2. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Introduction to Factoring Common Factors Factoring by Grouping 6.1

3. Slide 3 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Common Factors When factoring a polynomial, we first look for factors that are common to each term. By applying a distributive property we can often write a polynomial as a product. For example: 8x 2 = 2x  4x and 6x = 2x  3 And by the distributive property, 8x 2 + 6x = 2x(4x + 3)

4. Slide 4 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Finding common factors Factor. a.b.c.d. a. d. b. c.

5. Slide 5 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley When finding the greatest common factor for a polynomial, it is often helpful first to completely factor each term of the polynomial.

6. Slide 6 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Finding the greatest common factor Find the greatest common factor for the expression. Then factor the expression. 18x 2 + 3x 18x 2 = 2  3  3  x  x 3x = 3  x3  x The greatest common factor is 3x. 18x 2 + 3x = 3x(6x + 1)

7. Slide 7 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Factoring by Grouping Factoring by grouping is a technique that makes use of the associative and distributive properties.

8. Slide 8 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring out binomials Factor. a. 3x(x + 1) + 4(x + 1)b. 3x 2 (2x – 1) – x(2x – 1) a. Both terms in the expression contain the binomial x + 1. Use the distributive property to factor. b.

9. Slide 9 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring by grouping when the middle term is (+) Factor the polynomial.

10. Slide 10 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring by grouping when the middle term is (  ) Factor the polynomial.

11. Slide 11 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring out the GCF before grouping Completely factor the polynomial.

12. Slide 12 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

13. Factoring Trinomials I ( x 2 + bx + c ) Review of the FOIL Method Factoring Trinomials Having a Leading Coefficient of 1 6.2

14. Slide 14 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The product (x + 3) (x + 4) can be found as follows: x 2 + 4x + 3x + 12 x 2 + 7x + 12 The middle term is found by calculating the sum 4x and 3x, and the last term is found by calculating the product of 4 and 3. To solve equations using factoring, we use the zero- product property. It states that, if the product of two numbers is 0, then at least one of the numbers must equal 0.

15. Slide 15 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring a trinomial having only positive coefficients Factor each trinomial. a.b. a. b. Factors of 18 whose sum is 11. FactorsSum 1, 1819 2, 911 3, 69 Factors of 24 whose sum is 10. FactorsSum 1, 2425 2, 1214 3, 811 4, 610

16. Slide 16 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring a trinomial having a negative middle coefficient Factor each trinomial. a.b. a. b. Factors of 14 whose sum is −9. FactorsSum −1, −14−15 −2, −7−9 Factors of 48 whose sum is −19. FactorsSum −1, −48−49 −2, −24−26 −3, −16−19 −4, −12−16 −6, −8−14

17. Slide 17 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring a trinomial having a negative constant term Factor each trinomial. a.b. a. b. Factors of −15 whose sum is 2. FactorsSum 1, −15−14 − 1, 15+14 3, −5−2 −3, 52 Factors of −16 whose sum is −6. FactorsSum −1, 1615 1, −16−15 −2, 86 2, −8−6 −4, 40

18. Slide 18 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Discovering that a trinomial is prime Factor the trinomial. Factors of −8 whose sum is 6. FactorsSum 1, −8−7 −1, 87 2, −4−2 −2, 42 The table reveals that no such factor pair exists. Therefore, the trinomial is prime.

19. Slide 19 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring out the GCF before factoring further Factor each trinomial completely. a.b. a. b. Factor out common factor of 4. Factors (12)Sum (7) 1, 1213 2, 68 3, 47 Factor out common factor of 7. Factors (  10) Sum (3) 1,  10 99 2,  5 33  2, 5 3

20. Slide 20 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Finding the dimensions of a rectangle Find one possibility for the dimensions of a rectangle that has an area of x 2 + 12x + 35. The area of a rectangle equals length times width. If we can factor x 2 + 12x + 35, then the factors can represents its length and width. x x5 735 5x5x 7x7x x2x2 Because x 2 + 12x + 35 = (x + 5)(x + 7), one possibility for the rectangle’s dimensions is width x + 5 and length x + 7. Area = x 2 + 12x + 35

21. Slide 21 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

22. Factoring Trinomials II ( ax 2 + bx + c ) Factoring Trinomials by Grouping Factoring with FOIL in Reverse 6.3

23. Slide 23 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Factoring Trinomials by Grouping

24. Slide 24 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring ax 2 + bx + c by grouping Factor each trinomial. a.b. a. b. Multiply (2)(15) = 30 Factors of 30 whose sum is 13 10 and 3 Multiply (12)(  3) =  36 Factors of  36 whose sum is  5  9 and 4

25. Slide 25 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Discovering that a trinomial is prime Factor the trinomial. a. We need to find integers m and n such that mn = (3)(4) = 12 and m + n = 9. Because the middle term is positive, we consider only positive factors of 12. There are no factors whose sum is 9, the coefficient of the middle term. The trinomial is prime. Factors1, 122, 63, 4 Sum1387

26. Slide 26 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Factoring with FOIL in Reverse

27. Slide 27 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring the form ax 2 + bx + c Factor the trinomial. The factors of the last term are either 1 and 6 or 2 and 3. Try a set of factors. Try 1 and 6. Middle term is 13x not 7x.

28. Slide 28 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring the form ax 2 + bx + c—continued Factor the trinomial. Try 2 and 3 the factors of the last term. Middle term is 8x not 7x. Try another set of factors 3 and 2. Middle term is correct. The factors of the last term are either 1 and 6 or 2 and 3. Try a set of factors.

29. Slide 29 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

30. Slide 30 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

31. Special Types of Factoring Difference of Two Squares Perfect Square Trinomials 6.4

32. Slide 32 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley For any real numbers a and b, a 2 – b 2 = (a – b)(a + b). DIFFERENCE OF TWO SQUARES

33. Slide 33 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring the difference of two squares a.9x 2 – 16b.5x 2 + 8y 2 c.25x 4 – y 6 a.9x 2 – 16= (3x) 2 – (4) 2 = (3x – 4) (3x + 4) b.Because 5x 2 + 8y 2 is the sum of two squares, it cannot be factored. c. If we let a 2 = 25x 4 and b 2 = y 6, then a = 5x 2 and b = y 3. Thus, 25x 4 – y 6 = (5x 2 ) 2 – (y 3 ) 2 = (5x 2 – y 3 ) (5x 2 + y 3 ). Factor each difference of two squares.

34. Slide 34 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley For any real numbers a and b, a 2 + 2ab + b 2 = (a + b) 2 and a 2 − 2ab + b 2 = (a − b) 2. PERFECT SQUARE TRINOMIALS

35. Slide 35 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring perfect square trinomials a.x 2 + 8x + 16b. a. If possible, factor each trinomial as a perfect square. Let a 2 = x 2 and b 2 = 4 2. For a perfect square trinomial, the middle term must be 2ab. 2ab = 2(x)(4) = 8x, which equals the given middle term. Thus a 2 + 2ab + b 2 = (a + b) 2 implies x 2 + 8x + 16 = (x + 4) 2. 4x 2 − 12x + 9 x 2 + 8x + 16

36. Slide 36 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE continued b. Let a 2 = (2x) 2 and b 2 = 3 2. For a perfect square trinomial, the middle term must be 2ab. 2ab = 2(2x)(3) = 12x, which equals the given middle term. Thus a 2 − 2ab + b 2 = (a − b) 2 implies 4x 2 − 12x + 9 = (2x − 3) 2. 4x 2 − 12x + 9

37. Slide 37 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

38. Summary of Factoring Guidelines for Factoring Polynomials Factoring Polynomials 6.5

39. Slide 39 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

40. Slide 40 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring out a common factor Factor 6x 3 – 36x 2. Step 1:The greatest common factor is 6x 2. 6x 3 – 36x 2 = 6x 2 (x – 6) Step 2B:The binomial, x – 6, cannot be factored further. Step 3:The completely factored polynomial is 6x 2 (x – 6).

41. Slide 41 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring a difference of squares Factor 3x 3 – 75x. Step 1:The greatest common factor is 3x. 3x 3 – 75x = 3x(x 2 – 25) Step 2B:We can factor the binomial x 2 – 25 as a difference of squares. Step 3:The completely factored polynomial is 3x(x 2 – 25) = 3x(x – 5)(x + 5) 3x(x – 5)(x + 5).

42. Slide 42 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring a difference of squares Factor 2x 4 – 162. Step 1:The greatest common factor is 2. 2x 4 – 162 = 2(x 4 – 81) Step 2B:We can factor x 4 – 81 as the difference of squares, twice. Step 3:The completely factored polynomial is 2(x 4 – 81) = 2(x 2 – 9)(x 2 + 9) = 2(x – 3)(x + 3)(x 2 + 9) The sum of squares, x 2 + 9, cannot be factored further. 2(x – 3)(x + 3)(x 2 + 9).

43. Slide 43 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring a perfect square trinomial Factor 8x 3 + 40x 2 + 50x. Step 1:The greatest common factor is 2x. 8x 3 + 40x 2 + 50x = 2x(4x 2 + 20x + 25) Step 2C:The trinomial 4x 2 + 20x + 25 is a perfect square trinomial. Step 3:The completely factored polynomial is = 2x(2x + 5)(2x + 5)2x(4x 2 + 20x + 25) 2x(2x + 5) 2.

44. Slide 44 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring a sum of cubes Factor −2y 4 − 128y. Step 1:The greatest common factor is −2y. −2y 4 − 128y = −2y(y 3 + 64) Step 2B:The binomial y 3 + 64 can be factored as a sum of cubes. Step 3:The completely factored polynomial is −2y(y 3 + 64) −2y(y + 4)(y 2 – 4y + 16). = −2y(y + 4)(y 2 – 4y + 16)

45. Slide 45 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring by grouping Factor x 3 − 7x 2 – 4x + 28. Step 1:There are no common factors. Step 2A: Because there are four terms, we apply grouping. Step 3:The completely factored polynomial is (x − 2)(x + 2)(x − 7). = (x 3 − 7x 2 ) – (4x − 28)x 3 − 7x 2 – 4x + 28 = x 2 (x − 7) – 4(x − 7) = (x 2 − 4)(x − 7) difference of two squares

46. Slide 46 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Factoring a polynomial having two variables Factor 12x 2 − 75y 2. Step 1:The greatest common factor is 3. Step 2B:We can factor the binomial as the difference of squares. Step 3:The completely factored polynomial is 3(2x − 5y)(2x + 5y). = 3(2x − 5y)(2x + 5y) 3(4x 2 − 25y 2 )12x 2 − 75y 2 = 3(4x 2 − 25y 2 )

47. Slide 47 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Applying several techniques Factor 3x 5 − 27x 3 – 3x 2 + 27. Step 1: The greatest common factor is 3. Step 2A:Factor by grouping. = 3(x 5 − 9x 3 − x 2 + 9)3x 5 − 27x 3 – 3x 2 + 27 3(x 5 − 9x 3 − x 2 + 9) = 3(x 3 (x 2 – 9) − 1(x 2 – 9)) = 3(x 3 − 1)(x 2 – 9) Step 2B:Factor as the difference of two cubes and the difference of two squares. 3(x 3 − 1)(x 2 – 9) = 3(x − 1)(x 2 + x + 1)(x – 3)(x + 3) Step 3: The complete factorization is 3(x − 1)(x 2 + x + 1)(x – 3)(x + 3).

49. Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Equations by Factoring I (Quadratics) The Zero-Product Property Solving Quadratic Equations Applications 6.6

50. To solve equations we often use the zero-product property, which states that if the product of two numbers is 0, then at least one of the numbers must be 0.

51. Slide 51 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Applying the zero-product property Solve each equation. a.b. a. b.

52. Slide 52 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Any quadratic polynomial in the variable x can be written as ax 2 + bx + c with a ≠ 0. Any quadratic equation in the variable x can be written as ax 2 + bx + c = 0, with a ≠ 0. This form of quadratic equation is called the standard form of a quadratic equation.

53. Slide 53

54. Slide 54 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Solving equations by factoring a.36 + x 2 = –12x b. a. Solve each quadratic equation. Check your answers. x 2 − 49 = 0 36 + x 2 = –12x x 2 + 12x + 36 = 0 (x + 6)(x + 6) = 0 x = − 6 The only solution is −6. To check this value, substitute −6 for x in the given equation. 36 + (−6) 2 = –12(−6) 72 = 72

55. Slide 55 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution continued b.x 2 − 49 = 0 b.x 2 − 49 = 0 (x + 7)(x − 7) = 0 x = −7 x + 7 = 0x − 7 = 0 x = 7 To check these values, substitute 7 and −7 for x in the given equation. 7 2 − 49 = 0 0 = 0 (−7) 2 − 49 = 0 0 = 0 The solutions are −7 and 7. Solve each quadratic equation. Check your answers.

56. Slide 56 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Solving an equation by factoring Solve 2x 2 − 7x = −5 The solutions are (2x – 5)(x − 1) = 0 2x – 5 = 0or x – 1 = 0 or x = 1 2x 2 − 7x = −5 2x 2 − 7x + 5 = 0

57. Slide 57 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Modeling the flight of a model rocket If a model rocket is launched at 48 feet per second, then its height, h, after t seconds is h = 48t – 16t 2. After how long does the rocket strike the ground? The rocket strikes the ground when the height is 0. 48t – 16t 2 = 0 16t(3 – t) = 0 t = 0 16t = 03 – t = 0 t = 3 The rocket strikes the ground after 3 seconds.

58. Slide 58 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE Solution Finding the dimensions of a picture A frame surrounding a picture is 2 inches wide. The picture inside the frame is 7 inches longer than it is wide. If the overall area of the picture and frame is 198 square inches, find the dimensions of the picture inside the frame. Let x be the width of the picture and x + 7 be its length. 2 2 2 2 x + 7 x x + 11 x + 4

59. Slide 59 Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE continued (x + 4)(x + 11) =198 x 2 + 15x + 44 = 198 x 2 + 15x − 154 = 0 (x − 7)(x + 22) = 0 x − 7 = 0 or x + 22 = 0 x = 7 or x = −22 The only valid solution for x is 7 inches. Because the length is 7 inches more than the width, the dimensions are 7 inches and 14 inches.