1. Splash Screen

2. Lesson Menu Five-Minute Check (over Lesson 8–4) CCSS Then/Now New Vocabulary Example 1:Use the Distributive Property Key Concept: Factoring by Grouping Example 2:Factor by Grouping Example 3:Factor by Grouping with Additive Inverses Key Concept: Zero Product Property Example 4:Solve Equations Example 5:Real-World Example: Use Factoring

3. Over Lesson 8–4 5-Minute Check 1 A.16x 2 + 25 B. 16x 2 + 20x + 25 C. 16x 2 + 40x + 25 D. 4x 2 + 20x + 5 Find (4x + 5) 2.

4. Over Lesson 8–4 5-Minute Check 2 A.15a 2 – 30ab + 15b 2 B. 9a 2 – 30ab + 25b 2 C. 9a 2 – 15ab + 25b 2 D. 3a 2 – 15ab + 5b 2 Find (3a – 5b) 2.

5. Over Lesson 8–4 5-Minute Check 3 A.9x 2 + 24x – 16 B.9x 2 – 24x – 16 C.9x 2 + 16 D.9x 2 – 16 Find (3x + 4)(3x – 4).

6. Over Lesson 8–4 5-Minute Check 4 A.4c 2 – 36d 2 B.4c 2 + 36d 2 C.4c 2 + 24cd + 36d 2 D.4c 2 + 24cd – 36d 2 Find (2c 2 + 6d)(2c 2 – 6d).

7. Over Lesson 8–4 5-Minute Check 5 A.(x + 3) 2 (x – 6) 2 B.2x 2 – 6x + 45 C.(x + 3) 2 + (x – 6) 2 D.2x 2 + 45 Write a polynomial that represents the area of the figure at the right.

8. CCSS Content Standards A.SSE.2 Use the structure of an expression to identify ways to rewrite it. A.SSE.3a Factor a quadratic expression to reveal the zeros of the function it defines. Mathematical Practices 2 Reason abstractly and quantitatively. Common Core State Standards © Copyright 2010. National Governors Association Center for Best Practices and Council of Chief State School Officers. All rights reserved.

9. Then/Now Used the Distributive Property to evaluate expressions. Use the Distributive Property to factor polynomials.

10. Vocabulary factoring factoring by grouping Zero Product Property

11. Finding the Greatest Common Factor Objective: Factor using the greatest common factor (GCF)

12. Factors Factors (either numbers or polynomials) When an integer (polynomial) is written as a product of integers (polynomials), each of the integers (polynomials) in the product is a factor of the original number. Factoring – writing a polynomial as a product of polynomials.

13. Greatest common factor – largest quantity that is a factor of all the integers or polynomials involved. Greatest Common Factor Finding the GCF of a List of Integers or Terms 1) Prime factor the numbers and variables. 2) Identify common prime factors. 3) Take the product of all common prime factors. If there are no common prime factors, GCF is 1.

14. Find the GCF of each list of numbers. 1)12 and 8 12 = 2 · 2 · 3 8 = 2 · 2 · 2 So the GCF is 2 · 2 = 4. 2)7 and 20 7 = 1 · 7 20 = 2 · 2 · 5 There are no common prime factors so the GCF is 1. Greatest Common Factor Example

15. Find the GCF of each list of numbers. 1)6, 8 and 46 6 = 2 · 3 8 = 2 · 2 · 2 46 = 2 · 23 So the GCF is 2. 2)144, 256 and 300 144 = 2 · 2 · 2 · 2 · 3 · 3 256 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 300 = 2 · 2 · 3 · 5 · 5 So the GCF is 2 · 2 = 4. Greatest Common Factor Example

16. 1) x 3 and x 7 x 3 = x · x · x x 7 = x · x · x · x · x · x · x So the GCF is x · x · x = x 3 2) 6x 5 and 4x 3 6x 5 = 2 · 3 · x · x · x · x · x 4x 3 = 2 · 2 · x · x · x So the GCF is 2 · x · x · x = 2x 3 Find the GCF of each list of terms. Greatest Common Factor Example

17. Find the GCF of the following list of terms. a 3 b 2, a 2 b 5 and a 4 b 7 a 3 b 2 = a · a · a · b · b a 2 b 5 = a · a · b · b · b · b · b a 4 b 7 = a · a · a · a · b · b · b · b · b · b · b So the GCF is a · a · b · b = a 2 b 2 Notice that the GCF of terms containing variables will use the smallest exponent found amongst the individual terms for each variable. Greatest Common Factor Example

18. ASSIGNMENT Worksheet Kuta side all Page 62 odds (evens extra credit)

19. Factoring the Greatest Common Factor from Polynomials Objective: Factor using the greatest common factor (GCF)

20. The first step in factoring a polynomial is to find the GCF of all its terms. Then we write the polynomial as a product by factoring out the GCF from all the terms. The remaining factors in each term will form a polynomial. Factoring Polynomials

21. Steps for Factoring the GCF From Polynomials 1.Write down the problem. 2.Write each term on a separate line and then write the algebraic factorization of each term. 3.Pair up all factors that occur in each term and circle them.

22. Steps for Factoring the GCF From Polynomials 4.Multiply what is circled. This is the GCF. When writing the answer, put it outside parenthesis. 5.The “left-overs” for each term should be multiplied and put inside parenthesis. (What was left) 7. Check your answer by distributing. 6. GCF (LEFTOVERS)

23. Example Step 1 Step 2 3x 3 + 6x 2 y – 15xy 2 3x 3 = 3  x  x  x 6x 2 y= 2  3  x  x  y 15xy 2 = 3  5  x  y  y

24. Example Step 3 Step 4 Step 5 3x 3 = 3  x  x  x 6x 2 y= 2  3  x  x  y 15xy 2 = 3  5  x  y  y GCF=3x Leftovers

25. Example Step 6 Step 7 3x (x 2 + 2xy – 5y 2 ) Check your answer

26. 18x 2 =233xx -12x 3 = 223xxx 18x 2 - 12x 3 = 6x 2 ( ___ - ___ ) Check your answer by distributing. Factor 18x 2 - 12x 3. 32x

27. Factor 28a 2 b + 56abc 2. GCF = 28ab 28a 2 b + 56abc 2 = 28ab ( ___ + ___ ) Check your answer by distributing. 28ab(a + 2c 2 ) a2c 2

28. Factor 28a 2 + 21b - 35b 2 c 2 GCF = 7 28a 2 + 21b - 35b 2 c 2 = 7 ( ___ + ___ - ____ ) Check your answer by distributing. 7(4a 2 + 3b – 5b 2 c 2 ) 4a 2 5b 2 c 2 3b

29. Example 1 Use the Distributive Property A. Use the Distributive Property to factor 15x + 25x 2. First, find the GCF of 15x + 25x 2. 15x = 3 ● 5 ● x Factor each monomial. Circle the common prime factors. GCF = 5 ● x or 5x Write each term as the product of the GCF and its remaining factors. Then use the Distributive Property to factor out the GCF. 25x 2 = 5 ● 5 ● x ● x

30. Example 1 Use the Distributive Property = 5x(3 + 5x)Distributive Property Answer: The completely factored form of 15x + 25x 2 is 5x(3 + 5x). 15x + 25x 2 = 5x(3) + 5x(5 ● x) Rewrite each term using the GCF.

31. Example 1 Use the Distributive Property B. Use the Distributive Property to factor 12xy + 24xy 2 – 30x 2 y 4. 12xy=2 ● 2 ● 3 ● x ● y 24xy 2 =2 ● 2 ● 2 ● 3 ● x ● y ● y –30x 2 y 4 =–1 ● 2 ● 3 ● 5 ● x ● x ● y ● y ● y ● y GCF = 2 ● 3 ● x ● y or 6xy Circle common factors. Factor each term.

32. Example 1 Use the Distributive Property = 6xy(2 + 4y – 5xy 3 ) Distributive Property Answer: The factored form of 12xy + 24xy 2 – 30x 2 y 4 is 6xy(2 + 4y – 5xy 3 ). 12xy + 24xy 2 – 30x 2 y 4 = 6xy(2) + 6xy(4y) + 6xy(–5xy 3 ) Rewrite each term using the GCF.

33. Example 1 A.3xy(x + 4y) B.3(x 2 y + 4xy 2 ) C.3x(xy + 4y 2 ) D.xy(3x + 2y) A. Use the Distributive Property to factor the polynomial 3x 2 y + 12xy 2.

34. Example 1 A.3(ab 2 + 5a 2 b 2 + 9ab 3 ) B.3ab(b + 5ab + 9b 2 ) C.ab(b + 5ab + 9b 2 ) D.3ab 2 (1 + 5a + 9b) B. Use the Distributive Property to factor the polynomial 3ab 2 + 15a 2 b 2 + 27ab 3.

35. ASSIGNMENT Page 497 1-4,15-20

37. Concept

38. Example 2 Factor by Grouping Factor 2xy + 7x – 2y – 7. 2xy + 7x – 2y – 7 = (2xy – 2y) + (7x – 7)Group terms with common factors. = 2y(x – 1) + 7(x – 1)Factor the GCF from each group. = (2y + 7)(x – 1)Distributive Property Answer: (2y + 7)(x – 1) or (x – 1)(2y + 7)

39. Example 2 A.(4x – 5)(y + 3) B.(7x + 5)(2y – 3) C. (4x + 3)(y – 5) D. (4x – 3)(y + 5) Factor 4xy + 3y – 20x – 15.

40. Example 3 Factor by Grouping with Additive Inverses Factor 15a – 3ab + 4b – 20. 15a – 3ab + 4b – 20 = (15a – 3ab) + (4b – 20)Group terms with common factors. = 3a(5 – b) + 4(b – 5)Factor the GCF from each group. = 3a(–1)(b – 5) + 4(b – 5)5 – b = –1(b – 5) = –3a(b – 5) + 4(b – 5)3a(–1) = –3a = (–3a + 4)(b – 5) Distributive Property Answer: (–3a + 4)(b – 5) or (3a – 4)(5 – b)

41. Example 3 A.(2x – 3)(y – 5) B.(–2x + 3)(y + 5) C.(3 + 2x)(5 + y) D.(–2x + 5)(y + 3) Factor –2xy – 10x + 3y + 15.

42. Concept

43. Example 4 Solve Equations A. Solve (x – 2)(4x – 1) = 0. Check the solution. If (x – 2)(4x – 1) = 0, then according to the Zero Product Property, either x – 2 = 0 or 4x – 1 = 0. (x – 2)(4x – 1) = 0Original equation x – 2 = 0 or 4x – 1= 0 Zero Product Property x = 2 4x= 1Solve each equation. Divide.

44. Example 4 Solve Equations (x – 2)(4x – 1)=0 (x – 2)(4x – 1) = 0 Check Substitute 2 and for x in the original equation.(2 – 2)(4 ● 2 – 1) = 0 ? ? (0)(7) = 0 ?? 0 = 0 0 = 0

45. Example 4 Solve Equations B. Solve 4y = 12y 2. Check the solution. Write the equation so that it is of the form ab = 0. 4y=12y 2 Original equation 4y – 12y 2 =0Subtract 12y 2 from each side. 4y(1 – 3y)=0Factor the GCF of 4y and 12y 2, which is 4y. 4y = 0 or 1 – 3y=0Zero Product Property y = 0–3y=–1Solve each equation. Divide.

46. Example 4 Solve Equations Answer: The roots are 0 and. Check by substituting 0 and for y in the original equation. __ 1 3 1 3

47. Example 4 A.{3, –2} B.{–3, 2} C.{0, 2} D.{3, 0} A. Solve (s – 3)(3s + 6) = 0. Then check the solution.

48. Example 4 B. Solve 5x – 40x 2 = 0. Then check the solution. A.{0, 8} B. C.{0} D.

49. Example 5 Use Factoring FOOTBALL A football is kicked into the air. The height of the football can be modeled by the equation h = –16x 2 + 48x, where h is the height reached by the ball after x seconds. Find the values of x when h = 0. h= –16x 2 + 48x Original equation 0= –16x 2 + 48x h = 0 0= 16x(–x + 3)Factor by using the GCF. 16x= 0 or –x + 3= 0Zero Product Property x= 0 x= 3Solve each equation. Answer: 0 seconds, 3 seconds

50. Example 5 A.0 or 1.5 seconds B.0 or 7 seconds C.0 or 2.66 seconds D.0 or 1.25 seconds Juanita is jumping on a trampoline in her back yard. Juanita’s jump can be modeled by the equation h = –14t 2 + 21t, where h is the height of the jump in feet at t seconds. Find the values of t when h = 0.

51. End of the Lesson