In Grade 10, ou learned man was of fatoring: 1. GCF fatoring. Differene of Squares fatoring. Simple trinomial fatoring 4. Comple trinomial fatoring Fatoring Review This is a review of these methods of fatoring to aid us in the net topi of sinusoidal funtions, rational epressions and trigonemtir identities. GCF Fatoring The greatest ommon fator of two numbers ( and ) is defined as the largest fator that an be evenl divided into eah number ( and ). Eample 1: Find the GCF of the numbers 0 and 50. Solution: The fators of 0 are 1,, 4, 5, 10, 0. Beause 1 0 = 0; 10 = 0; 4 5 = 0 The fators of 50 are 1,, 5, 10, 5, 50. Beause 1 50 = 50; 5 = 50; 5 10 = 50 Therefore, 10 is the largest fator that is ommon to both 0 and 50, therefore it is the GCF. Eample : Find the GCF of the terms 9 10 8 z 9 and 1 9 8 z 10 and 18 8 9 Solution: The fators of 9 10 8 z 9 are 1,, 9, 10, 8, z 9 The fators of 1 9 8 z 10 are 1,,, 4, 6, 1, 9, 8, z 10 The fators of 18 8 9 are 1,,, 6, 9, 18, 8, 9 Therefore, the GCF is 8 8.
GCF fatoring is a method where ou find the GCF of all our terms and fator it out of the brakets (divide eah term b the GCF). Eample : Fator b a 6 1 15 Solution: Eample 4: Fator Solution: Eample 5: Fator 6 4 4 Solution: Eample 6: Fator 6 88 56 64 Solution: Eample 7: Fator 500 5 Solution: ) 4 (5 6 1 15 b a b a 1) ( 1) ( 6 4 4 11) 7 (8 8 8 88 8 56 8 64 8 4 6 100) 5( 5 500 5 5 5
Fatoring: greatest ommon fator (gf) Fator the following epressions b using the greatest ommon fator. 1) 5 + 6 ) 1 + 4 14 ) - + 4) -1-4 + 60 5) 96-19 + 4 6) 175 + 75 7) -189 + 105 8) 4 + 1 9) -6-9 10) 150 4-175 11) 9 + 60 1) 7 4 + 6 1) z 6 + z 14) 6 + 15) 70 00 0 16) 4-6 17) 5 5 + 4 + 10 18) 4 + 1 19) 18 + 6 90 0) 4 16 1) 9a - 18a ) 16a 5 b + a 4 b ) + 4 + 4) 5 + 4 4-5 5) - 6) a 5 - a 7) b + 16b 8) 5 7 9) 10 0) a 5n + a n 1) 5 ) 9 ) 5 4-1 4) + 5) 6 1 18 4 6) 4 + 7) 18b 9b 8) + 6 9) 1 + 4 40) 5 +
Answers: 1. ( 5 6). (6 7). ( 11) 4. 6 (1 9 10) 5. 4 (4 8 1) 6. 5 (7 ) 7. (6 5) 8. (8 7) 9. ( ) 10. 5 (6 7) 11. ( 0) 4 1. (1 7 ) 1. 14. z ( z 4 1) (6 1) 15. 0 (9 10 1) 16. 4( 8) 17. (5 10) 18. 4 ( ) 19. 18 ( 5) 0. 4 16 1. 9a ( a ) 4. 16a b( ab ). (1 ) 4. ( 4 5) 5. ( 1) 6. a (a 1) 7. 16b (b 1) 8. (5 7) 9. ( 10) 0. ( n a n a 1) 1.. ( 5) ( ). (5 1) 4. ( 1) 5. 6 (1 ) 6. ( 1) 7. 9b( b) 8. ( ) 9. 4 ( 1) 40. ( )
Differene of Squares Fatoring (DOS) The differene of squares method an onl be used if our epression/equation is a binomial and both terms are perfet squares subtrated from eah other, i.e. When using this method ou will end up with two brakets, one with a subtration sign and one with an addition sign, and the square roots of the terms in eah. Eample 1: Fator 49 49. Solution: 7 7 beause 49 7 Eample : Fator ( ) 4 Solution: beause 4 Eample : Fator m n h 100 mnh beause Solution: 10mnh 10 m n h 100 10 mnh Eample 4: Fator 49 Solution: annot use DOS fatoring beause is not a perfet square Eample 5: Fator 6 6 Solution: 6 6 beause 6 6 6
Eample 6: Fator 81 Solution: 9 9 beause 81 9 Eample 7: Fator 16 4 5 Solution: 4 54 5 beause 16 4 5 5 4
FACTORING: Differene of squares Fator the following epressions b using DOS. 1) 4 9 ) 5 11 ) 5 4) 1 5) 5-6) 100 7) 9 8) 196 1 9) 4a 11 10) 9 16 11) 1 100 1) 6 194 1) 0.01 16 14) 16 + 9 15) 16 16) 49 17) 4 1 18) 81 4 19) 16 11 0) 49 6 1) 1 9 ) 16 81 ) 100 4) 5 5) 4 6) 5 7) 64 8) 4 4 9) 49 16 0) a 1 1) 16 ) a 6 ) b 9 4) 81
Answers: 1.. 5 11 5 11. 5 5 4. 1 1 5. 5 5 6. 10 10 7. an t fator b DOS 8. 14 1 14 1 9. a 11 a 11 10. 4 4 11. 1 101 10 1. an t fator using DOS 1. 0.1 40.1 4 14. an t fator using DOS 15. 4 4 16. 7 7 17. 1 1 18. 9 9 19. 4 11 4 11 0. 7 67 6 1. 1 1. 4 94 9. 10 10 4. 5 5 5. 6. 5 5 7. 8 8 8. 9. 7 4 7 4 0. a 1 a 1 1. 4 4. a 6a 6. b b 4. 9 9
Simple Trinomial Fatoring This method of fatoring an onl be used if ou have trinomial epression where the leading oeffiient is equal to 1. i.e. 8 16. When using this method ou need two find two numbers that multipl to give ou the ending oeffiient and add to give ou the middle oeffiient. Those two numbers are plaed in two brakets with the square root of the first term. It ma be useful to set up a diagram to organize our thoughts. Eample 1: Fator 8 16 Solution: +16 therefore, 4 4 +4 +4 = +8 Eample : Fator z 7z 1 Solution: +1 therefore, z 4z -4 - = -7 Eample : Fator a a Solution: - therefore, a a 1 - +1 = - Eample 4: Fator 6 Solution: annot fator b simple trinomial method, there are no two numbers that multipl to give ou - and add to give ou +6.
Fatoring: simple trinomials Fator the following epressions. 1) + 10 + 1 ) + + ) - 6 + 8 4) - 5-14 5) - 5 + 4 6) 10 + 16 7) - 11 + 0 8) d + 0d + 80 9) + 15 + 54 10) a + 1a + 5 11) b 1b + 4 1) + 15 + 6 1) a 4a 1 14) + 18 77 15) - - 56 16) g + 9g + 14 17) + 110 18) + 11 + 4 19) + 4 + 1 0) + 4 1 1) 8 + 1 ) + 8-15 ) - - 0 4) 7 5) 4 5 6) 15 50 7) 4 8) 7 6 9) 1 11 0) 1 0 1) 5 ) 18 7 ) 15 56 4) 6 16 5) 8 15 6) 7 7) 16 9 8) 11 9) 1 1 40) 41) 14 4 4) 5 6
4) 44) 6 8 45) 8 16 46) 1 0 47) 1 11 48) 0 49) 1 6 50) 6 51) 1 5 5) 9 18 5) 1 4 54) 6 40 55) 1 56) 8 57) a 58) 10ab 4b m mn n 59) 15 44 60) t t 4 61) 1 6 6) b 4b 45 6) n n 18 64) 10 1 ANSWERS: 1. ( + )( + 7). ( + )( + 1). ( 4)( ) 4. ( 7)( + ) 5. ( 4)( 1) 6. ( ) ( 8) 7. ( 5)( 6) 8. ant fator 9. ( + 9)( + 6) 10. (a + 5)(a + 7) 11. (b 6)(b 7) 1. ( + )( + 1) 1. (a 7)(a + ) 14. ant fator 15. ( 8)( + 7) 16. (g + 7)(g + ) 17. ( 10)( + 11) 18. ( + 8)( + ) 19. ant fator 0. ( + 7)( ) 1. ( )( 6). ant fator. ( 5)( + 4) 4. ( 9)( + 8) 5. ( + 5)( - 1) 6. ( + 5)( + 10) 7. ( + 8)( 4) 8. ( + 6)( + 1) 9. ( + 11)( + 1) 0. ( + 10)( + ) 1. ( + 7)( 5). ( 6)( 1). ( 8)( 7) 4. ( 8)( + ) 5. ( )( 5) 6. ( 8)( + 9) 7. ( 1)( ) 8. ( + 11)( + 11) 9. ( + 1)( + 1) 40. ( )( ) 41. ( 1)( - ) 4. ( + )( + ) 4. ( 7)( + 9) 44. ( )( + 11) 45. ( 4)( 4) 46. ( 10)( ) 47. ( 11) ( 1) 48. ( + 5)( 4) 49. ( + 6)( + 6) 50. ( )( + ) 51. ( + 5)( + 7) 5. ( )( 6) 5. ( 6)( 7) 54. ( + 10)( 4) 55. ( + 1)( 11) 56. ant fator 57. (a 1b)(a + b)
58. (m n)(m n) 59. ( + 11)( + 4) 60. (t + )(t + 1) Comple Trinomial Fatoring 61. ( 6)( 6) 6. (b + 5)(b 9) 6. ( )( + 6) 64. ( 7)( ) This method of fatoring an onl be used if ou have trinomial epression where the leading oeffiient is greater than 1. i.e. 6. When using this method ou need two find two numbers that multipl to give ou the produt of the leading oeffiient and the ending oeffiient and add to give ou the middle oeffiient. Those two numbers replae the middle term in the original epression. You then fator out a number from the first two terms and another number from the last two terms. You should end up with the eat same two brakets, and ou fator out the ommon braket to give ou the final answer. It ma be useful to set up a diagram to organize our thoughts. Eample 1: Fator 6 Solution: -1 +4 - = +1 6 4 6 ( ) ( ) ( )( ) Eample : Fator 6 1 6 Solution: +6 +4 +9 = +1 6 6 1 6 4 9 6 ( ) ( ) ( )( )
Eample : Fator 4a 19a 1 Solution: +48 - -16 = -19 4a 4a 19a 1 a 16a 1 a(4a ) 4(4a ) (4a )( a 4) Eample 4: Fator 6 Solution: annot fator b omple trinomial method, there are no two numbers that multipl to give ou -4 and add to give ou +6. Eample 5: Fator 4z 19z 5 Solution: -0-0 1 = -19 4z 4z 19z 5 0z 1z 5 4z( z 5) 1( z 5) ( z 5)(4z 1)
Fatoring: Comple trinomials 1. 0 7. 5. 4 1 5 4. 5 8 1 5. 7 6. 5 64 7. 4 7 8. 4 1 9. 19 45 10. 1 56 11. 4 5 1 1. 5 9 54 1. 6 41 0 14. 7 15. 8 5 16. 4 7 17. 4 11 45 18. 5 1 0 19. 5 0. 11 4 1. 5 9 8. 4 1 9. 17 6 4. 5 5 6 5. 6 55 9 6. 7 7 7. 8 49 8. 5 9 8 9. 4 9 8 0. 0 1. 1 15. 5 11 1. 4 4. 8 9
ANSWERS: 1. ( - 1)( + 7). ( + 1)( ). (4 + 1)( + 5) 4. (5 + )( 6) 5. ( )( 1) 6. ( + 8)(5 8) 7. (4 7)( + 1) 8. ( + 1)( + 1) 9. ( + 9)( + 5) 10. ( + 8)( 7) 11. (4 + )( 7) 1. (5 + 6)( 9) 1. (6 5)( 6) 14. ( + )( 9) 15. ( + 1)( + 5) 16. (4 1)( + ) 17. (4 + 9)( 5) 18. ( + 5)(5 + 6) 19. ( + 7)( 5) 0. ( + 7)( 6) 1. (5 4)( ). (4 9)( 1). ( 1)( + 6) 4. (5 + 7)(5 + 9) 5. (6 1)( 9) 6. ( + 9)( 8) 7. ( + 7)( + 7) 8. (5 1)( + 8) 9. (4 + 7)( 4) 0. ( 5)( 6) 1. ( + )( + 5). ( )(5 + 4). (4 + )( 1) 4. ( + 1)( + 9)
Fatoring: All Tpes Most of the time the question will not tell ou what tpe of fatoring ou should use, and sometimes ou need to fator the epression using more than one method. A good rule of thumb is to tr to fator using the GCF method first, and then determine if an other tpe of fatoring an be used to further break down the epression. Eample 1: Fator ompletel 4 4 6 Solution: First, look for a GCF, whih there is of 4. 4( 6 9) Seond, see if the braket an be fatored using an other method, whih it an, b the simple trinomial method. 4( )( ) Third, see if the new brakets an be fatored, whih the annot. So, ou are done! The final answer is: 4( )( ) Eample : Fator ompletel 10 5 Solution: First, look for a GCF, whih there is none. Seond, see if it an be fatored using an other method, whih it an, b the simple trinomial method. ( 5)( 5) Third, see if the brakets an be fatored, whih the annot. So, ou are done! The final answer is: ( 5)( 5) Eample : Fator ompletel 49 1 Solution: First, look for a GCF, whih there is none. Seond, see if the braket an be fatored using an other method, whih it an, b the differene of squares method. ( 7 1)(7 1)
Third, see if the brakets an be fatored, whih the annot. So, ou are done! The final answer is: ( 7 1)(7 1) Eample 4: Fator ompletel 14 7 1 Solution: First, look for a GCF, whih there is of 7. 7 ( ) Seond, see if the braket an be fatored using an other method, whih the annot (it looks like a omple trinomial, but that method doesn t work). So, ou are done! The final answer is: 7 ( ) Eample 5: Fator ompletel 8 9 1 Solution: First, look for a GCF, whih there is none. Seond, see if the braket an be fatored using an other method, whih it an, b the omple trinomial method. ( 8 )( 4) Third, see if the new brakets an be fatored, whih the annot. So, ou are done! The final answer is: ( 8 )( 4) Eample 6: Fator ompletel 1z z 16 Solution: First, look for a GCF, whih there is of 4. 4(z 8z 4) Seond, see if the braket an be fatored using an other method, whih it an, b the omple trinomial method. (make sure to keep the 4 ou fatored out in step 1) 4( z )(z ) Third, see if the new brakets an be fatored, whih the annot. So, ou are done! The final answer is: 4( z )(z ) Eample 7: Fator ompletel 6 6 Solution: First, look for a GCF, whih there is of -. ( 18)
Fatoring: all tpes Seond, see if the braket an be fatored using an other method, whih it an, b the simple trinomial method. (make sure to keep the - ou fatored out in step 1) ( )( 6) Third, see if the new brakets an be fatored, whih the annot. So, ou are done! The final answer is: ( )( 6) Completel fator the following epressions using the appropriate method. 1) a b 4ab ) 4 ) a + 8a + 16 4) a + 6a ab b 5) 9 + 100 6) 18 7) z 7z + 1 8) ( ) + 4( ) 9) 16 6 10) + 5 4 11) m + 14m + 49 1) 8 4 4 + 1 1) 18t 4 t 14) 7h + 8 15) 8 + 6 9 16) 4 + 1 8 17) 14 7-1 18) + 5 19) 5 0) 18 + 81 1) 4 5 ) 6 16 ) 4 + 7 4) 5a 5a + 15a 5) 9 16 6) 64 7) a + 1a + 7 8) 6 + 1 + 6 9) - 0 48 0) 1 5 6 + 1) 6 54 ) 4 17-15 ) 5 5 4) 5 + + 4 + 1 5) 7-18 6) 8 + 16 7) 4-8) 4 + - 0 9) 0-7 40) 6 + - 1
Answers: 1. ab(a b). ( )( + ). (a + 4)(a + 4) 4. (a b)(a + ) 5. NF 6. ( 64) 7. (z 4)(z ) 8. ( 4)( ) 9. (4 6)(4 + 6) 10. ( + 8)( ) 11. (m+ 7)(m + 7) 1. 4 ( + ) 1. t (8t 1)(8t + 1) 14. NF 15. ( + )(4 ) 16. 4( + ) 17. 7( )( + 1) 18. ( 5)( + 7) 19. ( + 1)( ) 0. ( 9)( 9) 1. ( 5)( + 5). ( 8)( + ). (4 1)( +) 4. 5a(a 5a + ) 5. ( 4)( + 4) 6. ( 8)( + 8) 7. (a + )(a + 9) 8. 6( + 1)( + 1) 9. ( 1)( + ) 0. (4 + 1)