2 Identities and Factorization 2.1 Meaning of Identities 2.2 Difference of Two Squares 2.3 Perfect Square 2.4 Factorization by Using Identities

2.1 Meaning of Identities Consider an equation x  2  2x ...................... (1) Consider an equation x  2  (2  x) ............. (2) 3 2 1 1 x L.H.S.  x  2 5 4 3 2 1 R.H.S.  2x 6 4 2 2 L.H.S.  R.H.S.? No Yes 3 2 1 1 x L.H.S.  x  2 1 1 2 3 R.H.S.  (2  x) 1 1 2 3 L.H.S.  R.H.S.? Yes Only one value of x can satisfy equation (1). However, all values of x satisfy equation (2).

2.1 Meaning of Identities Consider the equations x  2  2x ...................... (1) x  2  (2  x) ............. (2) All values of x satisfy equation (2), that is, if we substitute any value of x into the equation (2), the L.H.S. is always equal to the R.H.S. We call such an equation an identity. Equation (1) is not an identity. ‘’ means ‘is identical to’.

2.1 Meaning of Identities The conclusion can also be written as ‘3(x – 2)  3x – 6’.

2.1 Meaning of Identities

2.1 Meaning of Identities The coefficients of the terms include the signs in front of the numbers.

Example 1T 2 Identities and Factorization Solution: Prove that the following equations are identities. (a) 6x  7  3(2x  3)  2 (b) (x  2)2  x2  4x  4 Solution: (a) R.H.S.  3(2x  3)  2  6x  9  2  6x  7  L.H.S. ∴ 6x  7  3(2x  3)  2 is an identity. 7

Example 1T 2 Identities and Factorization Solution: Prove that the following equations are identities. (a) 6x  7  3(2x  3)  2 (b) (x  2)2  x2  4x  4 Solution: (b) L.H.S.  (x  2)2  (x  2)(x  2)  x(x  2)  2(x  2)  x2  2x  2x  4  x2  4x  4  R.H.S. ∴ (x  2)2  x2  4x  4 is an identity.

Example 2T 2 Identities and Factorization Solution: Prove that the following equations are identities. (a) (b) (2x  1)(x  2)  x(x  4)  (x  2)(x  1) Solution: (a) L.H.S.  R.H.S.  L.H.S.  R.H.S. ∴ is an identity.

Example 2T 2 Identities and Factorization Solution: Prove that the following equations are identities. (a) (b) (2x  1)(x  2)  x(x  4)  (x  2)(x  1) Solution: (b) L.H.S.  (2x  1)(x  2)  x(x  4) R.H.S.  (x  2)(x  1)  2x(x  2)  (x  2)  x(x  4)  x(x  1)  2(x  1)  2x2  4x  x  2  x2  4x  x2  x  2x  2  x2  x  2  x2  x  2 ∵ L.H.S.  R.H.S. ∴ (2x  1)(x  2)  x(x  4)  (x  2)(x  1) is an identity.

Example 3T 2 Identities and Factorization Solution: Prove that 3(2x  1)  4(4x  3)  5(x  3) is not an identity. Solution: L.H.S.  3(2x  1)  4(4x  3)  6x  3  16x  12  10x  15 R.H.S.  5(x  3)  5x  15 ∵ L.H.S.  R.H.S. ∴ 3(2x  1)  4(4x  3)  5(x  3) is not an identity.

Example 4T 2 Identities and Factorization Solution: Determine whether is an identity. Solution: L.H.S.   R.H.S.  is an identity.

Example 5T 2 Identities and Factorization Solution: If Px + 10x2 + Q  7 + Rx2 , where P, Q and R are constants, find the values of P, Q and R. Solution: R.H.S.  By comparing the coefficients of the like terms, we have

Example 6T 2 Identities and Factorization Solution: If Ax2  B  Cx  (3x + 2)(3x  2), where A, B and C are constants, find the values of A, B and C. Solution: R.H.S.  (3x  2)(3x  2)  3x(3x  2)  2(3x  2)  9x2  6x  6x  4  9x2  4 ∴ Ax2  B  Cx  9x2  4 By comparing coefficients of like terms, we have

Example 7T 2 Identities and Factorization Solution: Let A, B and C be constants. If 2x2  5x  C  2(x  2)(Ax + 1) + Bx, find the values of A, B and C. Solution: R.H.S.  2(x  2)(Ax  1)  Bx  (2x  4)(Ax  1)  Bx  2x(Ax  1)  4(Ax  1)  Bx  2Ax2  2x  4Ax  4  Bx  2Ax2  (2  4A  B)x  4 ∴ 2x2  5x  C  2Ax2  (2  4A  B)x  4 By comparing the coefficients of like terms, we have 2A  2 2  4A  B  5 C  4 A  1 2  4(1)  B  5 C  4 B  7

2.2 Difference of Two Squares

2.2 Difference of Two Squares The identity can also be written as (a  b)(a  b)  a2  b2.

2.2 Difference of Two Squares

Example 8T 2 Identities and Factorization Solution: Expand the following expressions. (a) (5x  2)(5x  2) (b) (2x  7y)(7y + 2x) Solution: (a) (5x  2)(5x  2)  (5x)2  22 (b) (2x  7y)(7y  2x)  (2x  7y)(2x  7y)

Example 9T 2 Identities and Factorization Solution: Expand the following expressions. (a) (4  3x)(4  3x) (b) 2( 6x  5y)( 6x + 5y) Solution: (a) (b) 2(6x  5y)(6x + 5y) 

Example 10T 2 Identities and Factorization Solution: Expand the following expressions. (a) (5x  4y)(5x  4y) (b) Solution: (a) (5x  4y)(5x  4y)  (4y  5x)(4y  5x) (b)

Example 11T 2 Identities and Factorization Solution: Evaluate the following without using a calculator. (a) 1252  252 (b) 100.5  99.5 Solution: (a) 1252  252  (125  25)(125  25)  150  100 (b) 100.5  99.5  (100  0.5)(100  0.5)  1002  0.52  10 000  0.25

2.3 Perfect Square Try to prove these identities by expanding the L.H.S.

Example 12T 2 Identities and Factorization Solution: Expand the following expressions. (a) (3x  2y)2 (b) (2  5x)2 Solution: (a) (3x  2y)2  (3x)2  2(3x)(2y)  (2y)2 (b) (2  5x)2  22  2(2)(5x)  (5x)2

Example 13T 2 Identities and Factorization Solution: Expand the following expressions. (a) (2x  y)2 (b) 3(5x + 2y)2 Solution: (a) (b)

2 Identities and Factorization Example 14T Expand . Solution:

Example 15T 2 Identities and Factorization Solution: Evaluate the following without using a calculator. (a) 9952 (b) 1052 Solution: (a) 9952  (1000  5)2 (b) 1052  (100  5)2  10002  2(1000)(5)  52  1002  2(100)(5)  52  1 000 000  10 000  25  10 000  1000  25

2.4 Factorization by Using Identities

2.4 Factorization by Using Identities A. Difference of Two Squares

Example 16T 2 Identities and Factorization Solution: Factorize 25m2 – 121n2. Solution:

Example 17T 2 Identities and Factorization Solution: Factorize 27r2 – 75. Solution:

2.4 Factorization by Using Identities B. Perfect Square

Example 18T 2 Identities and Factorization Solution: Factorize 49a2 – 70ab + 25b2. Solution:

Example 19T 2 Identities and Factorization Solution: Factorize 2 + 16u + 32u2. Solution: