Your SlideShare is downloading.
×

×
Saving this for later?
Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.

Text the download link to your phone

Standard text messaging rates apply

Like this presentation? Why not share!

- 1 5 the least common multiple by math123b 680 views
- 1 6 addition and subtraction i by math123b 835 views
- 2 1 addition and subtraction ii by math123b 587 views
- 1.3 sign charts and inequalities by math123c 917 views
- 2 5 complex fractions by math123b 1167 views
- 4 4 more on algebra of radicals by math123b 378 views
- 2 3 solving rational equations by math123b 798 views
- 2 6 summary of usage of lcm by math123b 537 views
- 1 4 multiplication and division of ... by math123b 1789 views
- 1.4 the basic language of functions by math123c 851 views
- 4 3 algebra of radicals by math123b 1292 views
- 5 1 radical equations by math123b 541 views

Like this? Share it with your network
Share

928

views

views

Published on

No Downloads

Total Views

928

On Slideshare

0

From Embeds

0

Number of Embeds

0

Shares

0

Downloads

0

Comments

0

Likes

1

No embeds

No notes for slide

- 1. Review on Factoring Frank Ma © 2011
- 2. To factor means to rewrite an expression as a product in a nontrivial way. Review on Factoring
- 3. To factor means to rewrite an expression as a product in a nontrivial way. Following are the steps to factor an expression. Review on Factoring
- 4. To factor means to rewrite an expression as a product in a nontrivial way. Following are the steps to factor an expression. Review on Factoring I. Always pull out the greatest common factor first.
- 5. To factor means to rewrite an expression as a product in a nontrivial way. Following are the steps to factor an expression. Review on Factoring I. Always pull out the greatest common factor first. II. If the expression is a trinomial (three-term) ax2 + bx + c, use the reverse-FOIL method or the ac-method.
- 6. ± ± – + To factor means to rewrite an expression as a product in a nontrivial way. Following are the steps to factor an expression. Review on Factoring I. Always pull out the greatest common factor first. II. If the expression is a trinomial (three-term) ax2 + bx + c, use the reverse-FOIL method or the ac-method. III. Use the following factoring formulas if possible x2 – y2 = (x + y)(x – y). x3 y3 = ( x y )( x2 xy + y2 ).
- 7. ± ± – + To factor means to rewrite an expression as a product in a nontrivial way. Following are the steps to factor an expression. Review on Factoring Pulling out GCF I. Always pull out the greatest common factor first. II. If the expression is a trinomial (three-term) ax2 + bx + c, use the reverse-FOIL method or the ac-method. III. Use the following factoring formulas if possible x2 – y2 = (x + y)(x – y). x3 y3 = ( x y )( x2 xy + y2 ).
- 8. ± ± – + To factor means to rewrite an expression as a product in a nontrivial way. Following are the steps to factor an expression. Review on Factoring A common factor of two or more quantities is a factor that belongs to all the quantities. Pulling out GCF I. Always pull out the greatest common factor first. II. If the expression is a trinomial (three-term) ax2 + bx + c, use the reverse-FOIL method or the ac-method. III. Use the following factoring formulas if possible x2 – y2 = (x + y)(x – y). x3 y3 = ( x y )( x2 xy + y2 ).
- 9. ± ± – + To factor means to rewrite an expression as a product in a nontrivial way. Following are the steps to factor an expression. Review on Factoring Example A. a. Since 6 = (2)(3) and 15 = (3)(5), A common factor of two or more quantities is a factor that belongs to all the quantities. Pulling out GCF I. Always pull out the greatest common factor first. II. If the expression is a trinomial (three-term) ax2 + bx + c, use the reverse-FOIL method or the ac-method. III. Use the following factoring formulas if possible x2 – y2 = (x + y)(x – y). x3 y3 = ( x y )( x2 xy + y2 ).
- 10. ± ± – + To factor means to rewrite an expression as a product in a nontrivial way. Following are the steps to factor an expression. Review on Factoring Example A. a. Since 6 = (2)(3) and 15 = (3)(5), so 3 is a common factor. A common factor of two or more quantities is a factor that belongs to all the quantities. Pulling out GCF I. Always pull out the greatest common factor first. II. If the expression is a trinomial (three-term) ax2 + bx + c, use the reverse-FOIL method or the ac-method. III. Use the following factoring formulas if possible x2 – y2 = (x + y)(x – y). x3 y3 = ( x y )( x2 xy + y2 ).
- 11. ± ± – + To factor means to rewrite an expression as a product in a nontrivial way. Following are the steps to factor an expression. Review on Factoring Example A. a. Since 6 = (2)(3) and 15 = (3)(5), so 3 is a common factor. b. The common factors of 4ab, 6a are 2, a, 2a. A common factor of two or more quantities is a factor that belongs to all the quantities. Pulling out GCF I. Always pull out the greatest common factor first. II. If the expression is a trinomial (three-term) ax2 + bx + c, use the reverse-FOIL method or the ac-method. III. Use the following factoring formulas if possible x2 – y2 = (x + y)(x – y). x3 y3 = ( x y )( x2 xy + y2 ).
- 12. ± ± – + To factor means to rewrite an expression as a product in a nontrivial way. Following are the steps to factor an expression. Review on Factoring Example A. a. Since 6 = (2)(3) and 15 = (3)(5), so 3 is a common factor. b. The common factors of 4ab, 6a are 2, a, 2a. c. The common factors of 6xy2, 15x2y2 are 3, x, y2, xy2, .. A common factor of two or more quantities is a factor that belongs to all the quantities. Pulling out GCF I. Always pull out the greatest common factor first. II. If the expression is a trinomial (three-term) ax2 + bx + c, use the reverse-FOIL method or the ac-method. III. Use the following factoring formulas if possible x2 – y2 = (x + y)(x – y). x3 y3 = ( x y )( x2 xy + y2 ).
- 13. ± ± – + To factor means to rewrite an expression as a product in a nontrivial way. Following are the steps to factor an expression. Review on Factoring Example A. a. Since 6 = (2)(3) and 15 = (3)(5), so 3 is a common factor. b. The common factors of 4ab, 6a are 2, a, 2a. c. The common factors of 6xy2, 15x2y2 are 3, x, y2, xy2, .. d. The common factor of a(x+y), b(x+y) is (x+y). A common factor of two or more quantities is a factor that belongs to all the quantities. Pulling out GCF I. Always pull out the greatest common factor first. II. If the expression is a trinomial (three-term) ax2 + bx + c, use the reverse-FOIL method or the ac-method. III. Use the following factoring formulas if possible x2 – y2 = (x + y)(x – y). x3 y3 = ( x y )( x2 xy + y2 ).
- 14. The greatest common factor (GCF) is the common factor that has the largest coefficient and highest degree of each factor among all common factors. Review on Factoring
- 15. The greatest common factor (GCF) is the common factor that has the largest coefficient and highest degree of each factor among all common factors. Example B. Find the GCF of the given quantities. a. GCF{24, 36} Review on Factoring
- 16. The greatest common factor (GCF) is the common factor that has the largest coefficient and highest degree of each factor among all common factors. Example B. Find the GCF of the given quantities. a. GCF{24, 36} = 12. Review on Factoring
- 17. The greatest common factor (GCF) is the common factor that has the largest coefficient and highest degree of each factor among all common factors. Example B. Find the GCF of the given quantities. a. GCF{24, 36} = 12. b. GCF{4ab, 6a} Review on Factoring
- 18. The greatest common factor (GCF) is the common factor that has the largest coefficient and highest degree of each factor among all common factors. Example B. Find the GCF of the given quantities. a. GCF{24, 36} = 12. b. GCF{4ab, 6a} = 2a. Review on Factoring
- 19. The greatest common factor (GCF) is the common factor that has the largest coefficient and highest degree of each factor among all common factors. Example B. Find the GCF of the given quantities. a. GCF{24, 36} = 12. b. GCF{4ab, 6a} = 2a. c. GCF {6xy2, 15 x2y2} Review on Factoring
- 20. The greatest common factor (GCF) is the common factor that has the largest coefficient and highest degree of each factor among all common factors. Example B. Find the GCF of the given quantities. a. GCF{24, 36} = 12. b. GCF{4ab, 6a} = 2a. c. GCF {6xy2, 15 x2y2} = 3xy2. Review on Factoring
- 21. The greatest common factor (GCF) is the common factor that has the largest coefficient and highest degree of each factor among all common factors. Example B. Find the GCF of the given quantities. a. GCF{24, 36} = 12. b. GCF{4ab, 6a} = 2a. c. GCF {6xy2, 15 x2y2} = 3xy2. d. GCF{x3y5, x4y6, x5y4} = Review on Factoring
- 22. The greatest common factor (GCF) is the common factor that has the largest coefficient and highest degree of each factor among all common factors. Example B. Find the GCF of the given quantities. a. GCF{24, 36} = 12. b. GCF{4ab, 6a} = 2a. c. GCF {6xy2, 15 x2y2} = 3xy2. d. GCF{x3y5, x4y6, x5y4} = x3y4. Review on Factoring
- 23. The greatest common factor (GCF) is the common factor that has the largest coefficient and highest degree of each factor among all common factors. Example B. Find the GCF of the given quantities. a. GCF{24, 36} = 12. b. GCF{4ab, 6a} = 2a. c. GCF {6xy2, 15 x2y2} = 3xy2. d. GCF{x3y5, x4y6, x5y4} = x3y4. The Extraction Law Distributive law interpreted backwards gives the Extraction Law, that is, common factors may be extracted from sums or differences. Review on Factoring
- 24. The greatest common factor (GCF) is the common factor that has the largest coefficient and highest degree of each factor among all common factors. Example B. Find the GCF of the given quantities. a. GCF{24, 36} = 12. b. GCF{4ab, 6a} = 2a. c. GCF {6xy2, 15 x2y2} = 3xy2. d. GCF{x3y5, x4y6, x5y4} = x3y4. The Extraction Law Distributive law interpreted backwards gives the Extraction Law, that is, common factors may be extracted from sums or differences. AB ± AC A(B±C) Review on Factoring
- 25. The greatest common factor (GCF) is the common factor that has the largest coefficient and highest degree of each factor among all common factors. Example B. Find the GCF of the given quantities. a. GCF{24, 36} = 12. b. GCF{4ab, 6a} = 2a. c. GCF {6xy2, 15 x2y2} = 3xy2. d. GCF{x3y5, x4y6, x5y4} = x3y4. The Extraction Law Distributive law interpreted backwards gives the Extraction Law, that is, common factors may be extracted from sums or differences. AB ± AC A(B±C) This procedure is also called “factoring out common factor”. To factor, the first step always is to factor out the GCF. Review on Factoring
- 26. Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 Review on Factoring
- 27. (the GCF is 6x2y2) Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 Review on Factoring
- 28. (the GCF is 6x2y2) Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) Review on Factoring
- 29. (the GCF is 6x2y2) Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1) Review on Factoring
- 30. b. (2x – 3)3x – 2(2x – 3) (the GCF is 6x2y2) Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1) Review on Factoring
- 31. b. (2x – 3)3x – 2(2x – 3) Pull out the common factor (2x – 3), (2x – 3)3x – 2(2x – 3) (the GCF is 6x2y2) Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1) Review on Factoring
- 32. b. (2x – 3)3x – 2(2x – 3) Pull out the common factor (2x – 3), (2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2) (the GCF is 6x2y2) Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1) Review on Factoring
- 33. b. (2x – 3)3x – 2(2x – 3) Pull out the common factor (2x – 3), (2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2) (the GCF is 6x2y2) Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1) c. 3x – 3y + ax – ay Review on Factoring
- 34. b. (2x – 3)3x – 2(2x – 3) Pull out the common factor (2x – 3), (2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2) (the GCF is 6x2y2) Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1) c. 3x – 3y + ax – ay Group them into two groups. = (3x – 3y) + (ax – ay) Review on Factoring
- 35. b. (2x – 3)3x – 2(2x – 3) Pull out the common factor (2x – 3), (2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2) (the GCF is 6x2y2) Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1) c. 3x – 3y + ax – ay Group them into two groups. = (3x – 3y) + (ax – ay) Factor out the GCF of each group. = 3(x – y) + a(x – y) Review on Factoring
- 36. b. (2x – 3)3x – 2(2x – 3) Pull out the common factor (2x – 3), (2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2) (the GCF is 6x2y2) Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1) c. 3x – 3y + ax – ay Group them into two groups. = (3x – 3y) + (ax – ay) Factor out the GCF of each group. = 3(x – y) + a(x – y) Pull the factor (x – y) again. = (3 + a)(x – y) Review on Factoring
- 37. b. (2x – 3)3x – 2(2x – 3) Pull out the common factor (2x – 3), (2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2) (the GCF is 6x2y2) Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1) Reverse-FOIL (Guessing Method) c. 3x – 3y + ax – ay Group them into two groups. = (3x – 3y) + (ax – ay) Factor out the GCF of each group. = 3(x – y) + a(x – y) Pull the factor (x – y) again. = (3 + a)(x – y) Review on Factoring
- 38. b. (2x – 3)3x – 2(2x – 3) Pull out the common factor (2x – 3), (2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2) (the GCF is 6x2y2) Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1) Reverse-FOIL (Guessing Method) To factor a trinomial (three-term) of the form ax2 + bx + c, try an answer of the form (#x ± #) (#x ± #). c. 3x – 3y + ax – ay Group them into two groups. = (3x – 3y) + (ax – ay) Factor out the GCF of each group. = 3(x – y) + a(x – y) Pull the factor (x – y) again. = (3 + a)(x – y) Review on Factoring
- 39. b. (2x – 3)3x – 2(2x – 3) Pull out the common factor (2x – 3), (2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2) (the GCF is 6x2y2) Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1) Reverse-FOIL (Guessing Method) To factor a trinomial (three-term) of the form ax2 + bx + c, try an answer of the form (#x ± #) (#x ± #). To find the correct answer, we need to have(#)(#) = a, c. 3x – 3y + ax – ay Group them into two groups. = (3x – 3y) + (ax – ay) Factor out the GCF of each group. = 3(x – y) + a(x – y) Pull the factor (x – y) again. = (3 + a)(x – y) Review on Factoring
- 40. b. (2x – 3)3x – 2(2x – 3) Pull out the common factor (2x – 3), (2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2) (the GCF is 6x2y2) Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1) Reverse-FOIL (Guessing Method) To factor a trinomial (three-term) of the form ax2 + bx + c, try an answer of the form (#x ± #) (#x ± #). To find the correct answer, we need to have (#)(#) = a, (#)(#) = c c. 3x – 3y + ax – ay Group them into two groups. = (3x – 3y) + (ax – ay) Factor out the GCF of each group. = 3(x – y) + a(x – y) Pull the factor (x – y) again. = (3 + a)(x – y) Review on Factoring
- 41. b. (2x – 3)3x – 2(2x – 3) Pull out the common factor (2x – 3), (2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2) (the GCF is 6x2y2) Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1) Reverse-FOIL (Guessing Method) To factor a trinomial (three-term) of the form ax2 + bx + c, try an answer of the form (#x ± #) (#x ± #). To find the correct answer, we need to have (#)(#) = a, (#)(#) = c and that the inner and outer products combine to be bx. c. 3x – 3y + ax – ay Group them into two groups. = (3x – 3y) + (ax – ay) Factor out the GCF of each group. = 3(x – y) + a(x – y) Pull the factor (x – y) again. = (3 + a)(x – y) Review on Factoring
- 42. b. (2x – 3)3x – 2(2x – 3) Pull out the common factor (2x – 3), (2x – 3)3x – 2(2x – 3) = (2x – 3)(3x – 2) (the GCF is 6x2y2) Example C. Factor out the GCF. a. 12x2y3 + 6x2y2 = 6x2y2(2y) + 6x2y2(1) = 6x2y2(2y + 1) Reverse-FOIL (Guessing Method) To factor a trinomial (three-term) of the form ax2 + bx + c, try an answer of the form (#x ± #) (#x ± #). To find the correct answer, we need to have (#)(#) = a, (#)(#) = c and that the inner and outer products combine to be bx. That is (#x ± #) (#x ± #). combine to be bx c. 3x – 3y + ax – ay Group them into two groups. = (3x – 3y) + (ax – ay) Factor out the GCF of each group. = 3(x – y) + a(x – y) Pull the factor (x – y) again. = (3 + a)(x – y) Review on Factoring
- 43. Review on Factoring Example D. Factor x2 – 4x – 12 We are to fill in ( x ± ) ( x ± ).
- 44. Review on Factoring Example D. Factor x2 – 4x – 12 We are to fill in ( x ± ) ( x ± ). It must be (1x ± ) (1x ± ) to get the 1x2.
- 45. Review on Factoring Example D. Factor x2 – 4x – 12 We are to fill in ( x ± ) ( x ± ). It must be (1x ± ) (1x ± ) to get the 1x2. Furthermore, it must be (x + ) (x – ) to get the –12. Let's factor
- 46. Review on Factoring Example D. Factor x2 – 4x – 12 We are to fill in ( x ± ) ( x ± ). It must be (1x ± ) (1x ± ) to get the 1x2. Furthermore, it must be (x + ) (x – ) to get the –12. Let's factor 12 = (1)(12) = (2)(6) = (3)(4)
- 47. Review on Factoring Example D. Factor x2 – 4x – 12 We are to fill in ( x ± ) ( x ± ). It must be (1x ± ) (1x ± ) to get the 1x2. Furthermore, it must be (x + ) (x – ) to get the –12. Let's factor 12 = (1)(12) = (2)(6) = (3)(4) and the difference between 2 and 6 is ±4
- 48. Review on Factoring Example D. Factor x2 – 4x – 12 We are to fill in ( x ± ) ( x ± ). It must be (1x ± ) (1x ± ) to get the 1x2. Furthermore, it must be (x + ) (x – ) to get the –12. Let's factor 12 = (1)(12) = (2)(6) = (3)(4) and the difference between 2 and 6 is ±4 which matches the coefficient of the middle term –4x,
- 49. Review on Factoring Example D. Factor x2 – 4x – 12 We are to fill in ( x ± ) ( x ± ). It must be (1x ± ) (1x ± ) to get the 1x2. Furthermore, it must be (x + ) (x – ) to get the –12. Let's factor 12 = (1)(12) = (2)(6) = (3)(4) and the difference between 2 and 6 is ±4 which matches the coefficient of the middle term –4x, we see that x2 – 4x – 12 = (x + 2)(x – 6).
- 50. Review on Factoring Example D. Factor x2 – 4x – 12 We are to fill in ( x ± ) ( x ± ). It must be (1x ± ) (1x ± ) to get the 1x2. Furthermore, it must be (x + ) (x – ) to get the –12. Let's factor 12 = (1)(12) = (2)(6) = (3)(4) and the difference between 2 and 6 is ±4 which matches the coefficient of the middle term –4x, we see that x2 – 4x – 12 = (x + 2)(x – 6). If the number term c in the trinomial is a prime number, start by setting the number terms as 1 and c, i.e. ax2 + bx + c = (#x ± c)(#x ± 1)
- 51. Review on Factoring Example D. Factor x2 – 4x – 12 We are to fill in ( x ± ) ( x ± ). It must be (1x ± ) (1x ± ) to get the 1x2. Furthermore, it must be (x + ) (x – ) to get the –12. Let's factor 12 = (1)(12) = (2)(6) = (3)(4) and the difference between 2 and 6 is ±4 which matches the coefficient of the middle term –4x, we see that x2 – 4x – 12 = (x + 2)(x – 6). Example E. Factor 20x2 – 4x – 3 If the number term c in the trinomial is a prime number, start by setting the number terms as 1 and c, i.e. ax2 + bx + c = (#x ± c)(#x ± 1)
- 52. Review on Factoring Example D. Factor x2 – 4x – 12 We are to fill in ( x ± ) ( x ± ). It must be (1x ± ) (1x ± ) to get the 1x2. Furthermore, it must be (x + ) (x – ) to get the –12. Let's factor 12 = (1)(12) = (2)(6) = (3)(4) and the difference between 2 and 6 is ±4 which matches the coefficient of the middle term –4x, we see that x2 – 4x – 12 = (x + 2)(x – 6). Example E. Factor 20x2 – 4x – 3 Since 3 = (3)(1), it must be ( ± 3 ) ( ± 1 ) to get the –3. If the number term c in the trinomial is a prime number, start by setting the number terms as 1 and c, i.e. ax2 + bx + c = (#x ± c)(#x ± 1)
- 53. Review on Factoring Example D. Factor x2 – 4x – 12 We are to fill in ( x ± ) ( x ± ). It must be (1x ± ) (1x ± ) to get the 1x2. Furthermore, it must be (x + ) (x – ) to get the –12. Let's factor 12 = (1)(12) = (2)(6) = (3)(4) and the difference between 2 and 6 is ±4 which matches the coefficient of the middle term –4x, we see that x2 – 4x – 12 = (x + 2)(x – 6). Example E. Factor 20x2 – 4x – 3 Since 3 = (3)(1), it must be ( ± 3 ) ( ± 1 ) to get the –3. 20 = (1)(20) = (2)(10) = (4)(5), If the number term c in the trinomial is a prime number, start by setting the number terms as 1 and c, i.e. ax2 + bx + c = (#x ± c)(#x ± 1)
- 54. Review on Factoring Example D. Factor x2 – 4x – 12 We are to fill in ( x ± ) ( x ± ). It must be (1x ± ) (1x ± ) to get the 1x2. Furthermore, it must be (x + ) (x – ) to get the –12. Let's factor 12 = (1)(12) = (2)(6) = (3)(4) and the difference between 2 and 6 is ±4 which matches the coefficient of the middle term –4x, we see that x2 – 4x – 12 = (x + 2)(x – 6). Example E. Factor 20x2 – 4x – 3 Since 3 = (3)(1), it must be ( ± 3 ) ( ± 1 ) to get the –3. 20 = (1)(20) = (2)(10) = (4)(5), observe that the difference between (2)(3) and (10)(1) is ±4 If the number term c in the trinomial is a prime number, start by setting the number terms as 1 and c, i.e. ax2 + bx + c = (#x ± c)(#x ± 1)
- 55. Review on Factoring Example D. Factor x2 – 4x – 12 We are to fill in ( x ± ) ( x ± ). It must be (1x ± ) (1x ± ) to get the 1x2. Furthermore, it must be (x + ) (x – ) to get the –12. Let's factor 12 = (1)(12) = (2)(6) = (3)(4) and the difference between 2 and 6 is ±4 which matches the coefficient of the middle term –4x, we see that x2 – 4x – 12 = (x + 2)(x – 6). Example E. Factor 20x2 – 4x – 3 Since 3 = (3)(1), it must be ( ± 3 ) ( ± 1 ) to get the –3. 20 = (1)(20) = (2)(10) = (4)(5), observe that the difference between (2)(3) and (10)(1) is ±4 which matches the coefficient of the middle term –4x. If the number term c in the trinomial is a prime number, start by setting the number terms as 1 and c, i.e. ax2 + bx + c = (#x ± c)(#x ± 1)
- 56. Review on Factoring Example D. Factor x2 – 4x – 12 We are to fill in ( x ± ) ( x ± ). It must be (1x ± ) (1x ± ) to get the 1x2. Furthermore, it must be (x + ) (x – ) to get the –12. Let's factor 12 = (1)(12) = (2)(6) = (3)(4) and the difference between 2 and 6 is ±4 which matches the coefficient of the middle term –4x, we see that x2 – 4x – 12 = (x + 2)(x – 6). Example E. Factor 20x2 – 4x – 3 Since 3 = (3)(1), it must be ( ± 3 ) ( ± 1 ) to get the –3. 20 = (1)(20) = (2)(10) = (4)(5), observe that the difference between (2)(3) and (10)(1) is ±4 which matches the coefficient of the middle term –4x. Use 2x and 10x for the factors, we get 20x2 – 4x – 3 = (10x + 3)(2x – 1 ). If the number term c in the trinomial is a prime number, start by setting the number terms as 1 and c, i.e. ax2 + bx + c = (#x ± c)(#x ± 1)
- 57. The ac-Method Review on Factoring
- 58. The ac-Method The product of two binomials has four terms. Review on Factoring
- 59. The ac-Method The product of two binomials has four terms. We may use grouping-method to work backwards to obtain the two binomials. Review on Factoring
- 60. The ac-Method The product of two binomials has four terms. We may use grouping-method to work backwards to obtain the two binomials. Example F. Factor x2 – x – 6 by grouping. Review on Factoring
- 61. The ac-Method The product of two binomials has four terms. We may use grouping-method to work backwards to obtain the two binomials. Example F. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Review on Factoring
- 62. The ac-Method The product of two binomials has four terms. We may use grouping-method to work backwards to obtain the two binomials. Example F. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Review on Factoring
- 63. The ac-Method The product of two binomials has four terms. We may use grouping-method to work backwards to obtain the two binomials. Example F. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Pull out the common factor = x(x – 3) + 2(x – 3) Review on Factoring
- 64. The ac-Method The product of two binomials has four terms. We may use grouping-method to work backwards to obtain the two binomials. Example F. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Pull out the common factor = x(x – 3) + 2(x – 3) Pull out the common (x – 3) = (x – 3)(x + 2) Review on Factoring
- 65. The ac-Method The product of two binomials has four terms. We may use grouping-method to work backwards to obtain the two binomials. But how did we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in the first place? Example F. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Pull out the common factor = x(x – 3) + 2(x – 3) Pull out the common (x – 3) = (x – 3)(x + 2) Review on Factoring
- 66. The ac-Method The product of two binomials has four terms. We may use grouping-method to work backwards to obtain the two binomials. But how did we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in the first place? This leads to the ac-method. Example F. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Pull out the common factor = x(x – 3) + 2(x – 3) Pull out the common (x – 3) = (x – 3)(x + 2) Review on Factoring
- 67. The ac-Method The product of two binomials has four terms. We may use grouping-method to work backwards to obtain the two binomials. But how did we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in the first place? This leads to the ac-method. The ac-method tells us how to rewrite the trinomial so that we can use the grouping method. Example F. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Pull out the common factor = x(x – 3) + 2(x – 3) Pull out the common (x – 3) = (x – 3)(x + 2) Review on Factoring
- 68. The ac-Method The product of two binomials has four terms. We may use grouping-method to work backwards to obtain the two binomials. But how did we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in the first place? This leads to the ac-method. The ac-method tells us how to rewrite the trinomial so that we can use the grouping method. If it isn't possible to rewrite the trinomial according to the ac-method, then the trinomial is prime. Example F. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Pull out the common factor = x(x – 3) + 2(x – 3) Pull out the common (x – 3) = (x – 3)(x + 2) Review on Factoring
- 69. ac-Method: Review on Factoring
- 70. ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. Review on Factoring
- 71. ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, Review on Factoring
- 72. ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv = ac, Review on Factoring
- 73. ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv = ac, and u + v = b. Review on Factoring
- 74. ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv = ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c. Review on Factoring
- 75. ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv = ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) Review on Factoring
- 76. ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv = ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Review on Factoring
- 77. Example G. Factor 3x2 – 4x – 20 using the ac-method. ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv = ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Review on Factoring
- 78. Example G. Factor 3x2 – 4x – 20 using the ac-method. a = 3, c = –20. ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv = ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Review on Factoring
- 79. Example G. Factor 3x2 – 4x – 20 using the ac-method. a = 3, c = –20. Hence ac = 3(–20) = –60. ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv = ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Review on Factoring
- 80. Example G. Factor 3x2 – 4x – 20 using the ac-method. a = 3, c = –20. Hence ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4. ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv = ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Review on Factoring
- 81. Example G. Factor 3x2 – 4x – 20 using the ac-method. a = 3, c = –20. Hence ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4. By searching, we find that 6(–10) = –60 and 6 + (–10) = –4. or u, v are 6 and –10. ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv = ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Review on Factoring
- 82. Example G. Factor 3x2 – 4x – 20 using the ac-method. a = 3, c = –20. Hence ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4. By searching, we find that 6(–10) = –60 and 6 + (–10) = –4. or u, v are 6 and –10. Replace –4x by ux and vx ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv = ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Review on Factoring
- 83. Example G. Factor 3x2 – 4x – 20 using the ac-method. a = 3, c = –20. Hence ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4. By searching, we find that 6(–10) = –60 and 6 + (–10) = –4. or u, v are 6 and –10. Replace –4x by ux and vx then we get 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv = ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Review on Factoring
- 84. Example G. Factor 3x2 – 4x – 20 using the ac-method. a = 3, c = –20. Hence ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4. By searching, we find that 6(–10) = –60 and 6 + (–10) = –4. or u, v are 6 and –10. Replace –4x by ux and vx then we get 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 Put into two groups = (3x2 + 6x ) + (–10x – 20) ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv = ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Review on Factoring
- 85. Example G. Factor 3x2 – 4x – 20 using the ac-method. a = 3, c = –20. Hence ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4. By searching, we find that 6(–10) = –60 and 6 + (–10) = –4. or u, v are 6 and –10. Replace –4x by ux and vx then we get 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 Put into two groups = (3x2 + 6x ) + (–10x – 20) Pull out common factor = 3x(x + 2) – 10 (x + 2) ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv = ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Review on Factoring
- 86. Example G. Factor 3x2 – 4x – 20 using the ac-method. a = 3, c = –20. Hence ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4. By searching, we find that 6(–10) = –60 and 6 + (–10) = –4. or u, v are 6 and –10. Replace –4x by ux and vx then we get 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 Put into two groups = (3x2 + 6x ) + (–10x – 20) Pull out common factor = 3x(x + 2) – 10 (x + 2) ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv = ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Review on Factoring
- 87. Example G. Factor 3x2 – 4x – 20 using the ac-method. a = 3, c = –20. Hence ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4. By searching, we find that 6(–10) = –60 and 6 + (–10) = –4. or u, v are 6 and –10. Replace –4x by ux and vx then we get 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 Put into two groups = (3x2 + 6x ) + (–10x – 20) Pull out common factor = 3x(x + 2) – 10 (x + 2) Pull out common factor = (3x – 10)(x + 2) ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv = ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Review on Factoring
- 88. Review on Factoring Factoring Formula If it fits, use the Difference of Squares Formula x2 – y2 = (x + y)(x – y).
- 89. Review on Factoring Factoring Formula If it fits, use the Difference of Squares Formula x2 – y2 = (x + y)(x – y). Example H. a. 4x2 – 9y2
- 90. Review on Factoring Factoring Formula If it fits, use the Difference of Squares Formula x2 – y2 = (x + y)(x – y). Example H. a. 4x2 – 9y2 = (2x)2 – (3y)2
- 91. Review on Factoring Factoring Formula If it fits, use the Difference of Squares Formula x2 – y2 = (x + y)(x – y). Example H. a. 4x2 – 9y2 = (2x)2 – (3y)2 = (2x – 3y)(2x + 3y)
- 92. Review on Factoring Factoring Formula If it fits, use the Difference of Squares Formula x2 – y2 = (x + y)(x – y). Example H. a. 4x2 – 9y2 = (2x)2 – (3y)2 = (2x – 3y)(2x + 3y) b. 32x3 – 2x
- 93. Review on Factoring Factoring Formula If it fits, use the Difference of Squares Formula x2 – y2 = (x + y)(x – y). Example H. a. 4x2 – 9y2 = (2x)2 – (3y)2 = (2x – 3y)(2x + 3y) b. 32x3 – 2x = 2x(16x2 – 1)
- 94. Review on Factoring Factoring Formula If it fits, use the Difference of Squares Formula x2 – y2 = (x + y)(x – y). Example H. a. 4x2 – 9y2 = (2x)2 – (3y)2 = (2x – 3y)(2x + 3y) b. 32x3 – 2x = 2x(16x2 – 1) = 2x(4x + 1)(4x – 1)
- 95. Review on Factoring Factoring Formula If it fits, use the Difference of Squares Formula x2 – y2 = (x + y)(x – y). Example H. a. 4x2 – 9y2 = (2x)2 – (3y)2 = (2x – 3y)(2x + 3y) b. 32x3 – 2x = 2x(16x2 – 1) = 2x(4x + 1)(4x – 1)
- 96. Review on Factoring Factoring Formula If it fits, use the Difference of Squares Formula x2 – y2 = (x + y)(x – y). Example H. a. 4x2 – 9y2 = (2x)2 – (3y)2 = (2x – 3y)(2x + 3y) b. 32x3 – 2x = 2x(16x2 – 1) = 2x(4x + 1)(4x – 1) c. (89)(91) =
- 97. Review on Factoring Factoring Formula If it fits, use the Difference of Squares Formula x2 – y2 = (x + y)(x – y). Example H. a. 4x2 – 9y2 = (2x)2 – (3y)2 = (2x – 3y)(2x + 3y) b. 32x3 – 2x = 2x(16x2 – 1) = 2x(4x + 1)(4x – 1) c. (89)(91) = (90 – 1)(90 + 1)
- 98. Review on Factoring Factoring Formula If it fits, use the Difference of Squares Formula x2 – y2 = (x + y)(x – y). Example H. a. 4x2 – 9y2 = (2x)2 – (3y)2 = (2x – 3y)(2x + 3y) b. 32x3 – 2x = 2x(16x2 – 1) = 2x(4x + 1)(4x – 1) c. (89)(91) = (90 – 1)(90 + 1) = 902 – 12
- 99. Review on Factoring Factoring Formula If it fits, use the Difference of Squares Formula x2 – y2 = (x + y)(x – y). Example H. a. 4x2 – 9y2 = (2x)2 – (3y)2 = (2x – 3y)(2x + 3y) b. 32x3 – 2x = 2x(16x2 – 1) = 2x(4x + 1)(4x – 1) c. (89)(91) = (90 – 1)(90 + 1) = 902 – 12 = 8,100 – 1 = 7,099
- 100. Review on Factoring Factoring Formula If it fits, use the Difference of Squares Formula x2 – y2 = (x + y)(x – y). Example H. a. 4x2 – 9y2 = (2x)2 – (3y)2 = (2x – 3y)(2x + 3y) b. 32x3 – 2x = 2x(16x2 – 1) = 2x(4x + 1)(4x – 1) c. (89)(91) = (90 – 1)(90 + 1) = 902 – 12 = 8,100 – 1 = 7,099 The factors (x + y) and (x – y) are called the conjugate of each other.
- 101. Review on Factoring The Difference of Powers Formula (Optional)
- 102. Review on Factoring The Difference of Powers Formula (Optional) The Difference of Squares Formula is just a special case of a general Difference of Powers Formula.
- 103. Review on Factoring The Difference of Powers Formula (Optional) The Difference of Squares Formula is just a special case of a general Difference of Powers Formula. xn – yn = ( x – y )( xn–1 + xn–2y + xn–3y2 ... + xyn–2 + yn–1)
- 104. Review on Factoring The Difference of Powers Formula (Optional) The Difference of Squares Formula is just a special case of a general Difference of Powers Formula. This general formula is one of the most important factoring formula because it’s the foundation of modern accounting, finance and many other scientific disciplines. xn – yn = ( x – y )( xn–1 + xn–2y + xn–3y2 ... + xyn–2 + yn–1)
- 105. Review on Factoring The Difference of Powers Formula (Optional) The Difference of Squares Formula is just a special case of a general Difference of Powers Formula. This general formula is one of the most important factoring formula because it’s the foundation of modern accounting, finance and many other scientific disciplines. xn – yn = ( x – y )( xn–1 + xn–2y + xn–3y2 ... + xyn–2 + yn–1) xn – 1 = ( x – 1 )( xn–1 + xn–2 + xn–3... + x + 1) In particular that, if y = 1 then
- 106. Review on Factoring The Difference of Powers Formula (Optional) The Difference of Squares Formula is just a special case of a general Difference of Powers Formula. This general formula is one of the most important factoring formula because it’s the foundation of modern accounting, finance and many other scientific disciplines. xn – yn = ( x – y )( xn–1 + xn–2y + xn–3y2 ... + xyn–2 + yn–1) xn – 1 = ( x – 1 )( xn–1 + xn–2 + xn–3... + x + 1) In particular that, if y = 1 then For example, x6 – 1 = (x – 1)(x5 + x4 + .. + x +1).
- 107. Review on Factoring The Difference of Powers Formula (Optional) The Difference of Squares Formula is just a special case of a general Difference of Powers Formula. This general formula is one of the most important factoring formula because it’s the foundation of modern accounting, finance and many other scientific disciplines. xn – yn = ( x – y )( xn–1 + xn–2y + xn–3y2 ... + xyn–2 + yn–1) xn – 1 = ( x – 1 )( xn–1 + xn–2 + xn–3... + x + 1) In particular that, if y = 1 then Example I. Sum 29 + 28 + 27 + …+ 21 + 1 For example, x6 – 1 = (x – 1)(x5 + x4 + .. + x +1).
- 108. Review on Factoring The Difference of Powers Formula (Optional) The Difference of Squares Formula is just a special case of a general Difference of Powers Formula. This general formula is one of the most important factoring formula because it’s the foundation of modern accounting, finance and many other scientific disciplines. xn – yn = ( x – y )( xn–1 + xn–2y + xn–3y2 ... + xyn–2 + yn–1) xn – 1 = ( x – 1 )( xn–1 + xn–2 + xn–3... + x + 1) In particular that, if y = 1 then Example I. Sum 29 + 28 + 27 + …+ 21 + 1 29 + 28 + 27 + …+ 21 + 1 = For example, x6 – 1 = (x – 1)(x5 + x4 + .. + x +1).
- 109. Review on Factoring The Difference of Powers Formula (Optional) The Difference of Squares Formula is just a special case of a general Difference of Powers Formula. This general formula is one of the most important factoring formula because it’s the foundation of modern accounting, finance and many other scientific disciplines. xn – yn = ( x – y )( xn–1 + xn–2y + xn–3y2 ... + xyn–2 + yn–1) xn – 1 = ( x – 1 )( xn–1 + xn–2 + xn–3... + x + 1) In particular that, if y = 1 then Example I. Sum 29 + 28 + 27 + …+ 21 + 1 29 + 28 + 27 + …+ 21 + 1 = For example, x6 – 1 = (x – 1)(x5 + x4 + .. + x +1). By inspection, we get x = 2, n = 10.
- 110. Review on Factoring The Difference of Powers Formula (Optional) The Difference of Squares Formula is just a special case of a general Difference of Powers Formula. This general formula is one of the most important factoring formula because it’s the foundation of modern accounting, finance and many other scientific disciplines. xn – yn = ( x – y )( xn–1 + xn–2y + xn–3y2 ... + xyn–2 + yn–1) xn – 1 = ( x – 1 )( xn–1 + xn–2 + xn–3... + x + 1) In particular that, if y = 1 then Example I. Sum 29 + 28 + 27 + …+ 21 + 1 29 + 28 + 27 + …+ 21 + 1 = (2 – 1)(29 + 28 + 27 + …+ 21 + 1) For example, x6 – 1 = (x – 1)(x5 + x4 + .. + x +1). By inspection, we get x = 2, n = 10. this happens to be “1”
- 111. Review on Factoring The Difference of Powers Formula (Optional) The Difference of Squares Formula is just a special case of a general Difference of Powers Formula. This general formula is one of the most important factoring formula because it’s the foundation of modern accounting, finance and many other scientific disciplines. xn – yn = ( x – y )( xn–1 + xn–2y + xn–3y2 ... + xyn–2 + yn–1) xn – 1 = ( x – 1 )( xn–1 + xn–2 + xn–3... + x + 1) In particular that, if y = 1 then Example I. Sum 29 + 28 + 27 + …+ 21 + 1 29 + 28 + 27 + …+ 21 + 1 = (2 – 1)(29 + 28 + 27 + …+ 21 + 1) = 210 – 1 = 1023 Apply the formula For example, x6 – 1 = (x – 1)(x5 + x4 + .. + x +1). By inspection, we get x = 2, n = 10.
- 112. Review on Factoring The Difference of Powers Formula (Optional) The Difference of Squares Formula is just a special case of a general Difference of Powers Formula. This general formula is one of the most important factoring formula because it’s the foundation of modern accounting, finance and many other scientific disciplines. xn – yn = ( x – y )( xn–1 + xn–2y + xn–3y2 ... + xyn–2 + yn–1) xn – 1 = ( x – 1 )( xn–1 + xn–2 + xn–3... + x + 1) In particular that, if y = 1 then Example I. Sum 29 + 28 + 27 + …+ 21 + 1 29 + 28 + 27 + …+ 21 + 1 = (2 – 1)(29 + 28 + 27 + …+ 21 + 1) = 210 – 1 = 1023 Apply the formula For example, x6 – 1 = (x – 1)(x5 + x4 + .. + x +1). ± ± – +x3 y3 = ( x y )( x2 xy + y2 ). The cubic version is listed below for the sake of completeness By inspection, we get x = 2, n = 10.
- 113. Ex. A. Factor the following trinomials. use any method. If it’s prime, state so. 1. 3x2 – x – 2 2. 3x2 + x – 2 3. 3x2 – 2x – 1 4. 3x2 + 2x – 1 5. 2x2 – 3x + 1 6. 2x2 + 3x – 1 8. 2x2 – 3x – 27. 2x2 + 3x – 2 15. 6x2 + 5x – 6 10. 5x2 + 9x – 2 Ex. B. Factor. Factor out the GCF, the “–”, and arrange the terms in order first. 9. 5x2 – 3x – 2 12. 3x2 – 5x + 211. 3x2 + 5x + 2 14. 6x2 – 5x – 613. 3x2 – 5x + 2 16. 6x2 – x – 2 17. 6x2 – 13x + 2 18. 6x2 – 13x + 2 19. 6x2 + 7x + 2 20. 6x2 – 7x + 2 21. 6x2 – 13x + 6 22. 6x2 + 13x + 6 23. 6x2 – 5x – 4 24. 6x2 – 13x + 8 25. 6x2 – 13x – 8 26. 4x2 – 49 27. 25x2 – 4 28. 4x2 + 9 29. 25x2 + 9 30. – 6x2 – 5xy + 6y2 31. – 3x2 + 2x3– 2x 32. –6x3 – x2 + 2x 33. –15x2 – 25x2 – 10x 34. 12x2y2 –14x2y2 + 4xy2 Review on Factoring

Be the first to comment