Special Cases of Factoring

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)

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

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

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

Factoring by Grouping Objective: After completing this section, students should be able to factor polynomials by grouping.

Steps for factoring by grouping: 1. A polynomial must have 4 terms to factor by grouping. 2. We factor the first two terms and the second two terms separately. Use the rules for GCF to factor these.

Examples: These two terms must be the same.

Examples: These two terms must be the same. You must always check to see if the expression is factored completely. This expression can still be factored using the rules for difference of two squares. (see 6.2) This is a difference of two squares.

Examples: These two terms must be the same. You can rearrange the terms so that they are the same. These two terms must be the same. But they are not the same. So this polynomial is not factorable.

Try These: Factor by grouping.

Solutions: If you did not get these answers, click the green button next to the solution to see it worked out.

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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

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