Special Cases of Factoring Chapter 5.5 Perfect Square Trinomials a 2 + 2ab + b 2 (a + b) 2 = a 2 – 2ab + b 2 (a – b) 2 =
1. Check each term to see if there is a GCF of all terms. 2. Determine if the 1 st and 3 rd terms are perfect squares. 3. Determine if the 2 nd term is double the product of the values whose squares are the 1 st and 3 rd terms. 4. Write as a sum or difference squared. Perfect Square Trinomials a 2 + 2ab + b 2 (a + b) 2 = a 2 – 2ab + b 2 (a – b) 2 =
x2x2 2(x)(5) = 10x 1. GCF = x 2 = (x + 5) 2 25 = 2. Are the 1 st and 3 rd terms perfect squares 3. Is 2 nd term double the product of the values whose squares are the 1 st and 3 rd terms √ √ (x) 2 (5) 2 5. Factor. 4. Write as a sum squared x + 25
25x 2 2(5x)(-3) = -30x 1. GCF = 25x 2 = (5x – 3) 2 9 = 2. Are the 1 st and 3 rd terms perfect squares 3. Is 2 nd term double the product of the values whose squares are the 1 st and 3 rd terms √ √ (5x) 2 (-3) 2 6. Factor. 4. Write as a difference squared. 1 – 30 x + 9 Used -3 because the second term is – 30x
25x 2 2(5x)(6y) = 60xy 1. GCF = 25x 2 = (5x + 6y) 2 36y 2 = 2. Are the 1 st and 3 rd terms perfect squares 3. Is 2 nd term double the product of the values whose squares are the 1 st and 3 rd terms √ √ (5x) 2 (6y) 2 7a. Factor. 4. Write as a sum squared xy + 36y 2
64x 6 2(8x 3 )(-3) = -48x 3 1. GCF = 64x 6 = (8x 3 – 3) 2 9 = 2. Are the 1 st and 3 rd terms perfect squares 3. Is 2 nd term double the product of the values whose squares are the 1 st and 3 rd terms √ √ (8x 3 ) 2 (-3) 2 7b. Factor. 4. Write as a difference squared. 1 – 48 x3x3 + 9 Used -3 because the second term is – 48x 3
9x 2 2(3x)(2) = 12x 1. GCF = 9x 2 = Not a perfect square trinomial 4 = 2. Are the 1 st and 3 rd terms perfect squares 3. Is 2 nd term double the product of the values whose squares are the 1 st and 3 rd terms √ (3x) 2 (2) 2 8. Factor. 12x ≠ 15x x + 4 Use trial and error or the grouping method
9 9x 2 + 3x + 12x Factor. 1. GCF = 2. Grouping Number Split into 2 terms. (9)(4) 3. Find 2 integers whose product is 36 and sum is 15. 1, 36 2, 18 3, 12 x2x2 + 15x + 4 = 36
9x 2 8. Factor. (3x + 1)( ) 3x 4 3x + 4 3x 3x x + 1 (3x + 1) (3x + 1) 5. Factor by grouping. GCF = 3x GCF = 4 GCF = (3x + 1) + 3x + 12x + 4 ( ) ( ) 9 x2x2 + 15x + 4
3( ) 25x 2 2(5x)(-2) = -20x GCF = 25x 2 = 3(5x – 2) 2 4 = 2. Are the 1 st and 3 rd terms perfect squares 3. Is 2 nd term double the product of the values whose squares are the 1 st and 3 rd terms √ √ (5x) 2 (-2) Factor. 4. Write as a difference squared. 3 – 20 x + 4 Used -2 because the second term is – 20x 75x 2 – 60 x + 12
Special Cases of Factoring Chapter 5.5