 - 3x. (4x3 - 8x2) - (3x - 6) 4x2(x - 2) - 3(x - 2) (x - 2)(4x2 - 3) PRACTICE Factor each polynomial by grouping. 1. -3a3 - 9a2 - 6a - 18 2. -2c3 + 4c2 - 8c + 16 -3(a2 + 2)(a + 3) 3 2 3. 5s - 5s + 2 - 2s -2(c2 + 4)(c - 2) 4. 4z3 - 16z2 + 40 - 10z (5s2 - 2)(s - 1) Chapter 7 Chapter 7 2(2z2 - 5)(z - 4) 378 Lesson 2 378 Lesson 2 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. c(8c + 7) 2. 3n2(n + 4) 3. 3x(5x4 - 6) 4. 4(−2s4 + 5t3 - 7) 5. 6n(n5 + 3n3 − 4) 2 6. 5m2(−m - m + 1) 7. 16t(-t + 2) 8. 3x and 4x + 1 9. (m + 5)(3m + 4) 10. (b − 3)(16b + 1) 11. (x + 4)(2x2 + 3) 12. (4n + 3)(n2 + 1) 13. (5d − 3)(2d + 7) 14. (4n − 5)(3n2 − 2) 15. (b − 3)(5b3 − 1) 16. (t 2 − 2)(t − 5) Problem Solving © Houghton Mifflin Harcourt Publishing Company 1. 4x ft; (x + 1) ft 2. −3(x2 + 9x − 275) 3. 4(3x + 7); 31 feet 4. (5x + 4) m; (x2 − 2) m 5. C 6. G 7. C 8. F Chapter 7 379 Lesson 2 Name Class Date Notes 7-2 © Houghton Mifflin Harcourt Publishing Company Additional Practice 379 Chapter 7 Lesson 2 © Houghton Mifflin Harcourt Publishing Company © Houghton Mifflin Harcourt Publishing Company Problem Solving Chapter 7 Chapter 7 380 Lesson 2 380 Lesson 2

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