Special Cases of Factoring Chapter 5.5

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. Difference of Two Squares a 2 – b 2 (a + b)(a – b) =

64x 2 (8x + 1) (8x – 1) 1. GCF = ( ) 2 –1 ( ) 2 – 1. Factor Write as squares 8x1 3. (sum)(difference)

36x 2 (6x + 7) (6x – 7) 1. GCF = ( ) 2 –49 ( ) 2 – 2. Factor Write as squares 6x7 3. (sum)(difference)

100x 2 (10x + 9y) (10x – 9y) 1. GCF = ( ) 2 –81y 2 ( ) 2 – 3. Factor Write as squares 10x9y 3. (sum)(difference)

x8x8 (x 4 + 1) ( ) 1. GCF = ( ) 2 –1 ( ) 2 – 4. Factor Write as squares x (sum)(difference) x 4 – 1 ( ) 2 (x 4 + 1) (x 2 + 1) (x 2 – 1) (x 4 + 1) (x 2 + 1) (x + 1) (x – 1) x 2 1 ( ) 2 –

1. Check to see if there is a GCF. 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 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 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 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 GCF = 64x 6 = (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 √ √ (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 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

5( ) 4x 2 (2x + 3) (2x – 3) 1. GCF = ( ) 2 –9 ( ) 2 – 9. Factor Write as squares 2x3 3. (sum)(difference) 20x 2 –45 5

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