1. 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. 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).

3. 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’.

4. 2.1 Meaning of Identities
The conclusion can also be written as ‘3(x – 2)  3x – 6’.

5. 2.1 Meaning of Identities

6. 2.1 Meaning of Identities
The coefficients of the terms include the signs in front of the numbers.

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:
(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

8. 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.

9. 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.

10. 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.

11. 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.

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

13. 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

14. 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

15. 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

16. 2.2 Difference of Two Squares

17. 2.2 Difference of Two Squares
The identity can also be written as
(a  b)(a  b)  a2  b2.

18. 2.2 Difference of Two Squares

19. 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)

20. 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) 

21. 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)

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

23. 2.3 Perfect Square
Try to prove these identities by expanding the L.H.S.

24. 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

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

26. 2 Identities and Factorization
Example 14T
Expand
Solution:

27. 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
  2(1000)(5)  52
 1002  2(100)(5)  52
   25
  1000  25

28. 2.4 Factorization by Using Identities

29. 2.4 Factorization by Using Identities
A. Difference of Two Squares

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

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

32. 2.4 Factorization by Using Identities
B. Perfect Square

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

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