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Special Cases of Factoring

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. =. a 2 – b 2. (a + b)(a – b). 0. 1. Factor. 64x 2. –. 1. 1.

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Special Cases of Factoring

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  1. Chapter 5.5 Special Cases of Factoring

  2. 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)

  3. 0 1. Factor. 64x2 – 1 1 1. GCF = 2. Write as squares 8x 1 ( )2 – ( )2 3. (sum)(difference) (8x – 1) (8x + 1)

  4. 0 2. Factor. 36x2 – 49 1 1. GCF = 2. Write as squares 6x 7 ( )2 – ( )2 3. (sum)(difference) (6x – 7) (6x + 7)

  5. 0 3. Factor. 100x2 – 81y2 1 1. GCF = 2. Write as squares 10x 9y ( )2 – ( )2 3. (sum)(difference) (10x – 9y) (10x + 9y)

  6. 0 4. Factor. x8 – 1 1 1. GCF = x4 1 ( )2 – ( )2 2. Write as squares x4 – 1 ( ) (x4 + 1) 3. (sum)(difference) ( )2 ( )2 – x2 1 (x2 – 1) (x2 + 1) (x4 + 1) (x2 + 1) (x – 1) (x4 + 1) (x + 1)

  7. 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 2ndterm is double the product of the values whose squares are the 1st and 3rdterms. 4. Write as a sum or difference squared. (a + b)2 a2+ 2ab + b2 = a2– 2ab + b2 = (a – b)2

  8. 0 5. Factor. + 25 x + 10 1 x2 GCF = 2. Are the 1st and 3rd terms perfect squares (x + 5)2 (x)2 x2= √ 25 = (5)2 3. Is 2nd term double the product of the values whose squares are the 1st and 3rd terms √ 10x 2(x)(5) = 4. Write as a sum squared.

  9. 0 6. Factor. + 9 x – 30 1 25x2 GCF = 2. Are the 1st and 3rd terms perfect squares (5x – 3)2 (5x)2 25x2 = √ 9 = (-3)2 3. Is 2nd term double the product of the values whose squares are the 1st and 3rd terms Used -3 because the second term is – 30x √ -30x 2(5x)(-3) = 4. Write as a difference squared.

  10. 0 7a. Factor. + 36y2 xy + 60 1 25x2 GCF = 2. Are the 1st and 3rd terms perfect squares (5x + 6y)2 (5x)2 25x2 = √ 36y2 = (6y)2 3. Is 2nd term double the product of the values whose squares are the 1st and 3rd terms √ 60xy 2(5x)(6y) = 4. Write as a sum squared.

  11. 0 7b. Factor. + 9 x3 – 48 1 64x6 GCF = 2. Are the 1st and 3rd terms perfect squares (5x – 3)2 (8x3)2 64x6 = √ 9 = (-3)2 3. Is 2nd term double the product of the values whose squares are the 1st and 3rd terms Used -3 because the second term is – 48x3 √ -48x3 2(8x3)(-3) = 4. Write as a difference squared.

  12. 0 8. Factor. + 4 x + 15 1 9x2 GCF = 2. Are the 1st and 3rd terms perfect squares Not a perfect square trinomial (3x)2 9x2 = Use trial and error or the grouping method √ 4 = (2)2 3. Is 2nd term double the product of the values whose squares are the 1st and 3rd terms 12x 2(3x)(2) = 12x ≠ 15x

  13. 0 8. Factor. 1. GCF = 1 9 x2 + 15 x + 4 2. Grouping Number. (9)(4) = 36 9x2+ 3x + 12x + 4 3. Find 2 integers whose product is 36 and sum is 15. 1, 36 2, 18 3, 12 4. Split into 2 terms.

  14. 0 8. Factor. 5. Factor by grouping. 9 x2 + 15 x + 4 GCF = 3x 9x2 + 3x + 12x + 4 GCF = 4 GCF = (3x + 1) 4 3x + 1 ) ( 3x + 4 ) + 1 ( 3x 3x (3x + 1) (3x + 1) (3x + 1)( ) + 4 3x

  15. 0 9. Factor. 20x2 – 45 5 1. GCF = 2. Write as squares 5( ) 4x2 – 9 2x 3 ( )2 – ( )2 3. (sum)(difference) (2x – 3) (2x + 3) 5

  16. 0 10. Factor. + 12 x – 60 3 75x2 GCF = 2. Are the 1st and 3rd terms perfect squares + 4 x – 20 25x2 3( ) (5x)2 25x2 = √ 3(5x – 2)2 4 = (-2)2 3. Is 2nd term double the product of the values whose squares are the 1st and 3rd terms Used -2 because the second term is – 20x √ -20x 2(5x)(-2) = 4. Write as a difference squared.

  17. Chapter 5.5 Special Cases of Factoring

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