1 / 40

Topic 6.6.2

Factoring Quadratics — ax ² + bx + c. Topic 6.6.2. Lesson 1.1.1. Topic 6.6.2. Factoring Quadratics — ax ² + bx + c. California Standard:

borna
Download Presentation

Topic 6.6.2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Factoring Quadratics — ax² + bx + c Topic 6.6.2

  2. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c California Standard: 11.0 Students apply basic factoring techniques to second-and simple third-degree polynomials. These techniques include finding a common factor for all terms in a polynomial,recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: You’ll learn how to factor more complicated quadratic expressions. • Key words: • quadratic • polynomial • factor • trial and error

  3. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c The method in Topic 6.6.1 for factoring a quadratic expression only works if the x2-term has a coefficient of 1. It’s a little more complicated when the x2-coefficient (a) isn’t 1 — but only a little.

  4. ax2 + abx + aca(x2 + bx + c) Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c You Can Take Out a Common Factor from Each Term If you see a common factor in each term, such as a number or a variable, take it out first.

  5. Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 1 Factor 3x2 + 15x + 18. Solution 3x2 + 15x + 18 = 3(x2 + 5x + 6) Take out the common factor The expression in parentheses can be factored using the method in Topic 6.6.1: = 3(x + 2)(x + 3) But — if the expression in parentheses still has a¹ 1, then the expression will need to be factored using the second method, shown in Example 2. Solution follows…

  6. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each expression completely. 1. 3x2 + 15x + 12 2. 4y2 – 12y – 112 3. 2t2 – 22t + 60 4. 3r2 – 75 5. 4x2 + 32x + 64 6. 7p2 + 70p + 63 7. 5m2 + 20m + 15 8. 6x2 + 42x + 60 9. –2k2 – 20k – 32 10. –3m2 – 30m – 72 3(x + 1)(x + 4) 4(y + 4)(y – 7) 2(t – 5)(t – 6) 3(r – 5)(r + 5) 4(x + 4)(x + 4) 7(p + 1)(p + 9) 5(m + 1)(m + 3) 6(x + 2)(x + 5) –2(k + 2)(k + 8) –3(m + 4)(m + 6) Solution follows…

  7. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each expression completely. 11. 2 + x – x212. –3x2 – 84x – 225 13. 3x2y – 33xy – 126y14. 10x2 + 290x + 1000 15. 100 + 75x – 25x216. 100n2y + 100ny – 5600y 17. 32a2 – 8x2a218. 21a – 80 – a2 19. 2x2m2 + 28x2m – 102x220. 3a2b2x2 + 30a2bx + 63a2 –1(x + 1)(x – 2) –3(x + 3)(x + 25) 3y(x + 3)(x – 14) 10(x + 4)(x + 25) –25(x – 4)(x + 1) 100y(n – 7)(n + 8) 8a2(2 – x)(2 + x) –1(a – 5)(a – 16) 2x2(m + 17)(m – 3) 3a2(bx + 3)(bx + 7) Solution follows…

  8. a = a1a2 a1c2 + a2c1 = b c = c1c2 Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c You Can Factor ax2 + bx + c by Trial and Error If you can’tsee a common factor, then you need to get ax2 + bx + c into the form (a1x + c1)(a2x + c2), where a1 and a2 are factors of a, and c1 and c2 are factors of c. ax2 + bx + c = (a1x + c1)(a2x + c2)

  9. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Note that if a number is positivethen its two factors will be either both positiveor both negative. If a number is negative, then its two factors will have different signs— one positiveand one negative. These facts give important clues about the signs of c1 and c2.

  10. Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 2 Factor 3x2 + 11x + 6. Solution 3x2 + 11x + 6 Write down pairs of factors of a = 3: a1 = 1and a2 = 3 Write down pairs of factors of c = 6: c1 = 1and c2 = 6,c1 = 2and c2 = 3 Now find the combination of factors that gives a1c2 + a2c1 = 11. Put the x-terms into parentheses first, with the pair of coefficients 3 and 1: (3x )(x ). Now try all the pairs of c1 and c2 in the parentheses and find the possible values of a1c2 + a2c1 (and a1c2 – a2c1)… Solution continues… Solution follows…

  11. Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 2 Factor 3x2 + 11x + 6. Solution (continued) (3x 1)(x 6) multiplies to give 18x and x, which add/subtract to give 19x or 17x. (3x 6)(x 1) multiplies to give 3x and 6x, which add/subtract to give 9x or 3x. (3x 2)(x 3) multiplies to give 9x and 2x, which add/subtract to give 11x or 7x. (3x 3)(x 2) multiplies to give 6x and 3x, which add/subtract to give 9x or 3x. So (3x 2)(x 3) is the combination that gives 11x (so b = 11). Solution continues…

  12. Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 2 Factor 3x2 + 11x + 6. Solution (continued) Now fill in the + / – signs. Both c1 and c2 are positive (since c = c1c2 and b = a1c2 + a2c1 are positive), so the final factors are (3x + 2)(x + 3). Check by expanding the parentheses to make sure they give the original equation: (3x + 2)(x + 3) = 3x2 + 9x + 2x + 6 = 3x2 + 11x + 6 That’s what you started with, so (3x + 2)(x + 3) is the correct factorization.

  13. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 21. 2x2 + 5x + 2 22. 2a2 + 13a + 11 23. 2y2 + 7y + 3 Possible combinations: (2x 1)(x 2) multiply to 4x and x, which add/subtract to 5xor 3x(2x 2)(x 1) multiply to 2x and 2x, which add/subtract to 4x or 0xSo 2x2 + 5x + 2 = (2x + 1)(x + 2) Possible combinations: (2a 1)(a 11) multiply to 22a and 1a, which add/subtract to 23a or21a (2a 11)(a 1) multiply to 2a and 11a, which add/subtract to 13aor 9aSo 2a2 + 13a + 11 = (2a + 11)(a + 1) Possible combinations: (2y 1)(y 3) multiply to 6y and y, which add/subtract to 7yor5y (2y 3)(y 1) multiply to 2y and 3y, which add/subtract to 5y or ySo 2y2 + 7y + 3 = (2y + 1)(y + 3) Solution follows…

  14. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 24. 4x2 + 28x + 49 25. 4x2 + 12x + 9 26. 6x2 + 23x + 7 27. 3x2 + 13x + 12 (2x 7)(2x 7) multiply to 14x and 14x, which add/subtract to 28xor 0xSo 4x2 + 28x + 49 = (2x + 7)(2x + 7) (2x 3)(2x 3) multiply to 6x and 6x, which add/subtract to 12xor 0x So 4x2 + 12x + 9 = (2x + 3)(2x + 3) (2x 7)(3x 1) multiply to 2x and 21x, which add/subtract to 23xor19xSo 6x2 + 23x + 7 = (2x + 7)(3x + 1) (3x 4)(x 3) multiply to 9x and 4x, which add/subtract to 13xor5xSo 3x2 + 13x + 12 = (3x + 4)(x + 3) Solution follows…

  15. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 28. 2x2 + 11x + 5 29. 4a2 + 16a + 7 30. 2p2 + 14p + 12 31. 8a2 + 46a + 11 (2x 1)(x 5) multiply to 10x and x, which add/subtract to 11xor 9xSo 2x2 + 11x + 5 = (2x + 1)(x + 5) (2a 1)(2a 7) multiply to 14a and 2a, which add/subtract to 16aor 12aSo 4a2 + 16a + 7 = (2a + 1)(2a + 7) Take out a common factor of 2: 2(p2 + 7p + 6), then factor the rest.(p 6)(p 1) multiply to p and 6p, which add/subtract to 7p or 5pSo 2p2 + 14p + 12 = 2(p + 6)(p + 1) (4a 1)(2a 11) multiply to 44a and 2a, which add/subtract to 46aor 42aSo 8a2 + 46a + 11 = (4a + 1)(2a + 11) Solution follows…

  16. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 32. 3g2 + 51g + 216 33. 9x2 + 12x + 4 34. 4a2 + 36a + 81 35. 4b2 + 32b + 55 Take out a common factor of 3: 3(g2 + 17g + 72), then factor the rest.(g 8)(g 9) multiply to 9g and 8g, which add/subtract to 17g or gSo 3g2 + 51p + 216 = 3(g + 8)(g + 9) (3x 2)(3x 2) multiply to 6x and 6x, which add/subtract to 12xor 0xSo 9x2 + 12x + 4 = (3x + 2)2 (2a 9)(2a 9) multiply to 18a and 18a, which add/subtract to 36aor 0aSo 4a2 + 36a + 81 = (2a + 9)2 (2b 5)(2b 11) multiply to 22b and 10b, which add/subtract to 32bor 12bSo 4b2 + 32b + 55 = (2b + 5)(2b + 11) Solution follows…

  17. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 36. 3x2 + 22x + 24 37. 10a2 + 23a + 12 38. 6t2 + 23t + 20 (3x 4)(x 6) multiply to 18x and 4x, which add/subtract to 22xor 14xSo 3x2 + 22x + 24 = (3x + 4)(x + 6) (2a 3)(5a 4) multiply to 8a and 15a, which add/subtract to 23aor 7aSo 10a2 + 23a + 12 = (2a + 3)(5a + 4) (2t 5)(3t 4) multiply to 8t and 15t, which add/subtract to 23tor 7tSo 6t2 + 23t + 20 = (2t + 5)(3t + 4) Solution follows…

  18. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 39. 4x2 + 34x + 16 40. 15b2 + 96b + 36 Take out a common factor of 2: 2(2x2 + 17x + 8), then factor the rest.(2x 1)(x 8) multiply to 16x and x, which add/subtract to 17x or 15xSo 4x2 + 34x + 16 = 2(2x + 1)(x + 8) Take out a common factor of 3: 3(5b2 + 32b + 12), then factor the rest.(5b 2)(b 6) multiply to 30b and 2b, which add/subtract to 32b or 28bSo 15b2 + 96b + 36 = 3(5b + 2)(b + 6) Solution follows…

  19. Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 3 Factor 6x2 + 5x – 6. Solution Write down pairs of factors of a = 6: a1 = 6and a2 = 1, a1 = 2and a2 = 3, Write down pairs of factors of c = –6 (ignoring the minus sign for now): c1 = 1and c2 = 6, c1 = 2and c2 = 3 Put the x-terms into parentheses first, with the first pair of possible values for a1 and a2, 6 and 1: (6x )(1x ). Now try all the pairs of c1 and c2 in the parentheses and find the possible values of a1c2 + a2c1 and a1c2 – a2c1 as before… Solution continues… Solution follows…

  20. Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 3 Factor 6x2 + 5x – 6. Solution (continued) (6x 1)(x 6) multiplies to give 36x and x, which add/subtract to give 37x or 35x. (6x 6)(x 1) multiplies to give 6x and 6x, which add/subtract to give 12x or 0x. (6x 2)(x 3) multiplies to give 18x and 2x, which add/subtract to give 20x or 16x. (6x 3)(x 2) multiplies to give 12x and 3x, which add/subtract to give 15x or 9x. None of these combinations works, so try again with (2x )(3x )… Solution continues…

  21. Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 3 Factor 6x2 + 5x – 6. Solution (continued) (2x 1)(3x 6) multiplies to give 12x and 3x, which add/subtract to give 15x or 9x. (2x 6)(3x 1) multiplies to give 2x and 18x, which add/subtract to give 20x or 16x. (2x 3)(3x 2) multiplies to give 4x and 9x, which add/subtract to give 13x or 5x. 5x is what you want, so you can stop there — so (2x 3)(3x 2) is the right combination. Solution continues…

  22. Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 3 Factor 6x2 + 5x – 6. Solution (continued) Now fill in the + / – signs to get b = +5. One of c1 and c2 must be negative, to give c= –6, so the final factors are either (2x + 3)(3x – 2) or (2x – 3)(3x + 2). The x-term of (2x + 3)(3x – 2) will be 9x – 4x = 5x, whereas the x-term for (2x – 3)(3x + 2) will be 4x – 9x = –5x. So the correct factorization is (2x + 3)(3x – 2).

  23. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 41. 2x2 + 3x – 2 42. 3y2 – y – 2 43. 5k2 + 13k – 6 (2x 1)(x 2) multiply to 4x and x, which add/subtract to 5x or 3x(2x 2)(x 1) multiply to 2x and 2x, which add/subtract to 4x or 0xSo 2x2 + 3x – 2 = (2x – 1)(x + 2) (3y 1)(y 2) multiply to 6y and y, which add/subtract to 7y or 5y(3y 2)(y 1) multiply to 3y and 2y, which add/subtract to 5y or ySo 3y2 – y – 2 = (3y + 2)(y – 1) (5k 2)(k 3) multiply to 15k and 2k, which add/subtract to 17k or 13k(5k 3)(k 2) multiply to 10k and 3k, which add/subtract to 13k or 7kSo 5k2 + 13k – 6 = (5k – 2)(k + 3) Solution follows…

  24. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 44. 3x2 – x – 10 45. 6b2 – 23b + 7 46. 2x2 – 5x + 2 (3x 5)(x 2) multiply to 6x and 5x, which add/subtract to 11x or x(3x 2)(x 5) multiply to 15x and 2x, which add/subtract to 17x or 13xSo 3x2 – x – 10 = (3x + 5)(x – 2) (3b 1)(2b 7) multiply to 21b and 2b, which add/subtract to 23bor 19b(3b 7)(2b 1) multiply to 3b and 14b, which add/subtract to 17b or 11bSo 6b2 – 23b + 7 = (3b – 1)(2b – 7) (2x 2)(x 1) multiply to 2x and 2x, which add/subtract to 4x or 0x(2x 1)(x 2) multiply to 4x and x, which add/subtract to 5xor 3xSo 2x2 – 5x + 2 = (2x – 1)(x – 2) Solution follows…

  25. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 47. 3k2 – 2k – 1 48. 3v2 – 16v + 5 49. 18 + 5x – 2x2 (3k 1)(k 1) multiply to 3k and k, which add/subtract to 4k or 2kSo 3k2 – 2k – 1 = (3k + 1)(k – 1) (3v 5)(v 1) multiply to 3v and 5v, which add/subtract to 8v or 2v(3v 1)(v 5) multiply to 15v and v, which add/subtract to 16vor 10vSo 3v2 – 16v + 5 = (3v – 1)(v – 5) Take out a factor of –1: –(2x2 – 5x – 18), then factor the rest.(2x 9)(x 2) multiply to 4x and 9x, which add/subtract to 13x or 5x(2x 2)(x 9) multiply to 18x and 2x, which add/subtract to 20x or 16xSo 18 + 5x – 2x2 = –(2x – 9)(x + 2) Solution follows…

  26. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 50. 28 + x – 2x2 51. 9x2 + 12x + 4 52. 7a2 – 26a – 8 Take out a factor of –1: –(2x2 – x – 28), then factor the rest.(2x 7)(x 4) multiply to 8x and 7x, which add/subtract to 15x or x(2x 4)(x 7) multiply to 14x and 4x, which add/subtract to 18x or 10xSo 28 + x – 2x2 = –(2x + 7)(x – 4) (3x 2)(3x 2) multiply to 6x and 6x, which add/subtract to 12x or 0xSo 9x2 + 12x + 4 = (3x + 2)(3x + 2) (7a 2)(a 4) multiply to 28a and 2a, which add/subtract to 30a or 26a(7a 4)(a 2) multiply to 14a and 4a, which add/subtract to 18a or 10aSo 7a2 – 26a – 8 = (7a + 2)(a – 4) Solution follows…

  27. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 53. 3x2 – 7x – 6 54. 12x2 + 5x – 2 55. 6x2 + 2x – 20 (3x 3)(x 2) multiply to 6x and 3x, which add/subtract to 9x or 3x(3x 2)(x 3) multiply to 9x and 2x, which add/subtract to 11x or 7xSo 3x2 – 7x – 6 = (3x + 2)(x – 3) (4x 1)(3x 2) multiply to 8x and 3x, which add/subtract to 11x or 5x(4x 2)(3x 1) multiply to 4x and 6x, which add/subtract to 10x or 2xSo 12x2 + 5x – 2 = (4x – 1)(3x + 2) Take out a factor of 2: 2(3x2 + x – 10), then factor the rest.(3x 2)(x 5) multiply to 15x and 2x, which add/subtract to 17x or 13x(3x 5)(x 2) multiply to 6x and 5x, which add/subtract to 11x or xSo 6x2 + 2x – 20 = 2(3x – 5)(x + 2) Solution follows…

  28. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 56. 18x2 + x – 4 57. 6y2 + y – 12 58. 9m2 – 3m – 20 (2x 1)(9x 4) multiply to 8x and 9x, which add/subtract to 17x or x(2x 4)(9x 1) multiply to 2x and 36x, which add/subtract to 38x or 34xSo 18x2 + x – 4 = (2x + 1)(9x – 4) (3y 3)(2y 4) multiply to 12y and 6y, which add/subtract to 18y or 6y(3y 4)(2y 3) multiply to 9y and 8y, which add/subtract to 17y or ySo 6y2 + y – 12 = (3y – 4)(2y + 3) (3m 4)(3m 5) multiply to 15m and 12m, which add/subtract to 27m or 3mSo 9m2 – 3m – 20 = (3m + 4)(3m – 5) Solution follows…

  29. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Use the + and – Signs in the Quadratic to Work Faster Looking carefully at the signs in the quadratic that you are factoring can help to narrow down the choices for a1, a2, c1, and c2. The next three Examples show you how.

  30. Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 4 Factor 3x2 + 11x + 6. Solution Here c = 6, which is positive — so its factors c1 and c2 are either both positive or both negative. But since b = 11 is positive, you can tell that c1 and c2 must be positive (so that a1c2 + a2c1 is positive). Solution follows…

  31. Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 5 Factor 3x2 – 11x + 6. Solution Here c is also positive, so c1 and c2 are either both positive or both negative. But since b = –11 is negative, you can tell that c1 and c2 must be negative (so that a1c2 + a2c1 is negative). Solution follows…

  32. Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 6 Factor 6x2 + 5x – 6. Solution In this expression, c is negative, so one of c1 and c2 must be positive and the other must be negative. So instead of looking at both the sums and differences of all the different combinations a1c2 and a2c1, you only need to look at the differences. Solution follows…

  33. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each expression. 59. 2n2 + n – 3 60. 2x2 – 5x – 3 61. 4a2 + 4a + 1 c is negative, so one of c1 and c2 must be positive and the other negative(2n 3)(n 1) multiply to 2n and 3n. To add to get n, 3n must be positive and 2n negative. So 2n2 + n – 3 = (2n + 3)(n – 1) c is negative, so one of c1 and c2 must be positive and the other negative(2x 1)(x 3) multiply to 6x and x. To add to get –5x, x must be positive and 6x negative. So 2x2 – 5x – 3 = (2x + 1)(x – 3) b and c are positive, so c1 and c2 must both be positive.(4a 1)(a 1) multiply to 4a and a and add to 5a(2a 1)(2a 1) multiply to 2a and 2a and add to 4aSo 4a2 + 4a + 1 = (2a + 1)(2a + 1) Solution follows…

  34. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each expression. 62. 3x2 – 4x + 1 63. 9y2 + 6y + 1 64. 4t2 + t – 3 b is negative and c is positive, so c1 and c2 must both be negative.(3x 1)(x 1) multiply to 3x and x and add to 4xSo 3x2 – 4x + 1 = (3x – 1)(x – 1) b and c are positive, so c1 and c2 must both be positive.(3y 1)(3y 1) multiply to 3y and 3y and add to 6ySo 9y2 + 6y + 1 = (3y + 1)2 c is negative, so one of c1 and c2 must be positive and the other negative(4t 3)(t 1) multiply to 4t and 3t. To add to get t, 4t must be positive and 3t negative. So 4t2 + t – 3 = (4t – 3)(t + 1) Solution follows…

  35. Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each expression. 65. 5x2 – x – 18 66. 9x2 – 6x + 1 67. 6t2 + t – 1 68. b2 + 10b + 21 c is negative, so one of c1 and c2 must be positive and the other negative(5x 9)(x 2) multiply to 10x and 9x. To add to get –x, 9x must be positive and 10x negative. So 5x2 – x – 18 = (5x + 9)(x – 2) b is negative and c is positive, so c1 and c2 must both be negative.(3x 1)(3x 1) multiply to 3x and 3x and add to 6x. So 9x2 – 6x + 1 = (3x – 1)2 c is negative, so one of c1 and c2 must be positive and the other negative(3t 1)(2t 1) multiply to 3t and 2t. To add to get t, 3t must be positive and 2t negative. So 6t2 + t – 1 = (3t – 1)(2t + 1) b and c are positive, so c1 and c2 must both be positive. (b 7)(b 3) multiply to 3b and 7b and add to 10b.So 9y2 + 6y + 1 = (3y + 1)2 Solution follows…

  36. Topic 6.6.2 Factoring Quadratics — ax² + bx + c Independent Practice Factor each polynomial. 1. 5k2 – 7k + 2 2. 4k2 – 15k + 9 3. 12t2 – 11t + 2 4. 9 – 7k – 2k2 5. 10 + h – 3h26. 15x2 – 14x – 8 7. 6 + 5x – 6x28. 3x4 + 8x2 – 3 (5k – 2)(k – 1) (4k – 3)(k – 3) (4t – 1)(3t – 2) –(2k + 9)(k – 1) –(3h + 5)(h – 2) (5x + 2)(3x – 4) –(3x + 2)(2x – 3) (3x2 – 1)(x2 + 3) Solution follows…

  37. Topic 6.6.2 Factoring Quadratics — ax² + bx + c Independent Practice 9. If the area of a rectangle is (6x2 + 25x + 14) square units and the length is (3x + 2) units, find the width w in terms of x. 10. The area of a parallelogram is (12x2 + 7x – 10) cm2, where x is positive. If the area is given by the formula Area = base × height, find the base length and the height of the parallelogram, given that they are both linear factors of the area. (2x + 7) units (4x + 5) cm, (3x – 2) cm Solution follows…

  38. Topic 6.6.2 Factoring Quadratics — ax² + bx + c Independent Practice Factor each of these expressions completely. 11. x2ab – 3abx – 18ab 12. (d + 2)x2 – 7x(d + 2) – 18(d + 2) 13. (x + 1)m2 – 2m(x + 1) + (x + 1) 14. –2dk2 – 14dk+ 36d ab(x + 3)(x – 6) (d + 2)(x – 9)(x + 2) (x + 1)(m – 1)2 –2d(k + 9)(k– 2) Solution follows…

  39. Topic 6.6.2 Factoring Quadratics — ax² + bx + c Independent Practice 15. Five identical rectangular floor tiles have a total area of (15x2 + 10x – 40) m2. Find the dimensions of each floor tile, if the length of each side can be written in the form ax + b, where aand b are integers. (3x– 4) m by (x + 2) m. Solution follows…

  40. Topic 6.6.2 Factoring Quadratics — ax² + bx + c Round Up Now you can factor lots of different types of polynomials. In the next Section you’ll learn about another type — quadratic expressions containing two different variables.

More Related