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Chapter 5.5 Special Cases of Factoring Difference of Two Squares 1. Check to see if there is a GCF. 2. Write each term as a square. 3. Write those values that are squared as the product of a sum and a difference. a2 – b2 = (a + b)(a – b) 1. Factor. 64x2 – 1 ( 8x )2 – ( 1 )2 1. GCF = 1 2. Write as squares 3. (sum)(difference) (8x + 1)(8x – 1) 2. Factor. 36x2 – 49 ( 6x )2 – ( 7 )2 1. GCF = 1 2. Write as squares 3. (sum)(difference) (6x + 7)(6x – 7) 3. Factor. 100x2 – 81y2 (10x )2 – (9y)2 1. GCF = 1 2. Write as squares 3. (sum)(difference) (10x + 9y)(10x – 9y) 4. Factor. x8 – 1 2 4 (x ) – (1) 1. GCF = 1 2 (x4 + 1)(x4 – 1) 2. Write as squares 3. (sum)(difference) ( x2 )2– (1)2 (x4 + 1)(x2 + 1)(x2 – 1) (x4 + 1)(x2 + 1)(x + 1)(x – 1) Perfect Square Trinomials 1. Check to see if there is a GCF. 2. Determine if the 1st and 3rd terms are perfect squares. 3. Determine if the 2nd term is double the product of the values whose squares are the 1st and 3rd terms. 4. Write as a sum or difference squared. a2 + 2ab + b2 = (a + b)2 a2 – 2ab + b2 = (a – b)2 5. Factor. x2 + 10x + 25 (x + 5)2 GCF = 1 2. Are the 1st and 3rd terms perfect squares x2 = (x)2 25 = (5)2 √ 3. Is 2nd term double the product of the values whose squares are the 1st and 3rd terms 2(x)(5) = 10x √ 4. Write as a sum squared. 6. Factor. 25x2 – 30x + 9 (5x – 3)2 GCF = 1 2. Are the 1st and 3rd terms perfect squares 25x2 = (5x)2 9= Used -3 because the second term is – 30x (-3)2 √ 3. Is 2nd term double the product of the values whose squares are the 1st and 3rd terms 2(5x)(-3) = -30x √ 4. Write as a difference squared. 7a. Factor. 25x2 + 60xy + 36y2 (5x + 6y)2 GCF = 1 2. Are the 1st and 3rd terms perfect squares 25x2 = (5x)2 36y2 = (6y)2 √ 3. Is 2nd term double the product of the values whose squares are the 1st and 3rd terms 2(5x)(6y) = 60xy √ 4. Write as a sum squared. 7b. Factor. 64x6 – 48x3 + 9 (5x – 3)2 Used -3 because the second term is – 48x3 GCF = 1 2. Are the 1st and 3rd terms perfect squares 2 64x6 = (8x3) √ 2 9 = (-3) 3. Is 2nd term double the product of the values whose squares are the 1st and 3rd terms 2(8x3)(-3) = -48x3 √ 4. Write as a difference squared. 8. Factor. 9x2 + 15x + 4 Not a perfect square trinomial Use trial and error or the grouping method GCF = 1 2. Are the 1st and 3rd terms perfect squares 9x2 = (3x)2 4= (2)2 √ 3. Is 2nd term double the product of the values whose squares are the 1st and 3rd terms 2(3x)(2) = 12x 12x ≠ 15x 8. Factor. 9x2 + 15x + 4 1. GCF = 1 2. Grouping Number. 9x2 + 3x + 12x + 4 (9)(4) = 36 3. Find 2 integers whose product is 36 and sum is 15. 1, 36 2, 18 3, 12 4. Split into 2 terms. 8. Factor. 9x2 + 15x + 4 5. Factor by grouping. GCF = 3x 9x2 + 3x + 12x + 4 4 3x 3x( 3x + 1) + 4(3x + 1 ) (3x + 1) (3x + 1) (3x + 1)( 3x + 4) GCF = 4 GCF = (3x + 1) 9. Factor. 20x2 – 45 5(4x2 – 9) ( 2x )2 – ( 3 )2 5(2x + 3)(2x – 3) 1. GCF = 5 2. Write as squares 3. (sum)(difference) 10. Factor. 75x2 – 60x + 12 3( 25x2 – 20x + 4) 3(5x – 2)2 Used -2 because the second term is – 20x GCF = 3 2. Are the 1st and 3rd terms perfect squares 25x2 = (5x)2 4= (-2)2 √ 3. Is 2nd term double the product of the values whose squares are the 1st and 3rd terms 2(5x)(-2) = -20x √ 4. Write as a difference squared. Chapter 5.5 Special Cases of Factoring