1. Copyright © 2011 Pearson Education, Inc. Factoring CHAPTER 6.1Greatest Common Factor and Factoring by Grouping 6.2Factoring Trinomials of the Form x 2 + bx + c 6.3Factoring Trinomials of the Form ax 2 + bx + c, where a  1 6.4Factoring Special Products 6.5Strategies for Factoring 6.6Solving Quadratic Equations by Factoring 6.7Graphs of Quadratic Equations and Functions 6

2. Copyright © 2011 Pearson Education, Inc. Greatest Common Factor and Factoring by Grouping 6.1 1.List all possible factors for a given number. 2.Find the greatest common factor of a set of numbers or monomials. 3.Write a polynomial as a product of a monomial GCF and a polynomial. 4.Factor by grouping.

3. Slide 6- 3 Copyright © 2011 Pearson Education, Inc. Factored form: A number or an expression written as a product of factors. Following are some examples of factored form: An integer written in factored form with integer factors: 28 = 2 14 A monomial written in factored form with monomial factors: 8x 5 = 4x 2 2x 3 A polynomial written in factored form with a monomial factor and a polynomial factor: 2x + 8 = 2(x + 4) A polynomial written in factored form with two polynomial factors: x 2 + 5x + 6 = (x + 2)(x + 3)

4. Slide 6- 4 Copyright © 2011 Pearson Education, Inc. Example 1 List all natural number factors of 36. Solution: To list all the natural number factors, we can divide 36 by 1, 2, 3, and so on, writing each divisor and quotient pair as a product until we have all possible combinations. 1 36 2 18 3 12 4 9 6 The natural number factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.

5. Slide 6- 5 Copyright © 2011 Pearson Education, Inc. Greatest common factor (GCF): The largest natural number that divides all given numbers with no remainder. Listing Method for Finding GCF To find the GCF of a set of numbers by listing, 1. List all possible factors for each given number. 2. Search the lists for the largest factor common to all lists.

6. Slide 6- 6 Copyright © 2011 Pearson Education, Inc. Example 2 Find the GCF of 48 and 54. Solution: Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 Factors of 54: 1, 2, 3, 6, 9, 18, 27, 54 The GCF of 48 and 54 is 6.

7. Slide 6- 7 Copyright © 2011 Pearson Education, Inc. Prime Factorization Method for Finding GCF To find the GCF of a given set of numbers by prime factorization, 1. Write the prime factorization of each given number in exponential form. 2. Create a factorization for the GCF that includes only those prime factors common to all factorizations, each raised to its smallest exponent in the factorizations. 3.Multiply the factors in the factorization created in Step 2. Note: If there are no common prime factors, then the GCF is 1.

8. Slide 6- 8 Copyright © 2011 Pearson Education, Inc. Example 3a Find the GCF of 45 and 72. Solution: Write the prime factorization 45 and 72 in factored form. 45 = 3 2 5 72 = 2 3 3 2 The common prime factor is 3. GCF = 3 2 = 9

9. Slide 6- 9 Copyright © 2011 Pearson Education, Inc. Example 3b Find the GCF of 3240 and 8316. Solution: Write the prime factorization of each number. 3240 = 2 3 3 4 5 8316 = 2 2 3 3 7 11 The common prime factors are 2 and 3. Take the term with the smallest exponent. GCF = 2 2 3 3 = 108

10. Slide 6- 10 Copyright © 2011 Pearson Education, Inc. Example 4a Find the GCF of a 4 b 4 c 5 and a 3 b 7. Solution: Each monomial is already written as a prime factorization. The common primes are a and b. The smallest exponent for a is 3. The smallest exponent for b is 4. GCF = a 3 b 4

11. Slide 6- 11 Copyright © 2011 Pearson Education, Inc. Example 4b Find the GCF of 45a 3 b and 30a 2. Solution: Write the prime factorization of each monomial, treating the variables like prime factors. 45a 3 b = 3 2 5 a 3 b 30a 2 = 2 3 5 a 2 The common prime factors are 3, 5, and a. GCF = 3 5 a 2 = 15a 2

12. Slide 6- 12 Copyright © 2011 Pearson Education, Inc. Factoring a Monomial GCF Out of a Polynomial To factor a monomial GCF out of a given polynomial, 1. Find the GCF of the terms in the polynomial. 2. Rewrite the given polynomial as a product of the GCF and the quotient of the polynomial and the GCF. polynomial = GCF

13. Slide 6- 13 Copyright © 2011 Pearson Education, Inc. Example 5 Factor Solution 1.Find the GCF of 18x 2 and 24x. The GCF is 6x. 2. Write the given polynomial as a product of the GCF and the quotient of the polynomial and the GCF.

14. Slide 6- 14 Copyright © 2011 Pearson Education, Inc. continued Factor Solution Check We can check by multiplying the factored form using the distributive property. This is left to you.

15. Slide 6- 15 Copyright © 2011 Pearson Education, Inc. Example 6 Factor Solution 1.Find the GCF of both terms. The GCF is 2. Write the given polynomial as a product of the GCF and the quotient of the polynomial and the GCF.

16. Slide 6- 16 Copyright © 2011 Pearson Education, Inc. continued Factor Solution Check We can check by multiplying the factored form using the distributive property. This is left to you.

17. Slide 6- 17 Copyright © 2011 Pearson Education, Inc. Example 7 Factor. Solution 1.Find the GCF of Because the first term in the polynomial is negative, we will factor out the negative of the GCF to avoid a negative first term inside the parentheses. We will factor out  3x 2 y.

18. Slide 6- 18 Copyright © 2011 Pearson Education, Inc. continued 2.Write the given polynomial as the product of the GCF and the parentheses containing the quotient of the given polynomial and the GCF.

19. Slide 6- 19 Copyright © 2011 Pearson Education, Inc. Example 8 Factor. Solution: Notice that this expression is a sum of two products, a and (b + 5), and 8 and (b + 5). Further, note that (b + 5) is the GCF of the two products.

20. Slide 6- 20 Copyright © 2011 Pearson Education, Inc. Factoring by Grouping To factor a four-term polynomial by grouping, 1. Factor out any monomial GCF (other than 1) that is common to all four terms. 2. Group together pairs of terms and factor the GCF out of each pair or group. 3.If there is a common binomial factor, then factor it out. 4.If there is no common binomial factor, then interchange the middle two terms and repeat the process. If there is still no common binomial factor, then the polynomial cannot be factored by grouping.

21. Slide 6- 21 Copyright © 2011 Pearson Education, Inc. Example 9a Factor. Solution First we look for a monomial GCF (other than 1). This polynomial does not have one. Because the polynomial has four terms, we now try to factor by grouping.

22. Slide 6- 22 Copyright © 2011 Pearson Education, Inc. Example 9b Factor. Solution There is a monomial GCF, 2a; factor this from all four terms.

23. Slide 6- 23 Copyright © 2011 Pearson Education, Inc. Factor by factoring out the GCF. a) b) c) d) 6.1

24. Slide 6- 24 Copyright © 2011 Pearson Education, Inc. Factor by factoring out the GCF. a) b) c) d) 6.1

25. Slide 6- 25 Copyright © 2011 Pearson Education, Inc. Factor by grouping. a) b) c) d) 6.1

26. Slide 6- 26 Copyright © 2011 Pearson Education, Inc. Factor by grouping. a) b) c) d) 6.1

27. Copyright © 2011 Pearson Education, Inc. Factoring Trinomials of the Form x 2 + bx + c 6.2 1.Factor trinomials of the form x 2 + bx + c. 2.Factor out a monomial GCF, then factor the trinomial of the form x 2 + bx + c.

28. Slide 6- 28 Copyright © 2011 Pearson Education, Inc. Following are some examples of trinomials of the form x 2 + bx + c. x 2 + 5x + 6 or x 2 –7x + 12 or x 2 – 5x – 24 Products in the form x 2 + bx + c are the result of the product of two binomials. When we factor a trinomial of the form x 2 + bx + c, we reverse the FOIL process, using the fact that b is the sum of the last terms in the binomials and c is the product of the last terms in the binomials.

29. Slide 6- 29 Copyright © 2011 Pearson Education, Inc. Factoring x 2 + bx + c To factor a trinomial of the form x 2 + bx + c, 1. Find two numbers with a product equal to c and a sum equal to b. 2. The factored trinomial will have the form: (x + first number) (x + second number). Note: The signs in the binomial factors can be minus signs, depending on the signs of b and c.

30. Slide 6- 30 Copyright © 2011 Pearson Education, Inc. Example 1a Factor. x 2 – 6x + 8 Solution: We must find a pair of numbers whose product is 8 and whose sum is –6. If two numbers have a positive product and negative sum, they must both be negative. Following is a table listing the products and sums: ProductSum (–1)(–8) = 8–1 + (–8) = –9 (–2)(–4) = 8–2 + (–4) = –6 This is the correct combination.

31. Slide 6- 31 Copyright © 2011 Pearson Education, Inc. continued Answer Check We can check by multiplying the binomial factors to see if their product is the original polynomial. x 2 – 6x + 8 = (x – 2)(x – 4) (x – 2)(x – 4) = x 2 – 4x – 2x + 8 = x 2 – 6x + 8 Multiply the factors using FOIL. The product is the original polynomial.

32. Slide 6- 32 Copyright © 2011 Pearson Education, Inc. Example 1b Factor. x 2 + 2x – 24 Solution: We must find a pair of numbers whose product is –24 and whose sum is 2. Because the product is negative, the two numbers have different signs. ProductSum (–2)(12) =  24 –2 + 12 = 10 (–4)(6) =  24 –4 + 6 = 2 (–8)(3) = –24–8 + 3 = –5 This is the correct combination.

33. Slide 6- 33 Copyright © 2011 Pearson Education, Inc. continued Answer Check We can check by multiplying the binomial factors to see if their product is the original polynomial. x 2 + 2x – 24 = (x – 4)(x + 6) (x – 4)(x + 6) = x 2 + 6x – 4x – 24 = x 2 + 2x – 24 Multiply the factors using FOIL. The product is the original polynomial.

34. Slide 6- 34 Copyright © 2011 Pearson Education, Inc. Example 1c Factor. x 2 – 6x – 27 Solution: We must find a pair of numbers whose product is –27 and whose sum is  6. Because the product is negative, the two numbers have different signs. ProductSum (–1)(27) =  27 –2 + 27 = 25 (–9)(3) =  27–9 + 3 =  6 This is the correct combination.

35. Slide 6- 35 Copyright © 2011 Pearson Education, Inc. continued Answer Check We can check by multiplying the binomial factors to see if their product is the original polynomial. x 2 – 6x – 27 = (x – 9)(x + 3) (x – 9)(x + 3) = x 2 + 3x – 9x – 27 = x 2 – 6x – 27 Multiply the factors using FOIL. The product is the original polynomial.

36. Slide 6- 36 Copyright © 2011 Pearson Education, Inc. Example 1d Factor. x 2 + 2x + 5 Solution: We must find a pair of numbers whose product is 5 and whose sum is 2. If two numbers have a positive product and a positive sum, both must be positive ProductSum (1)(5) = 51 + 5 = 6 The polynomial cannot be factored. The polynomial is prime. There is no combination of factors whose product is 5 and sum is 2.

37. Slide 6- 37 Copyright © 2011 Pearson Education, Inc. Example 2a Factor. a 2 – ab – 20b 2 Solution: We must find a pair of terms whose product is  20b 2 and whose sum is –1b. These terms would have to be –5b and 4b. Answer a 2 – ab – 20b 2 =(a – 5b)(a + 4b) Check (a – 5b)(a + 4b) = a 2 + 4ab – 5ab – 20b 2 = a 2 – ab – 20b 2

38. Slide 6- 38 Copyright © 2011 Pearson Education, Inc. Example 3a Factor. 4xy 3 + 12xy 2 – 72xy Solution First, we look for a monomial GCF (other than 1). Notice that the GCF of the terms is 4xy. Factoring out the monomial, we have 4xy 3 + 12xy 2 – 72xy = 4xy(y 2 + 3y – 18) Now try to factor the trinomial to two binomials. We must find a pair of numbers whose product is –18 and whose sum is 3.

39. Slide 6- 39 Copyright © 2011 Pearson Education, Inc. continued Answer ProductSum (–1)(18) = –18–1 + 18 = 17 (–2)(9) = – 18–2 + 9 = 7 (–3)(6) = – 18–3 + 6 = 3 This is the correct combination. 4xy 3 + 12xy 2 – 72xy =4xy(y – 3)(y + 6)

40. Slide 6- 40 Copyright © 2011 Pearson Education, Inc. Example 3b Factor. x 4 + 2x 3 + 5x 2 Solution First, factor out the GCF, x 2. x 2 ( x 2 + 2x + 5) Now try to factor the trinomial to two binomials. We must find a pair of numbers whose product is 5 and whose sum is 2. From a previous example, we learned this polynomial is prime. The final factored form is x 2 ( x 2 + 2x + 5)

41. Slide 6- 41 Copyright © 2011 Pearson Education, Inc. Factor. x 2 + 5x – 36 a) (x + 3)(x – 12 ) b) (x – 3)(x + 12 ) c) (x + 9)(x – 4 ) d) (x – 9)(x + 4 ) 6.2

42. Slide 6- 42 Copyright © 2011 Pearson Education, Inc. Factor. x 2 + 5x – 36 a) (x + 3)(x – 12 ) b) (x – 3)(x + 12 ) c) (x + 9)(x – 4 ) d) (x – 9)(x + 4 ) 6.2

43. Slide 6- 43 Copyright © 2011 Pearson Education, Inc. Factor completely. 5rs 3 – 10rs 2 – 40rs a) 5rs(s 2 – 2s – 8 ) b) 5rs(s 2 + 2s – 8 ) c) 5rs(s + 2)(s – 4 ) d) 5rs(s – 2)(s + 4 ) 6.2

44. Slide 6- 44 Copyright © 2011 Pearson Education, Inc. Factor completely. 5rs 3 – 10rs 2 – 40rs a) 5rs(s 2 – 2s – 8 ) b) 5rs(s 2 + 2s – 8 ) c) 5rs(s + 2)(s – 4 ) d) 5rs(s – 2)(s + 4 ) 6.2

45. Copyright © 2011 Pearson Education, Inc. Factoring Trinomials of the Form ax 2 + bx + c, where a  1 6.3 1.Factor trinomials of the form ax 2 + bx + c, where a  1, by trial. 2.Factor trinomials of the form ax 2 + bx + c, where a  1, by grouping.

46. Slide 6- 46 Copyright © 2011 Pearson Education, Inc. We will focus on factoring trinomials in which the coefficient of the squared term is other than 1, such as the following: 3x 2 + 17x + 108x 2 + 29x – 12 In general, like trinomials of the form x 2 + bx + c, trinomials of the form ax 2 + bx + c, where a 1, also have two binomial factors.

47. Slide 6- 47 Copyright © 2011 Pearson Education, Inc. Factoring by Trial and Error To factor a trinomial of the form ax 2 + bx + c, where a ≠ 1, by trial and error, 1. Look for a monomial GCF in all the terms. If there is one, factor it out. 2. Write a pair of first terms whose product is ax 2. ax 2

48. Slide 6- 48 Copyright © 2011 Pearson Education, Inc. 3. Write a pair of last terms whose product is c. 4. Verify that the sum of the inner and outer products is bx (the middle term of the trinomial). c + Outer bx Inner

49. Slide 6- 49 Copyright © 2011 Pearson Education, Inc. If the sum of the inner and outer products is not bx, try the following: a. Exchange the last terms of the binomials from step 3, then repeat step 4. b. For each additional pair of last terms, repeat steps 3 and 4. c. For each additional pair of first terms, repeat steps 2 - 4.

50. Slide 6- 50 Copyright © 2011 Pearson Education, Inc. Example 1 Factor. Solution The first terms must multiply to equal 6x 2. These could be x and 6x, or 2x and 3x. The last terms must multiply to equal –5. Because –5 is negative, the last terms in the binomials must have different signs. This factor pair must be 1 and 5.

51. Slide 6- 51 Copyright © 2011 Pearson Education, Inc. continued Now we multiply binomials with various combinations of these first and last terms until we find a combination whose inner and outer products combine to equal 13x. Correct combination. Incorrect combinations. Answer

52. Slide 6- 52 Copyright © 2011 Pearson Education, Inc. Example 2 Factor. Solution First, we factor out the monomial GCF, 3x. The last terms must multiply to equal 3. Because 3 is a prime number, its factors are 1 and 3. Now we factor the trinomial within the parentheses. The first terms must multiply to equal 7x 2. These could be x and 7x.

53. Slide 6- 53 Copyright © 2011 Pearson Education, Inc. continued Now we multiply binomials with various combinations of these first and last terms until we find a combination whose inner and outer products combine to equal –20x. Correct combination. Answer

54. Slide 6- 54 Copyright © 2011 Pearson Education, Inc. Objective 2 Factor trinomials of the form ax 2 + bx + c, where a  1, by grouping.

55. Slide 6- 55 Copyright © 2011 Pearson Education, Inc. Factoring ax 2 + bx + c, where a ≠ 1, by Grouping To factor a trinomial of the form ax 2 + bx + c, where a ≠ 1, by grouping, 1. Look for a monomial GCF in all the terms. If there is one, factor it out. 2. Multiply a and c. 3. Find two factors of this product whose sum is b. 4. Write a four-term polynomial in which bx is written as the sum of two like terms whose coefficients are the two factors you found in step 3. 5. Factor by grouping.

56. Slide 6- 56 Copyright © 2011 Pearson Education, Inc. Example 3a Factor. Solution Notice that for this trinomial, a = 2, b = –15, and c = 7. We begin my multiplying a and c: (2)(7) = 14. Now we find two factors of 14 whose sum is –15. Notice that these two factors must both be negative. Factors of acSum of Factors of ac (–2)(–7) = 14–2 + (–7) = –9 (–1)(–14) = 14–1 + (– 14) = –15 Correct

57. Slide 6- 57 Copyright © 2011 Pearson Education, Inc. continued –15x–x – 14x

58. Slide 6- 58 Copyright © 2011 Pearson Education, Inc. Example 3b Factor. Solution Notice that for this trinomial, a = 4, b = 17, and c = 4. We begin my multiplying a and c: (4)(4) = 16. Now we find two factors of 16 whose sum is 17. Notice that these two factors must both be positive. Factors of acSum of Factors of ac (1)(16) = 161 + 16 = 17 (2)(8) = 162 + 8 = 10 (4)(4) = 164 + 4 = 8 Correct

59. Slide 6- 59 Copyright © 2011 Pearson Education, Inc. continued +17x + x + 16x

60. Slide 6- 60 Copyright © 2011 Pearson Education, Inc. Example 3c Factor. Solution Notice that there is a GCF, 2x 2, that we can factor out. 2x 2 (6x 2 + 7x – 3) Now we factor the trinomial within the parentheses. a = 6, c =  3; ac = (6)(  3) =  18 Find two factors of  18 whose sum is 7. Because the product is negative, the two factors will have different signs.

61. Slide 6- 61 Copyright © 2011 Pearson Education, Inc. Example 3c Factor. Solution Factors of acSum of Factors of ac (  1)(18) =  18  1 + 18 = 17 (  2)(9) =  18  2 + 9 = 7 (  3)(6) =  18  3 + 6 = 3 Correct 2x 2 (6x 2 + 7x – 3) Now write 7x as  2x + 9x and then factor by grouping.

62. Slide 6- 62 Copyright © 2011 Pearson Education, Inc. continued +7x  2x + 9x

63. Slide 6- 63 Copyright © 2011 Pearson Education, Inc. Factor completely. 6x 2 –33x – 63 a) 3(2x + 7)(x – 3) b) 3(2x + 3)(x – 7) c) 3(2x – 3)(x + 7) d) 3(2x – 7)(x + 3) 6.3

64. Slide 6- 64 Copyright © 2011 Pearson Education, Inc. Factor completely. 6x 2 –33x – 63 a) 3(2x + 7)(x – 3) b) 3(2x + 3)(x – 7) c) 3(2x – 3)(x + 7) d) 3(2x – 7)(x + 3) 6.3

65. Slide 6- 65 Copyright © 2011 Pearson Education, Inc. Factor completely. 2x 2 +3x – 20 a) (2x + 2)(x – 10) b) (2x + 4)(x – 5) c) (2x – 5)(x + 4) d) (2x – 10)(x + 2) 6.3

66. Slide 6- 66 Copyright © 2011 Pearson Education, Inc. Factor completely. 2x 2 +3x – 20 a) (2x + 2)(x – 10) b) (2x + 4)(x – 5) c) (2x – 5)(x + 4) d) (2x – 10)(x + 2) 6.3

67. Copyright © 2011 Pearson Education, Inc. Factoring Special Products 6.4 1.Factor perfect square trinomials. 2.Factor a difference of squares. 3.Factor a difference of cubes. 4.Factor a sum of cubes.

68. Slide 6- 68 Copyright © 2011 Pearson Education, Inc. Factoring Perfect Square Trinomials a 2 + 2ab + b 2 = (a + b) 2 a 2 – 2ab + b 2 = (a – b) 2

69. Slide 6- 69 Copyright © 2011 Pearson Education, Inc. Example 1a Factor. 9a 2 + 6a + 1 Solution This trinomial is a perfect square because the first and the last terms are perfect squares and twice the product of their roots is the middle term. 9a 2 + 6a + 1 The square root of 9a 2 is 3a.The square root of 1 is 1. Twice the product of 3a and 1 is (2)(3a)(1) = 6a, which is the middle term. Answer 9a 2 + 6a + 1 = (3a + 1) 2

70. Slide 6- 70 Copyright © 2011 Pearson Education, Inc. Example 1b Factor. 16x 2 – 56x + 49 Solution This trinomial is a perfect square. The square root of 16x 2 is 4x.The square root of 49 is 7. Twice the product of 4x and 7 is (2)(4x)(7) = 56x, which is the middle term. Answer 16x 2 – 56x + 49 = (4x – 7) 2 16x 2 – 56x + 49 Use a 2 – 2ab + b 2 = (a – b) 2, where a = 4x and b = 7.

71. Slide 6- 71 Copyright © 2011 Pearson Education, Inc. Example 2a Factor. 9x 2 – 6xy + y 2 Solution = (3x – y) 2 9x 2 – 6xy + y 2 Use a 2 – 2ab + b 2 = (a – b) 2, where a = 3x and b = y.

72. Slide 6- 72 Copyright © 2011 Pearson Education, Inc. Example 2b Factor. Solution Use a 2 – 2ab + b 2 = (a – b) 2, where a = 3a and b = 7. Factor out the monomial GCF, ab.

73. Slide 6- 73 Copyright © 2011 Pearson Education, Inc. Factoring a Difference of Squares a 2 – b 2 = (a + b)(a – b) Warning: A sum of squares a 2 + b 2 is prime and cannot be factored.

74. Slide 6- 74 Copyright © 2011 Pearson Education, Inc. Example 3a Factor. 9x 2 – 16y 2 Solution This binomial is a difference of squares because 9x 2 – 16y 2 = (3x) 2 – (4y) 2. To factor it, we use the rule a 2 – b 2 = (a + b)(a – b). a 2 – b 2 = (a + b)(a – b) 9x 2 – 16y 2 = (3x) 2 – (4y) 2 = (3x + 4y)(3x – 4y)

75. Slide 6- 75 Copyright © 2011 Pearson Education, Inc. Example 3b Factor. 16n 4 – 25 Solution This binomial is a difference of squares, where a = 4n 2 and b = 5. (4n 2 + 5)(4n 2 – 5)16n 4 – 25 = Use a 2 – b 2 = (a + b)(a – b).

76. Slide 6- 76 Copyright © 2011 Pearson Education, Inc. Example 3c Factor. Solution The terms in this binomial have a monomial GCF, 2a 7. Factor out the GCF. Factor 36a 2 – 49, using a 2 – b 2 = (a + b)(a – b) with a = 6a and b = 7.

77. Slide 6- 77 Copyright © 2011 Pearson Education, Inc. Example 4 Factor. Solution The binomial is a difference of squares, where a = x 2 and b = 25. Factor x 2 – 25, using a 2 – b 2 = (a + b)(a – b) with a = x and b = 5.

78. Slide 6- 78 Copyright © 2011 Pearson Education, Inc. Factoring a Difference of Cubes a 3 – b 3 = (a – b)(a 2 + ab + b 2 )

79. Slide 6- 79 Copyright © 2011 Pearson Education, Inc. Example 5a Factor. 125x 3 – 27 Solution This binomial is a difference of cubes. a 3 – b 3 = (a – b) (a 2 + a b + b 2 ) 125x 3 – 27 = (5x) 3 – (3) 3 = (5x – 3)((5x) 2 + (5x)(3) + (3) 2 ) = (5x – 3)(25x 2 + 15x + 9) Note: The trinomial may seem like a perfect square. However, to be a perfect square, the middle term should be 2ab. In this trinomial, we only have ab, so it cannot be factored.

80. Slide 6- 80 Copyright © 2011 Pearson Education, Inc. Example 5b Factor. Solution The binomial has a GCF, a.

81. Slide 6- 81 Copyright © 2011 Pearson Education, Inc. Factoring a Sum of Cubes a 3 + b 3 = (a + b)(a 2 – ab + b 2 )

82. Slide 6- 82 Copyright © 2011 Pearson Education, Inc. Example 6a Factor. 125x 3 + 64 Solution This binomial is a difference of cubes. a 3 + b 3 = (a + b) (a 2  a b + b 2 ) 125x 3 + 64 = (5x) 3 + (4) 3 = (5x + 4)((5x) 2  (5x)(4) + (4) 2 ) = (5x + 4)(25x 2  20x + 16)

83. Slide 6- 83 Copyright © 2011 Pearson Education, Inc. Example 6b Factor. 6x +162xy 3 Solution The terms in this binomial have a monomial GCF, 6x. = 6x(1 + 27y 3 ) 6x +162xy 3 = 6x(1 + 3y)((1) 2 – (1)(3y) + (3y) 2 ) = 6x(1 + 3y)(1 – 3y + 9y 2 )

84. Slide 6- 84 Copyright © 2011 Pearson Education, Inc. Factor completely. 4a 2 – 20a + 25 a) (2a + 5) 2 b) (2a – 5) 2 c) (4a + 5) 2 d) (4a – 5) 2 6.4

85. Slide 6- 85 Copyright © 2011 Pearson Education, Inc. Factor completely. 4a 2 – 20a + 25 a) (2a + 5) 2 b) (2a – 5) 2 c) (4a + 5) 2 d) (4a – 5) 2 6.4

86. Slide 6- 86 Copyright © 2011 Pearson Education, Inc. Factor completely. 9x 2 – 49 a) (3x + 5) 2 b) (3x + 7)(3x – 7) c) (3x – 7) 2 d) (7x + 3)(7x – 3) 6.4

87. Slide 6- 87 Copyright © 2011 Pearson Education, Inc. Factor completely. 9x 2 – 49 a) (3x + 5) 2 b) (3x + 7)(3x – 7) c) (3x – 7) 2 d) (7x + 3)(7x – 3) 6.4

88. Slide 6- 88 Copyright © 2011 Pearson Education, Inc. Factor completely. 2n 2 + 24n + 72 a) 2(n + 6) 2 b) 2(n + 6)(n – 6) c) 2(n – 6) 2 d) (2n + 6)(2n – 6) 6.4

89. Slide 6- 89 Copyright © 2011 Pearson Education, Inc. Factor completely. 2n 2 + 24n + 72 a) 2(n + 6) 2 b) 2(n + 6)(n – 6) c) 2(n – 6) 2 d) (2n + 6)(2n – 6) 6.4

90. Copyright © 2011 Pearson Education, Inc. Strategies for Factoring 6.5 1.Factor polynomials.

91. Slide 6- 91 Copyright © 2011 Pearson Education, Inc. Factoring a Polynomial To factor a polynomial, first factor out any monomial GCF, then consider the number of terms in the polynomial. If the polynomial has I.Four terms, try to factor by grouping. II. Three terms, determine whether the trinomial is a perfect square. A. If the trinomial is a perfect square, then consider its form. 1. If in form a 2 + 2ab + b 2, the factored form is (a + b) 2. 2. If in form a 2  2ab + b 2, the factored form is (a  b) 2. B.If the trinomial is not a perfect square, consider its form. 1. If in form x 2 + bx + c, find two factors of c whose sum is b, and write the factored form as (x + first number)(x + second number).

92. Slide 6- 92 Copyright © 2011 Pearson Education, Inc. Factoring a Polynomial continued 2. If in form ax 2 + bx + c, where a  1, then use trial and error. Or, find two factors of ac whose sum is b; write these factors as coefficients of two like terms that, when combined, equal bx; and then factor by grouping. III.Two terms, determine if the binomial is a difference of squares, sum of cubes, or difference of cubes. A. If given a binomial that is a difference of squares, a 2 – b 2, the factors are conjugates and the factored form is (a + b)(a – b). Note that a sum of squares cannot be factored. B. If given a binomial that is a sum of cubes, a 3 + b 3, the factored form is (a + b)(a 2 – ab + b 2 ). C. If given a binomial that is a difference of cubes, a 3 – b 3, the factored form is (a – b)(a 2 + ab + b 2 ). Note: Always check to see if any of the factors can be factored.

93. Slide 6- 93 Copyright © 2011 Pearson Education, Inc. Example 1a Factor. 12x 2 – 8x – 15 Solution There is no GCF. Not a perfect square, since the first and last terms are not perfect squares. Use trial and error or grouping. (x – 3)(12x + 5) = 12x 2 + 5x – 36x – 15 No (6x – 3)(2x + 3) = 12x 2 + 18x – 6x – 9 No (6x + 5)(2x – 3) = 12x 2 – 18x + 10x – 15 12x 2 – 8x – 15 Correct

94. Slide 6- 94 Copyright © 2011 Pearson Education, Inc. Example 1b Factor. 5x 3 – 10x 2 – 120x Solution 5x(x 2 – 2x – 24) Factored out the monomial GCF, 5x. Look for two numbers whose product is –24 and whose sum is  2. 5x(x + 4)(x – 6) ProductSum (  1)(24) =  24  1 + 24 = 23 4(  6) =  244 + (  6) =  2 Correct combination.

95. Slide 6- 95 Copyright © 2011 Pearson Education, Inc. Example 1c Factor. 8a 4 – 72n 2 Solution 8a 4 – 72n 2 = 8(a 4 – 9n 2 ) Factor out the monomial GCF, 8. a 4 – 9n 2 is a difference of squares = 8(a 2 – 3n)(a 2 + 3n)

96. Slide 6- 96 Copyright © 2011 Pearson Education, Inc. Example 1d Factor. 12y 5 + 84y 3 Solution 12y 3 (y 2 + 7) Factor out the monomial GCF, 12y 3.

97. Slide 6- 97 Copyright © 2011 Pearson Education, Inc. Example 1e Factor. 150x 3 y – 120x 2 y 2 + 24xy 3 Solution 6xy(25x 2 – 20xy + 4y 2 ) 6xy(5x – 2y) 2 Factor out the monomial GCF, 6xy. Factor the perfect square trinomial.

98. Slide 6- 98 Copyright © 2011 Pearson Education, Inc. Example 1f Factor. x 5 – 2x 3 – 27x 2 + 54 Solution No common monomial, factor by grouping. (x 5 – 2x 3 ) + (– 27x 2 + 54) x 3 (x 2 – 2) – 27(x 2 – 2) (x 2 – 2)(x 3 – 27) Difference of cubes (x 2 – 2)(x – 3)(x 2 + 3x + 9)

99. Slide 6- 99 Copyright © 2011 Pearson Education, Inc. Factor. 6x 2 + 17x + 5 a) (6x + 1)(x + 5) b) (3x + 1)(2x + 5) c) (6x + 1)(x – 5) d) (3x – 1)(2x – 5) 6.5

100. Slide 6- 100 Copyright © 2011 Pearson Education, Inc. Factor. 6x 2 + 17x + 5 a) (6x + 1)(x + 5) b) (3x + 1)(2x + 5) c) (6x + 1)(x – 5) d) (3x – 1)(2x – 5) 6.5

101. Slide 6- 101 Copyright © 2011 Pearson Education, Inc. Factor. 7y 4 + 49y 2 a) 7y(y 3 + 7y) b) 7y 2 (y 2 + 49) c) y 2 (7y 2 + 49) d) 7y 2 (y 2 + 7) 6.5

102. Slide 6- 102 Copyright © 2011 Pearson Education, Inc. Factor. 7y 4 + 49y 2 a) 7y(y 3 + 7y) b) 7y 2 (y 2 + 49) c) y 2 (7y 2 + 49) d) 7y 2 (y 2 + 7) 6.5

103. Copyright © 2011 Pearson Education, Inc. Solving Quadratic Equations by Factoring 6.6 1.Use the zero-factor theorem to solve equations containing expressions in factored form. 2.Solve quadratic equations by factoring. 3.Solve problems involving quadratic equations. 4.Use the Pythagorean theorem to solve problems.

104. Slide 6- 104 Copyright © 2011 Pearson Education, Inc. Zero-Factor Theorem If a and b are real numbers and ab = 0, then a = 0 or b = 0.

105. Slide 6- 105 Copyright © 2011 Pearson Education, Inc. Example 1 Solve. (x + 4)(x + 5) = 0 Solution According to the zero-factor theorem, one of the two factors, or both factors, must equal 0. x + 4 = 0 or x + 5 = 0 Solve each equation. x =  4 x =  5 Check For x =  4: For x =  5: (x + 4)(x + 5) = 0 (x + 4)(x + 5) = 0 (  4 + 4)(  4 + 5) = 0 (  5 + 4)(  5 + 5) = 0 0(1) = 0 (  1)(0) = 0

106. Slide 6- 106 Copyright © 2011 Pearson Education, Inc. Solving Equations with Two or More Factors Equal to 0 To solve an equation in which two or more factors are equal to 0, use the zero-factor theorem. 1. Set each factor equal to zero. 2. Solve each of those equations.

107. Slide 6- 107 Copyright © 2011 Pearson Education, Inc. Example 2 Solve. a. y(5y + 2) = 0b. x(x + 2)(5x – 4) = 0 Solution a. y(5y + 2) = 0 y = 0 or 5y + 2 = 0 5y =  2 This equation is already solved. b. To check, we verify that the solutions satisfy the original equations.

108. Slide 6- 108 Copyright © 2011 Pearson Education, Inc. Quadratic equation in one variable: An equation that can be written in the form ax 2 + bx + c = 0, where a, b, and c are all real numbers and a  0.

109. Slide 6- 109 Copyright © 2011 Pearson Education, Inc. Solving Quadratic Equations Using Factoring To solve a quadratic equation using factoring, 1. Write the equation in standard form (ax 2 + bx + c = 0). 2. Write the variable expression in factored form. 3. Use the zero-factor theorem to solve.

110. Slide 6- 110 Copyright © 2011 Pearson Education, Inc. Example 3 Solve. 2x 2 – 5x – 3 = 0 Solution The equation is in standard form, so we can factor. 2x 2 – 5x – 3 = 0 (2x + 1)(x – 3) = 0 Use the zero-factor theorem to solve. 2x + 1 = 0 or x – 3 = 0 To check, we verify that the solutions satisfy the original equations.

111. Slide 6- 111 Copyright © 2011 Pearson Education, Inc. Example 4 Solve. x 3 + 4x 2 – 21x = 0 Solution x 3 + 4x 2 – 21x = 0 Factor out the monomial GCF, x. x(x 2 + 4x – 21) = 0 Use the zero-factor theorem to solve. x(x – 3)(x + 7) = 0 x = 0orx – 3 = 0orx + 7 = 0 To check, we verify that the solutions satisfy the original equations.

112. Slide 6- 112 Copyright © 2011 Pearson Education, Inc. Example 5a Solve. 6y 2 + 11y = 10 + 4y Solution Write the equation in standard form. 6y 2 + 11y = 10 + 4y 6y 2 + 7y = 10 Subtract 4y from both sides. 6y 2 + 7y – 10 = 0 Subtract 10 from both sides. (6y – 5)(y + 2) = 0 Factor. 6y – 5 = 0 or y + 2 = 0 Use the zero-factor theorem.

113. Slide 6- 113 Copyright © 2011 Pearson Education, Inc. Example 5b Solve. y(y – 7) = –12 Solution Write the equation in standard form.

114. Slide 6- 114 Copyright © 2011 Pearson Education, Inc. Example 6 The product of two consecutive odd natural numbers is 323. Find the numbers. Understand Odd numbers are 1, 3, 5,… Let x = the first odd number Let x + 2 = consecutive odd number The word product means that two numbers are multiplied to equal 323. PlanTranslate to an equation, then solve.

115. Slide 6- 115 Copyright © 2011 Pearson Education, Inc. continued Executex(x + 2) = 323 x(x + 2) – 323 = 0 x 2 + 2x – 323 = 0 (x + 19)(x – 17) = 0 x + 19 = 0x – 17 = 0 x = –19x = 17 Answer Because –19 is not a natural number and 17 is, the first number is 17. This means that the consecutive odd natural number is 19. Check 17 and 19 are consecutive odd natural numbers and their product is 323.

116. Slide 6- 116 Copyright © 2011 Pearson Education, Inc. The Pythagorean Theorem Given a right triangle, where a and b represent the lengths of the legs and c represents the length of the hypotenuse, then a 2 + b 2 = c 2. c (hypotenuse) b (leg) a (leg)

117. Slide 6- 117 Copyright © 2011 Pearson Education, Inc. Example 9 Find the length of the missing side. Solution Use the Pythagorean theorem, a 2 + b 2 = c 2 15 2 + 36 2 = c 2 Substitute. 225 + 1296 = c 2 Simplify exponential forms. 1521 = c 2 Add. c 2 – 1521 = 0 Standard form. (c – 39)(c + 39) = 0 Factor. c – 39 = 0 or c + 39 = 0 c = 39 or c = –39 Only the positive solution is sensible. ? 36 15

118. Slide 6- 118 Copyright © 2011 Pearson Education, Inc. Solve. x 2 = 6x – 8 a) 2 and 4 b) 2 and  4 c)  2 and 4 d) 1 and  8 6.6

119. Slide 6- 119 Copyright © 2011 Pearson Education, Inc. Solve. x 2 = 6x – 8 a) 2 and 4 b) 2 and  4 c)  2 and 4 d) 1 and  8 6.6

120. Slide 6- 120 Copyright © 2011 Pearson Education, Inc. Solve. One natural number is four times another. The product of the two numbers is 900. Find the larger number. a) 15 b) 30 c) 35 d) 60 6.6

121. Slide 6- 121 Copyright © 2011 Pearson Education, Inc. Solve. One natural number is four times another. The product of the two numbers is 900. Find the larger number. a) 15 b) 30 c) 35 d) 60 6.6

122. Slide 6- 122 Copyright © 2011 Pearson Education, Inc. Find the length of the hypotenuse. a) 15 b) 46 c) 50 d) 62 ? 48 14 6.6

123. Slide 6- 123 Copyright © 2011 Pearson Education, Inc. Find the length of the hypotenuse. a) 15 b) 46 c) 50 d) 62 ? 48 14 6.6

124. Copyright © 2011 Pearson Education, Inc. Graphs of Quadratic Equations and Functions 6.7 1.Graph quadratic equations in the form y = ax 2 + bx + c. 2.Graph quadratic functions.

125. Slide 6- 125 Copyright © 2011 Pearson Education, Inc. Quadratic equation in two variables: An equation that can be written in the form y = ax 2 + bx + c, where a, b, and c are real numbers and a  0. Axis of symmetry: A line that divides a graph into two symmetrical halves. Vertex: The lowest point on a parabola that opens up or the highest point on a parabola that opens down. vertex (0, 0) axis of symmetry x = 0 (y-axis)

126. Slide 6- 126 Copyright © 2011 Pearson Education, Inc. Graphing Quadratic Equations To graph a quadratic equation, 1. Find ordered pair solutions and plot them in the coordinate plane. Continue finding and plotting solutions until the shape of the parabola can be clearly seen. 2. Connect the points to form a parabola.

127. Slide 6- 127 Copyright © 2011 Pearson Education, Inc. Example 1a Graph. y = 2x 2 + 1 Solution Complete a table of solutions. xy 22 9 11 3 01 13 29 Plot the points. Connect the points.

128. Slide 6- 128 Copyright © 2011 Pearson Education, Inc. Example 1b Graph. y = x 2 – 2x – 3 Solution Complete a table of solutions. xy 22 5 11 0 0 33 1 44 2 33 30 Plot the points. Connect the points.

129. Slide 6- 129 Copyright © 2011 Pearson Education, Inc. Example 1c Graph. y =  3x 2 + 4 Solution Complete a table of solutions. xy 22 88 11 1 04 11 2 88 Plot the points. Connect the points.

130. Slide 6- 130 Copyright © 2011 Pearson Education, Inc. Opening of a Parabola Given an equation in the form y = ax 2 + bx + c, if a > 0, then the parabola opens upward; if a < 0, then the parabola opens downward.

131. Slide 6- 131 Copyright © 2011 Pearson Education, Inc. Graphing Quadratic Functions To graph a quadratic function, 1. Find enough ordered pairs by evaluating the function for various values of x so that when those ordered pairs are plotted, the shape of the parabola can be clearly seen. 2. Connect the points to form the parabola.

132. Slide 6- 132 Copyright © 2011 Pearson Education, Inc. Example 3a Graph. f(x) = 2x 2 – 1 Solution Complete a table of solutions. xy 22 7 11 1 0 11 11 27 Plot the points. Connect the points. This parabola opens upward since a > 0.

133. Slide 6- 133 Copyright © 2011 Pearson Education, Inc. Example 3b Graph. f(x) =  2x 2 + 8x – 1 Solution Complete a table of solutions. xy 11  11 0 11 15 27 35 4 11 Plot the points. Connect the points. This parabola opens downward since a < 0.

134. Slide 6- 134 Copyright © 2011 Pearson Education, Inc. Example 4a Determine whether the graph is the graph of a function. Give the domain and range. Solution This is a function because any vertical line intersects the graph in at most one point. Domain: all real numbers or ( ,  ) Range: {y|y ≥  3} or [  3,  )

135. Slide 6- 135 Copyright © 2011 Pearson Education, Inc. Example 4b Determine whether the graph is the graph of a function. Give the domain and range. Solution This is NOT a function because a vertical line can be drawn that intersects the graph in more than one point. Domain: {x|x ≥ 1} or [1,  ) Range: all real numbers or ( ,  )

136. Slide 6- 136 Copyright © 2011 Pearson Education, Inc. Graph. y = x 2 – 2 a)b) c)d) 6.7

137. Slide 6- 137 Copyright © 2011 Pearson Education, Inc. Graph. y = x 2 – 2 a)b) c)d) 6.7

138. Slide 6- 138 Copyright © 2011 Pearson Education, Inc. Graph. f(x) =  x 2 + 2x – 2 a)b) c)d) 6.7

139. Slide 6- 139 Copyright © 2011 Pearson Education, Inc. Graph. f(x) =  x 2 + 2x – 2 a)b) c)d) 6.7