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FACTORIZATION USING IDENTITY

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  FACTORIZATION - USING IDENTITY  An identity is an equality that remains true regardless of the values chosen for its variables. ALGEBRAIC IDENTITIES  EXAMPLE 1  Factorize:   9X 2  + 12XY + 4Y 2  [  a 2  + 2ab + b 2  = (a + b) 2  ]    9X 2  + 12XY + 4Y 2  = (3x) 2  + 2(3x)(2y) + (2y) 2   9X 2  + 12XY + 4Y 2  = (3x + 2y) 2  EXAMPLE 2  Factorize:  25a 2  – 10a + 1 [ a 2  – 2ab + b 2  = ( a – b) 2  ] 25a 2  – 10a + 1 = (5a) 2  – 2(5a)(1) + 1 2   25a 2  – 10a + 1 = (5a – 1) 2   EXAMPLE 3 Factorize:  36m 2  – 49n 2 [ a 2  –b 2  = (a + b) (a - b) ] 36m 2  – 49n 2   = (6m) 2  – (7n) 2 36m 2  – 49n 2  = (6m + 7n) (6m – 7n)  EXAMPLE 4 Factorize: 4x 2  + 9y 2  + 25z 2  + 12xy + 30yz + 20xz [(a + b + c) 2  = a 2  + b 2  + c 2  + 2ab +2bc +2ca ]   4x 2  + 9y 2  + 25z 2  + 12xy + 30yz + 20xz =   (2x+3y+5z) 2
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FACTORIZATION - USING PRODUCT AND SUM

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  FACTORIZATION - USING PRODUCT AND SUM EXAMPLE 1 Factorize: X 2  + 8x + 15   a= 1, b = 8, c = 15   Product = a × c                = 1 × 15                 = 15 Sum = b          = 8   To get the product 15 we can multiply 3 and 5 and to get the sum of 15 we can add 3 and 5. Therefore, the middle term 8x can be written as 3x + 5x.   X 2  + 8x + 15 =  X 2  + 3x + 5x + 15                      = (x 2  + 3x) + (5x + 15)                       = x(x + 3) + 5(x + 3)                       = (x + 5) (x + 3) are the two factors EXAMPLE 2 Factorize: 7X 2  + 2x - 5   a= 7, b = 2, c = -5   Product = a ×  c                = 7× -5                 = -35   Sum = b          = 2   To get the product -35 we can multiply -5 and 7 and to get the sum of 15 we can add -5 and 7. Therefore, the middle term 2x can be written as -5x + 7x.   7X 2  + 2x – 5 =  7X 2  – 5x + 7x -5                      = (7X 2  + 7x) – ( 5X+ 5)                       = 7X(X + 1) – 5(X +1)                       = (7X – 5)(X+1)  are the t
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ALGEBRAIC IDENTITIES

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  ALGEBRAIC IDENTITIES An identity is an equality that remains true regardless of the values chosen for its variables. Algebraic Identities     (a+b) 2  = a 2  + 2ab + b 2 (a-b) 2  = a 2  - 2ab + b 2   (a+b)(a-b) = a 2  – b 2   (x+a)(x+b) = x 2  + (a+b)x +ab 1. Prove (a+b) 2  = a 2  + 2ab + b 2   Solution:    (a+b)  2  = (a+b) (a+b)    (a+b)  2  = a (a+b) + b (a+b)    (a+b)  2  = a 2  + ab + ab + b 2          (a+b)  2  = a 2  +2ab + b 2 EXAMPLE:1  EXPAND ( 3x + 4y) 2   Solution: a =  3x and b =  4y = (3x) 2  + 2(3x)(4y) + (4y) 2 = 9x 2  + 24xy + 16y 2 2. Prove  (a-b) 2  = a 2  - 2ab + b 2 Solution: (a - b) 2   = (a-b) (a-b) (a - b) 2   = a(a-b) – b(a-b) (a - b) 2   =  a 2  – ab – ab + b 2 (a - b) 2   =  a 2  - 2ab + b 2 EXAMPLE: 2 Expand (2a – 3b) 2  Solution: = (2a) 2  – 2(2a)(3b) + (3b) 2 = 4a 2  – 12ab + 9b 2 3. Prove (a+b)(a-b) = a 2  – b 2   Solution: (a+b)(a-b) =  a (a-b) + b (a-b) (a+b)(a-b) =  a 2  – ab + ab – b 2 (a+b)(a-b) =  a 2  – b 2 EXAMPLE : 3 Expand    (5x + 4y)(5x – 4y
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