1. Identities and Factorization 4 4.1Meaning of Identity 4.2Difference of Two Squares 4.3Perfect Square Chapter Summary Case Study 4.4Factorization by Taking out Common Factors 4.5Factorization by Grouping Terms

2. P. 2 Case Study Rachael wants to calculate the area of a square. She measures the length of the square is 103 cm. We have the following formula for the perfect square: (a  b) 2  a 2  2ab  b 2 We can consider 103  100  3, that is a  100 and b  3. Without using a calculator, Rachael can find the area by the use of the perfect square. Thus, the area  103 2 cm 2  (100  3) 2 cm 2  [100 2  2(100)(3)  3 2 ] cm 2  (10000  600  9) cm 2  10609 cm 2 Is there any easier way to find the area other than direct multiplication? You can use the perfect square to find the area.

3. P. 3 4.1 Meaning of Identity Consider an equation x  2  2x.............................. (1) NoYesNo L.H.S.  R.H.S.? 6420 22R.H.S.  2x 54321 L.H.S.  x  2 3210 11 x Consider an equation x  2 =  (2  x)..................... (2) Yes L.H.S.  R.H.S.? 10 11 22 33R.H.S.   (2  x) 10 11 22 33L.H.S.  x  2 3210 11 x Only one value of x can satisfy equation (1). However, all values of x satisfy equation (2).

4. P. 4 However, to prove that an equation is an identity, it is impossible for us to check with all values of unknown(s). Thus, we can simplify the algebraic expressions on both sides and match up the like terms. If all the like terms and the corresponding coefficients on both sides of the equation are equal, then the given equation is an identity. 4.1 Meaning of Identity Consider the equations: x  2  2x.............................. (1) x  2 =  (2  x)..................... (2) 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. If any value of the unknown(s) can satisfy an equation, then the equation is called an identity.

5. P. 5 Prove that the following equations are identities. (a)6x  7  3(2x  3)  2 (b)(5x  3)  2(x  4)  3(x  3)  2 (a)R.H.S.  3(2x  3)  2 4.1 Meaning of Identity Example 4.1T Solution:  6x  9  2  6x  7  L.H.S. ∴ 6x  7  3(2x  3)  2 is an identity. (b)L.H.S.  (5x  3)  2(x  4)  5x  3  2x  8  3x  11 R.H.S.  3(x  3)  2  3x  9  2  3x  11 ∵ L.H.S.  R.H.S. ∴ (5x  3)  2(x  4)  3(x  3)  2 is an identity.

6. P. 6 4.1 Meaning of Identity Example 4.2T Prove that the following equations are identities. (a)(x  2) 2  x 2  4x  4 (b)(2x  1)(x  2)  x(x  4)  (x  2)(x  1) (c) Solution: (a)L.H.S.  (x  2) 2  (x  2)(x  2)  x 2  4x  4  R.H.S. ∴ (x  2) 2  x 2  4x  4 is an identity.  x 2  2x  2x  4  x(x  2)  2(x  2)

7. P. 7 4.1 Meaning of Identity Example 4.2T Prove that the following equations are identities. (a)(x  2) 2  x 2  4x  4 (b)(2x  1)(x  2)  x(x  4)  (x  2)(x  1) (c) Solution: (b)L.H.S.  (2x  1)(x  2)  x(x  4)  2x(x  2)  (x  2)  x(x  4)  x 2  x  2 ∴ (2x  1)(x  2)  x(x  4)  (x  2)(x  1) is an identity. It is easier to expand the complicated expression first.  2x 2  4x  x  2  x 2  4x R.H.S.  (x  2)(x  1)  x(x  1)  2(x  1)  x 2  x  2x  2  x 2  x  2 ∵ L.H.S.  R.H.S.

8. P. 8 4.1 Meaning of Identity Example 4.2T Prove that the following equations are identities. (a)(x  2) 2  x 2  4x  4 (b)(2x  1)(x  2)  x(x  4)  (x  2)(x  1) (c) Solution: (c)L.H.S.    x  4  x  2  6 6  R.H.S. ∴   6 is an identity.

9. P. 9 4.1 Meaning of Identity Prove that 3(2x  1)  4(4x  3)   5(x  3) is not an identity. Solution: The symbol ‘  ’ means ‘is not equal to’. L.H.S.  3(2x  1)  4(4x  3)  6x  3  16x  12   10x  15 ∵ L.H.S.  R.H.S. ∴ 3(2x  1)  4(4x  3)   5(x  3) is not an identity. R.H.S.   5(x  3)   5x  15 Example 4.3T Alternative Solution: Try to find a value of x such that it does not satisfy the given equation. We may try with small values like 0 or 1 first. When x  0, L.H.S.  3[2(0)  1]  4[4(0)  3] R.H.S.   5[(0)  3] ∵ L.H.S.  R.H.S. ∴ 3(2x  1)  4(4x  3)   5(x  3) is not an identity.  15   15

10. P. 10 For example, consider Ax  B  2x  3. Ax and 2x are like terms. B and 3 are also like terms. By comparing the coefficients of the like terms, we have A  2 and B  3. 4.1 Meaning of Identity By using the fact that the like terms on both sides of an identity are equal, we can find the value of the unknown coefficients or constants in an identity. The symbol ‘  ’ means ‘is identical to’.

11. P. 11 4.1 Meaning of Identity If Ax 2  Bx  C  (3x  2)(3x  2), find the values of A, B and C. Solution: Example 4.4T R.H.S.  (3x  2)(3x  2)  3x(3x  2)  2(3x  2)  9x 2  6x  6x  4  9x 2  4 By comparing coefficients of like terms, we have A  9, B  0 and C   4. ∴ Ax 2  Bx  C  9x 2  4

12. P. 12 4.1 Meaning of Identity If 2x 2  5x  C  2(x  2)(Ax  1)  Bx, find the values of A, B and C. Solution: Example 4.5T R.H.S.  2(x  2)(Ax  1)  Bx  (2x  4)(Ax  1)  Bx  2x(Ax  1)  4(Ax  1)  Bx  2Ax 2  2x  4Ax  4  Bx By comparing the coefficients of like terms, we have 2A  2  2Ax 2  (2  4A  B)x  4 A  1 2  4A  B  5 2  4(1)  B  5 B  7 C 4C 4 C  4 ∴ 2x 2  5x  C  2Ax 2  (2  4A  B)x  4

13. P. 13 When x  0, 2(0) 2  5(0)  C  2(0  2)[A(0)  1]  B(0) 4.1 Meaning of Identity If 2x 2  5x  C  2(x  2)(Ax  1)  Bx, find the values of A, B and C. Alternative Solution: Example 4.5T Substitute different values of x into the identity to find the unknowns. There are no restrictions on the choice of the values of x, but for easier calculation, it is better to choose smaller values. –C  2(–2)(1) + 0 C  4 When x  2, 2(2) 2  5(2)  C  2(2  2)[A(2)  1]  B(2) 8  10  4  2B When x  1, 2(1) 2  5(1)  C  2(1  2)[A(1)  1]  B(1) 2  5  4  2(  1)(A  1)  7 B  7  4   2(A  1) A  1 x  2 is selected as the term 2(x  2)(Ax  1) will vanish.

14. P. 14 4.2 Difference of Two Squares (a  b)(a  b)  a 2  b 2 The above identity is called the difference of two squares. We can prove the above identity algebraically. For any values of a and b, L.H.S.  (a  b)(a  b)  a(a  b)  b(a  b)  a 2  ab  ba  b 2  a2  b2 a2  b2  R.H.S. ∴ (a  b)(a  b)  a 2  b 2

15. P. 15 (a)(5x  2)(5x  2)  (5x) 2  2 2 (b)(  4  3x)(  4  3x)  (  4) 2  (3x) 2 (c)(  2x  7y)(  2x  7y)  (  2x) 2  (7y) 2 4.2 Difference of Two Squares Expand the following expressions. (a)(5x  2)(5x  2) (b)(  4  3x)(  4  3x) (c)(  2x  7y)(  2x  7y) Solution:  25x 2  4  16  9x 2  4x 2  49y 2 Example 4.6T

16. P. 16 4.2 Difference of Two Squares Solution: Example 4.7T Expand.

17. P. 17 4.2 Difference of Two Squares Example 4.8T Evaluate the following without using a calculator. (a)125 2  25 2 (b)100.5  99.5 Solution: (a)125 2  25 2  (125  25)(125  25)  150  100  15 000 (b)100.5  99.5  (100  0.5)(100  0.5)  100 2  0.5 2  10 000  0.25  9999.75

18. P. 18 4.3 Perfect Square (a  b) 2  a 2  2ab  b 2 The above identities are called the perfect square. We can prove the above identities algebraically. For any values of a and b, L.H.S.  (a  b) 2  (a  b)(a  b)  a 2  ab  ba  b 2  a 2  2ab  b 2  R.H.S. ∴ (a  b) 2  a 2  2ab  b 2  a(a  b)  b(a  b) (a  b) 2  a 2  2ab  b 2 and L.H.S.  (a  b) 2  (a  b)(a  b)  a 2  ab  ba  b 2  a 2  2ab  b 2  R.H.S. ∴ (a  b) 2  a 2  2ab  b 2  a(a  b)  b(a  b)

19. P. 19 4.3 Perfect Square (a)(3x  2y) 2  (3x) 2  2(3x)(2y)  (2y) 2 (b)(2  5x) 2  2 2  2(2)(5x)  (5x) 2 Expand the following expressions. (a)(3x  2y) 2 (b)(2  5x) 2 Solution:  9x 2  12xy  4y 2  4  20x  25x 2 Example 4.9T

20. P. 20 4.3 Perfect Square Example 4.10T Expand the following expressions. (a)3(  2x  y) 2 (b) Solution: (a)3(  2x  y) 2  3[(  2x) 2  2(  2x)(y)  y 2 ] (b)  12x 2  12xy  3y 2  3(4x 2  4xy  y 2 )

21. P. 21 4.3 Perfect Square Example 4.11T Evaluate the following without using a calculator. (a)995 2 (b)105 2 Solution: (a)995 2  (1000  5) 2  1000 2  2(1000)(5)  5 2  1 000 000  10 000  25  990 025 (b)105 2  (100  5) 2  100 2  2(100)(5)  5 2  10 000  1000  25  11 025

22. P. 22 4.4 Factorization by Taking out Common Factors This method is called factorization by taking out common factors. Consider the polynomial 2xy  xz. x is the common factor for the terms 2xy and xz. We can take this common factor out: 2xy  xz  x(2y  z) We already know how to expand the polynomial a(b  c) into ab  ac and the process is called expansion. The process of expressing ab  ac into a(b  c) is called factorization. Factorization is the reverse process of expansion. a(b  c)  ab  ac Expansion Factorization If we want to know whether the results of factorization are correct, then we can expand the expressions to check the results.

23. P. 23 4.4 Factorization by Taking out Common Factors Example 4.12T (a)5a 2 x  15a 2 x 2  5a 2 x(1)  5a 2 x(3x) (b)45pq  60pqr  15pq(3)  15pq(4r) (c)18x 3 y  24x 2 y 2  30xy 3  6xy(3x 2 )  6xy(4xy)  6xy(5y 2 ) Factorize the following expressions. (a)5a 2 x  15a 2 x 2 (b)45pq  60pqr (c)18x 3 y  24x 2 y 2  30xy 3 Solution:  5a 2 x(1  3x)  6xy(3x 2  4xy  5y 2 )  15pq(3  4r)

24. P. 24 4.4 Factorization by Taking out Common Factors Example 4.13T Factorize the following expressions. (a)(2a  b)c  (2a – b)d(b)2m(x  2y)  4n(2y  x) (c)  5rt  2t(3r  4s)(d)18m 2 n(p  q) 2  27mn 2 (q  p) Solution: a  b   (b  a) (a)(2a  b)c  (2a – b)d  (2a  b)(c  d) (b)2m(x  2y)  4n(2y  x)  2m(x  2y)  4n(x  2y)  2(x  2y)(m  2n) (c)  5rt  2t(3r  4s)   t[5r  2(3r  4s)]   t(5r  6r  8s)   t(11r  8s)which is equivalent to t(8s  11r) (d)18m 2 n(p  q) 2  27mn 2 (q  p)  18m 2 n(p  q) 2  27mn 2 (p  q)  9mn(p  q)[2m(p  q)  3n]  9mn(p  q)(2mp  2mq)  3n)

25. P. 25 ac  ad  bc  bd  (ac  ad)  (bc  bd) 4.5 Factorization by Grouping Terms In the previous section, we learnt how to factorize polynomials by taking out common factors. However, for some expressions like ac  ad  bc  bd, we may be unable to find any common factors for all terms. In such a situation, we can group the terms first and then take out the common factors of each group to factorize the expression.  a(c  d)  b(c  d) a(c  d)  b(c  d)  (a  b)(c  d) Grouping terms Taking common factors out in each group Taking out common factor (c  d) This method is called factorization by grouping terms.

26. P. 26 (a)1  y  5xy  5x  (1  y)  (5xy  5x) 4.5 Factorization by Grouping Terms Example 4.14T Factorize the following expressions. (a)1  y  5xy  5x (b)6xyz  6wyz  5w  5x Solution:  (1  y)  5x(1  y)  (1  y)(1  5x) (a)1  y  5xy  5x  (1  5x)  (5xy  y)  (1  5x)  y(1  5x)  (1  5x)(1  y)  (1  y)  5x(y  1)  (1  5x)  y(5x  1) Alternative Solution:  5x(y  1)  (1  y)  5x(y  1)  (y  1)  (y  1)(5x  1) Group the x terms.Group the y terms. OR

27. P. 27 4.5 Factorization by Grouping Terms Example 4.14T Factorize the following expressions. (a)1  y  5xy  5x (b)6xyz  6wyz  5w  5x Solution: (b) 6xyz  6wyz  5w  5x  (6xyz  6wyz)  (5x  5w)  6yz(x  w)  5(x  w)  (x  w)(6yz  5) (b) 6xyz  6wyz  5w  5x  (6xyz  5x)  (6wyz  5w)  x(6yz  5)  w(6yz  5)  (6yz  5)(x  w) Group the positive and the negative terms separately. Alternative Solution:

28. P. 28 4.5 Factorization by Grouping Terms Example 4.15T Factorize the following expressions. (a) 3az  6bz  12b  6a(b) x 2 y 2  y 2  x 2  1 Solution: (a)3az  6bz  12b  6a  3z(a  2b)  6(2b  a)  3(a  2b)(z  2)  3z(a  2b)  6(a  2b) (b)x 2 y 2  y 2  x 2  1  (x 2 y 2  y 2 )  (x 2  1)  y 2 (x 2  1)  (x 2  1) Group the y 2 terms. (b)x 2 y 2  y 2  x 2  1  (x 2 y 2  x 2 )  (y 2  1)  x 2 (y 2  1)  (y 2  1) Alternative Solution: Group the x 2 terms.  (x 2  1)(y 2  1)  (y 2  1)(x 2  1)

29. P. 29 Chapter Summary 4.1 Meaning of Identity 1.If any value of the unknown(s) can satisfy an equation, then the equation is called an identity. 2.If all the like terms and their corresponding coefficients on both sides of the equation are equal, then the given equation is an identity. 3.The like terms on the two sides of an identity are always equal.

30. P. 30 4.2 Difference of Two Squares Chapter Summary For any values of a and b, (a  b)(a  b)  a 2  b 2

31. P. 31 4.3 Perfect Square Chapter Summary For any values of a and b, (a  b) 2  a 2  2ab  b 2 and (a  b) 2  a 2  2ab  b 2

32. P. 32 4.4 Factorization by Taking out Common Factors Chapter Summary 1.The process of rewriting a polynomial as a product of its factors is called factorization. 2.Factorization is the reverse process of expansion. 3.When each term of a polynomial has one or more common factors, we can factorize the polynomial by taking out common factors.

33. P. 33 4.5 Factorization by Grouping Terms Chapter Summary Divide a polynomial into several groups first, then take out the common factors of each group to factorize the polynomial.

34. 4.1 Meaning of Identity Follow-up 4.1 Prove that the following equations are identities. (a)3(3x  9)   9(3  x) (b)5  (3x  13)  3(6  x) (a)L.H.S.  3(3x  9) Solution:  9x  27 ∴ 3(3x  9)   9(3  x) is an identity. (b)L.H.S.  5  (3x  13)  5  3x  13   3x  18 R.H.S.  3(6  x)  18  3x   3x  18 ∵ L.H.S.  R.H.S. ∴ 5  (3x  13)  3(6  x) is an identity. R.H.S.   9(3  x)   27  9x  9x  27 ∵ L.H.S.  R.H.S.

35. 4.1 Meaning of Identity Solution: (a)L.H.S.  (x  4)(x  1)  x(x  1)  4(x  1)  x 2  x  4x  4  R.H.S. ∴ (x  4)(x  1)  x 2  3x  4 is an identity. Follow-up 4.2 Prove that the following equations are identities. (a)(x  4)(x  1)  x 2  3x  4 (b)(x  2)(x  3)  4x  (x  6)(x  1) (c)  x 2  3x  4

36. 4.1 Meaning of Identity Solution: Follow-up 4.2 Prove that the following equations are identities. (a)(x  4)(x  1)  x 2  3x  4 (b)(x  2)(x  3)  4x  (x  6)(x  1) (c) (b)L.H.S.  (x  2)(x  3)  4x  x(x  3)  2(x  3)  4x  x 2  5x  6 ∴ (x  2)(x  3)  4x  (x  6)(x  1) is an identity.  x 2  3x  2x  6  4x R.H.S.  (x  6)(x  1)  x(x  1)  6(x  1)  x 2  x  6x  6  x 2  5x  6 ∵ L.H.S.  R.H.S.

37. 4.1 Meaning of Identity Solution: Follow-up 4.2 Prove that the following equations are identities. (a)(x  4)(x  1)  x 2  3x  4 (b)(x  2)(x  3)  4x  (x  6)(x  1) (c)  2x  5  2x  3 (c)L.H.S.   4x  2 R.H.S.  2(2x  1)  4x  2 ∵ L.H.S.  R.H.S. ∴  2(2x  1) is an identity.

38. 4.1 Meaning of Identity Solution: L.H.S.  4x  8(x  2)  4x  8x  16  12x  16 ∵ L.H.S.  R.H.S. ∴ 4x  8(x  2)  12(x  1) is not an identity. R.H.S.  12(x  1)  12x  12 Alternative Solution: When x  0, L.H.S.  4(0)  8(0  2) R.H.S.  12[(0)  1] ∵ L.H.S.  R.H.S. ∴ 4x  8(x  2)  12(x  1) is not an identity.  16  12 Follow-up 4.3 Prove that 4x  8(x  2)  12(x  1) is not an identity.

39. 4.1 Meaning of Identity If Ax 2  Bx  C  (2x  1)(x  4), find the values of A, B and C. R.H.S.  (2x  1)(x  4)  2x(x  4)  (x  4)  2x 2  8x  x  4  2x 2  7x  4 By comparing coefficients of like terms, we have A  2, B  7 and C   4. Follow-up 4.4 Solution: ∴ Ax 2  Bx  C  2x 2  7x  4

40. 4.1 Meaning of Identity Follow-up 4.5 If 9x 2  Bx  10  (Ax  1)(x  2)  C, find the values of A, B and C. Solution: R.H.S.  (Ax  1)(x  2)  C  Ax(x  2)  (x  2)  C  Ax 2  2Ax  x  2  C  Ax 2  (2A  1)x  (2  C) By comparing the coefficients of like terms, we have A  9 B  2A  1  2(9)  1  19 2  C   10 C   12 ∴ 9x 2  Bx  10  Ax 2  (2A  1)x  (2  C) Also try to substitute different values of x into the identity to find the unknowns, such as x  0,  2 and 1.

41. 4.2 Difference of Two Squares Follow-up 4.6 (a)(3x  1)(3x  1)  (3x) 2  1 2 (b)(  2x  5)(  2x  5)  (  2x) 2  5 2 (c)(6x  5y)(6x  5y)  (6x) 2  (5y) 2 Expand the following expressions. (a)(3x  1)(3x  1) (b)(  2x  5)(  2x  5) (c)(6x  5y)(6x  5y) Solution:  9x 2  1  4x 2  25  36x 2  25y 2

42. Expand. 4.2 Difference of Two Squares Follow-up 4.7 Solution:

43. 4.2 Difference of Two Squares Follow-up 4.8 Evaluate the following without using a calculator. (a)995 2  5 2 (b)302  298 Solution: (a)995 2  5 2  (995  5)(995  5)  1000  990  990 000 (b)302  298  (300  2)(300  2)  300 2  2 2  90 000  4  89 996

44. 4.3 Perfect Square Follow-up 4.9 (a)(3x  4y) 2  (3x) 2  2(3x)(4y)  (4y) 2 (b)(2x  y) 2  (2x) 2  2(2x)(y)  y 2 Expand the following expressions. (a)(3x  4y) 2 (b)(2x  y) 2 Solution:  9x 2  24xy  16y 2  4x 2  4xy  y 2

45. 4.3 Perfect Square Follow-up 4.10 Solution: (a)4(  1  4x) 2  4[(  1) 2  2(  1)(4x)  (4x) 2 ]  4  32x  64x 2  4(1  8x  16x 2 ) Expand the following expressions. (a)4(  1  4x) 2 (b) (b)

46. 4.3 Perfect Square Follow-up 4.11 Evaluate the following without using a calculator. (a)102 2 (b)89 2 Solution: (a)102 2  (100  2) 2  100 2  2(100)(2)  2 2  10 000  400  4 (b)89 2  (90  1) 2  90 2  2(90)(1)  1 2  8100  180  1  7921  10 404

47. 4.4 Factorization by Taking out Common Factors Follow-up 4.12 (a)3a 2 x  12ax 2  3ax(a)  3ax(4x) (b) 25a 2 c  40abc 2  5ac(5a)  5ac(8bc) (c)14m 3 n  28m 2 n 2  21mn 3  7mn(2m 2 )  7mn(4mn)  7mn (3n 2 ) Factorize the following expressions. (a)3a 2 x  12ax 2 (b)25a 2 c  40abc 2 (c)14m 3 n  28m 2 n 2  21mn 3 Solution:  3ax(a  4x)  7mn(2m 2  4mn  3n 2 )  5ac(5a  8bc)

48. 4.4 Factorization by Taking out Common Factors Follow-up 4.13 Factorize the following expressions. (a)(a  c)b  (a – c)d(b)5m(p  q)  n(q  p) (c)(4x  3y)z  5yz(d)21p 2 q(a  b)  28pq 2 (a  b) 2 Solution: (a)(a  c)b  (a – c)d  (a  c)(b  d) (b)5m(p  q)  n(q  p)  5m(p  q)  n(p  q)  (p  q)(5m  n) (c)(4x  3y)z  5yz  z[(4x  3y)  5y]  z(4x  2y)  2z(2x  y) (d)21p 2 q(a  b)  28pq 2 (a  b) 2  7pq(a  b)[3p  4q(a  b)]  7pq(a  b)(3p  4aq  4bq) Simplify the like terms after taking out the common factor.

49. 4.5 Factorization by Grouping Terms Follow-up 4.14 Factorize the following expressions. (a) 6a  6b  ac  bc(b) 4rst  6rsu  2t  3u Solution: (a)6a  6b  ac  bc  (6a  6b)  (ac  bc)  (a  b)(6  c)  6(a  b)  c(a  b) (b)4rst  6rsu  2t  3u  (4rst  6rsu)  (2t  3u)  (2t  3u)(2rs  1)  2rs(2t  3u)  (2t  3u)

50. 4.5 Factorization by Grouping Terms Follow-up 4.15 Factorize the following expressions. (a)  2x 2 y  6xz  12xy  36z (b) ab  1  b  a Solution: (a)  2x 2 y  6xz  12xy  36z   2(x 2 y  3xz  6xy  18z)   2[(x 2 y  3xz)  (6xy  18z)]   2[x(xy  3z)  6(xy  3z)]   2(xy  3z)(x  6) (b)ab  1  b  a  (ab  a)  (1  b)  a(b  1)  (b  1)  (b  1)(a  1)