Expanding Brackets and Factorising

Chapter 7

Expanding brackets and factorising

This chapter will show you how to ✔ expand and simplify expressions with brackets ✔ solve equations and inequalities involving brackets ✔ factorise by removing a common factor ✔ expand two brackets

7.1 Expanding brackets You will need to know how to ● multiply positive and negative numbers ● add and subtract negative numbers ● collect like terms When multiplying algebraic terms remember that x × 3 5 3 × x 5 3x y × y 5 y2 gh 5 g × h More complicated multiplications can also be simplified.

EXAMPLE 1

Simplify 3f × 4g.

To multiply algebraic terms, 3f × 4g 5 3 × f × 4 × g multiply the numbers then multiply 5 3 × 4 × f × g the letters. 5 12 × fg 5 12fg

Multiplying a bracket You can work out 6 × 34 by thinking of 34 as 30 1 4. 6 × 34 5 6 × (30 1 4) 30 4 5 6 × 30 1 6 × 4 5 1 180 24 6 6 × 30 6 × 4 5 204

Algebra 101

M07_CME_SB_IGCSE_6867_U07.indd 101 28/8/09 11:35:25 Brackets are often used in algebra. 6(x 1 4) means 6 × (x 1 4) As in the 6 × 34 example, you have to multiply each term inside the brackets by 6. 6(x 1 4) 5 6 × x 1 6 × 4 5 6x 1 24 It is like working out the area of a rectangle that has length x 1 4 and width 6

x 4

area = 6 × x area = 6 × 4 6 = 6x = 24

Total area 5 6(x 1 4) 5 6 × x 1 6 × 4 You find the total area by adding 5 6x 1 24 the area of the two smaller When you do this it is called expanding the brackets. rectangles. It is also known as removing the brackets or multiplying out the brackets.

When you remove the brackets you must multiply each term inside the brackets by the term outside the bracket.

EXAMPLE 2

Simplify these by multiplying out the brackets. (a) 5(a 1 6) (b) 2(x 2 8) (c) 3(2c 2 d)

You must multiply each term inside (a) 5(a 1 6) 5 5 × a 1 5 × 6 the bracket by the term outside the 5 5a 1 30 bracket. (b) 2(x 2 8) 5 2 × x 2 2 × 8 5 2x 2 16 3 × 2c 5 3 × 2 × c 5 6 × c 5 6c (c) 3(2c 2 d) 5 3 × 2c 2 3 × d 5 6c 2 3d

102 Algebra

M07_CME_SB_IGCSE_6867_U07.indd 102 28/8/09 11:35:26 Expanding brackets and factorising

EXERCISE 7A

1 Simplify these expressions. (a) 2 × 5k (b) 3 × 6b (c) 4a × 5 (d) 3a × 2b (e) 4c × 3d (f) x × 5y

2 Expand the brackets to find the value of these expressions. Remember, you must multiply each Check your answers by working out the brackets first. term inside the brackets by the term (a) 2(50 1 7) (b) 5(40 1 6) (c) 6(70 1 3) outside the bracket. (d) 3(40 2 2) (e) 7(50 2 4) (f) 8(40 2 3) A common mistake is to forget to multiply the second term. 3 Remove the brackets from these. (a) 5(p 1 6) (b) 3(x 1 y) (c) 4(u 1 v 1 w) (d) 2(y 2 8) (e) 7(9 2 z) (f) 8(a 2 b 1 6) 4 Expand the brackets in these expressions. Remember (a) 3(2c 1 6) (b) 5(4t 1 3) (c) 2(5p 1 q) 3 × 2c 5 3 × 2 × c 5 6c (d) 3(2a 2 b) (e) 6(3c 2 2d) (f) 7(2x 1 y 2 3) (g) 6(3a 2 4b 1 c) (h) 2(x2 1 3x 1 2) (i) 4(y2 2 3y 2 10) 5 Write down the 6 pairs of cards which show equivalent expressions. 4(x 1 2y) 4x 1 2y 2(4x 1 y) 4(2x 2 y)

A B C D

8x 2 8y 4x 1 8y 8(x 2 y) 2x 2 8y

E F G H

8x 1 2y 2(x 2 4y) 2(2x 1 y) 8x 2 4y

I J K L

You can use the same method for expressions that have an algebraic term instead of a number term outside the bracket.

EXAMPLE 3

Expand the brackets in these expressions. (a) a(a 1 4) (b) x(2x 2 y) (c) 3t(t2 1 1)

(a) a(a 1 4) 5 a × a 1 a × 4 Remember a × a 5 a2 5 a2 1 4a (b) x(2x 2 y) 5 x × 2x 2 x × y x × 2x 5 x × 2 × x 5 2 × x × x 5 2x2 5 2x2 2 xy Remember x × y 5 xy (c) 3t(t2 1 1) 5 3t × t2 1 3t × 1 3 5 3t 1 3t 3t × t2 5 3 × t × t × t 5 3t3

Algebra 103

M07_CME_SB_IGCSE_6867_U07.indd 103 28/8/09 11:35:27 EXERCISE 7B

Expand the brackets in these expressions. 1 b(b 1 4) 2 a(5 1 a) 3 k(k 2 6) 4 m(9 2 m) 5 a(2a 1 3) 6 g(4g 1 1) 7 p(2p 1 q) 8 t(t 1 5w) 9 m(m 1 3n) 10 x(2x 2 y) 11 r(4r 2 t) 12 a(a 2 4b) 13 2t(t 1 5) 14 3x(x 2 8) 15 5k(k 1 l) Remember 3x × 4x 5 4 × 3 × x × x 16 3a(2a 1 4) 17 2g(4g 1 h) 18 5p(3p 2 2q) 5 12x2 19 3x(2y 1 5z) 20 r(r2 1 1) 21 a(a2 1 3) 22 t(t2 –7) 23 2p(p2 1 3q) 24 4x(x2 1 x)

Adding and subtracting expressions with brackets Adding To add expressions with brackets, expand the brackets first, then collect like terms to simplify your answer. Collecting like terms means adding all the terms in x, all the terms in y and so on.

EXAMPLE 4

Expand then simplify these expressions. (a) 3(a 1 4) 1 2a 1 10 (b) 3(2x 1 5) 1 2(x 2 4)

Expand the brackets first. Then (a) 3(a 1 4) 1 2a 1 10 5 3a 1 12 1 2a 1 10 collect like terms. 5 3a 1 2a 1 12 1 10 5 5a 1 22 Expand both sets of brackets first. (b) 3(2x 1 5) 1 2(x 2 4) 5 6x 1 15 1 2x 2 8 5 6x 1 2x 1 15 2 8 5 8x 1 7

Subtracting Multiplying If you have an expression like 23(2x 2 5), multiply both terms in the brackets by 23.  ×    23 × 2x 5 26x and 23 × 25 5 15  ×    So 23(2x 2 5) 5 23 × 2x 1 23 × 25  ×    5 26x 1 15  ×   

104 Algebra

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EXAMPLE 5

Expand these expressions. 2 1 2 2 (a) 2(3t 4) (b) 3(4x 1) 22 × 3 5 26 22 × 4 5 28

(a) 22(3t 1 4) 5 22 × 3t 1 22 × 4 5 26t 1 28 5 26t 2 8 (b) 23(4x 2 1) 5 23 3 4x 1 23 3 21 23 × 4 5 212 23 × 21 5 13 5 212x 1 3

EXAMPLE 6

Expand then simplify these expressions. (a) 3(2t 1 1) 2 2(2t 1 4) (b) 8(x 1 1) 2 3(2x 2 5)

(a) 3(2t 1 1) 2 2(2t 1 4) 5 6t 1 3 2 4t 2 8 Remember to multiply both terms 5 6t 2 4t 1 3 2 8 in the second bracket by 22. 5 2t 2 5 1 2 2 5 1 2 1 (b) 8(x 1) 3(2x 5) 8x 8 6x 15 Expand the brackets first. 5 8x 2 6x 1 8 1 15 Remember that 23 × 25 5 115. 5 2x 1 23 Then collect like terms.

EXERCISE 7C

Expand these expressions. 1 22(2k 1 4) 2 23(2x 1 6) 3 25(3n 2 1) 4 24(3t 1 5) 5 23(4p 2 1) 6 22(3x 2 7) Expand then simplify these expressions. 7 3(y 1 4) 1 2y 1 10 8 2(k 1 6) 1 3k 1 9 9 4(a 1 3) 2 2a 1 6 10 3(t 2 2) 1 4t 2 10 11 3(2y 1 3) 1 2(y 1 5) 12 4(x 1 7) 1 3(x 1 4) 13 3(2x 1 5) 1 2(x 2 4) 14 2(4n 1 5) 1 5(n 2 3) 15 3(x 2 5) 1 2(x 2 3) 16 4(2x 2 1) 1 2(3x 2 2) 17 3(2b 1 1) 2 2(2b 1 4) 18 4(2m 1 3) 2 2(2m 1 5) 19 2(5t 1 3) 2 2(3t 1 1) 20 5(2k 1 2) 2 4(2k 1 6) 21 8(a 1 1) 2 3(2a 2 5) 22 2(4p 1 1) 2 4(p 2 3) 23 5(2g 2 4) 2 2(4g 2 6) 24 2(w 2 4) 2 3(2w 2 1) 25 x(x 1 3) 1 4(x 1 2) 26 x(2x 1 1) 2 3(x 2 4) Algebra 105

M07_CME_SB_IGCSE_6867_U07.indd 105 28/8/09 11:35:29 7.2 Solving equations involving brackets

Equations sometimes involve brackets. When dealing with equations involving brackets, you usually expand the brackets first.

EXAMPLE 7

Solve 4(c 1 3) 5 20.

Expand the bracket by multiplying Method A both terms inside the bracket by the term outside the bracket. 4(c 1 3) 5 20 4c 1 12 5 20 You must subtract 12 from both 4 c 1 12 2 12 5 20 2 12 sides before dividing both sides by 4. 4c 5 8 c 4 5 8 4 4 c 5 2 Method B 4(c 1 3) 5 20 Since 4 divides exactly into 20 you c 1 3 5 20 can divide both sides by 4 first. 4 c 1 3 5 5 c 5 5 2 3 c 5 2

EXAMPLE 8

Solve 2(3p 2 4) 5 7.

5 2(3p – 4) 7 Expand the bracket. 6p – 8 5 7 6 p – 8 1 8 5 7 1 8 You must add 8 to both sides before dividing both sides by 6. 6p 5 15 p 6 5 15 6 6 p 5 2.5

106 Algebra

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EXERCISE 7D

1 Solve these equations. (a) 4(g 1 6) 5 32 (b) 7(k 1 1) 5 21 (c) 5(s 1 10) 5 65 (d) 2(n 2 4) 5 6 (e) 3(f 2 2) 5 24 (f) 6(v 2 3) 5 42 (g) 4(m 2 3) 5 14 (h) 2(w 1 7) 5 19 2 Solve these equations. (a) 4(5t 1 2) 5 48 (b) 3(2r 1 4) 5 30 (c) 2(2b 1 2) 5 22 (d) 2(3w 2 6) 5 27 (e) 3(4x 2 2) 5 24 (f) 5(2y 1 11) 5 40 (g) 6(2k 2 1) 5 36 (h) 3(2a 2 13) 5 18

When two brackets are involved, expand both brackets then collect Like terms are terms of the same kind. In Example 9 there are only like terms before solving. terms in m and number terms.

EXAMPLE 9

Solve 2(2m 1 10) 5 12(m 2 1). Expand the brackets on both sides of the equation and collect like terms. 2(2m 1 10) 5 12(m 2 1) 4m 1 20 5 12m 2 12 Collect terms in m on the RHS 20 1 12 5 12m 2 4m because 12m on the RHS is greater 32 5 8m than 4m on the LHS. This keeps the m 5 4 m term positive.

When an equation involves fractions, it can be transformed into an LCM means Lowest Common equation without fractions by multiplying all terms by the LCM of Multiple. the numbers in the denominators.

EXAMPLE 10

x 1 Solve the equation 17 5 x 1 2. 4

x 1 Multiply both sides by 4, collect like 17 5 x 1 2 4 terms and then finally divide by 3. (x 1 17) 4 5 4(x 1 2) 4 Note the use of brackets. x 1 17 5 4x 1 8 17 2 8 5 4x 2 x Collect the terms in x on the RHS 9 5 3x and the numbers on the LHS. 4x on the RHS is greater than x on x 5 3 the LHS.

Algebra 107

M07_CME_SB_IGCSE_6867_U07.indd 107 28/8/09 11:35:32 EXAMPLE 11 Look at the denominators. x 2 x 1 3 and 5 have a LCM of 15 so Solve the equation 6 5 4. 3 5 multiply both sides of the equation by 15. x 2 6 5 x 1 4 3 5 Note the use of brackets. Always 15(x 2 6) 15(x 1 4) 5 put them in when you multiply in 3 5 5(x – 6) 5 3(x 1 4) this way.

5x 2 30 5 3x 1 12 Then solve using the method shown 5x 2 3x 5 12 1 30 in Example 9. 2x 5 42 x 5 21

EXAMPLE 12

2x 1 3 x 2 2 5 Solve the equation 1 5 . 6 3 2

The LCM here is 6. 6(2x 1 3) 1 6(x 2 2) 5 6(5) 6 3 2 2x 1 3 1 2(x 2 2) 5 15 Note the use of brackets. 2x 1 3 1 2x 2 4 5 15 The most common mistake is to 4x 2 1 5 15 forget to multiply all terms by the 4x 5 16 LCM. x 5 4 This means the number on the RHS as well as the terms on the LHS.

EXERCISE 7E

1 Solve these equations. (a) 2a 1 4 5 5(a 2 1) (b) 3(d 2 2) 5 2d 2 1 (c) 5(x 1 3) 5 11x 1 3 (d) 12p 1 3 5 3(p 1 7) (e) 4t 1 3 5 3(2t 2 3) (f) 3b 2 4 5 2(2b 2 7) (g) 8(3g 2 1) 5 15g 1 10 (h) 3(2k 1 6) 5 17k 1 7 (i) 2(y 1 5) 5 3y 1 12 (j) 5r 1 3 5 4(2r 1 3) 2 Solve the following equations by expanding both brackets. Use Example 9 to help. (a) 2(b 1 1) 5 8(2b 2 5) (b) 5(4a 1 7) 5 3(8a 1 9) (c) 6(x 2 2) 5 3(3x 2 8) (d) 5(2p 1 2) 5 6(p 1 5) (e) 9(3s 2 4) 5 5(4s 2 3) (f) 4(10t 2 7) 5 3(6t 2 2) (g) 4(2w 1 2) 5 2(5w 1 7) (h) 3(3y 2 2) 5 7(y 2 2)

108 Algebra

M07_CME_SB_IGCSE_6867_U07.indd 108 28/8/09 11:35:33 Expanding brackets and factorising

3 Solve these equations. Use Example 10 to help. d 1 3 6y 2 5 (a) 5 3 2 d (b) 5 y 1 3 15 5 3x 2 1 6 1 a (c) 5 x 2 2 (d) 5 a 1 4 8 2 c 2 8 10 2 b (e) 5 c 1 1 (f) 5 12 1 b 4 3 4 Solve these equations. Use Example 11 to help. x 1 1 x 2 1 2x 2 1 x (a) 5 (b) 5 3 4 3 2 3x 1 1 2x x 1 3 x 2 3 (c) 5 (d) 5 5 3 2 5 x 1 2 3x 1 6 8 2 x 2x 1 2 (e) 5 (f) 5 7 5 2 5 Solve these equations. 5 Use Example 12 to help. x 1 1 x 1 2 x 1 2 x 1 1 (a) 1 5 3 (b) 1 5 3 2 5 4 7 3x 1 2 x 1 2 3x 2 1 x 1 2 1 (c) 1 5 2 (d) 2 5 5 3 5 3 5 x 2 3 x 1 3 2x 1 5 x 1 4 (e) 2 5 1 (f) 2 5 2 4 3 4 3

7.3 Solving inequalities involving brackets

Inequalities can also involve brackets.

Remember that there is usually more than one answer when you solve an inequality and you need to state all possible values of the solution set.

EXAMPLE 13

Solve these inequalities. You must remember to keep the (a) 9 < 3(y 2 1) (b) 3(2x 2 5) . 2(x 1 4) inequality sign in your answer. For example, if you leave (a) as (a) 9 < 3(y 2 1) (b) 3(2x 2 5) . 2(x 1 4) 4 5 y you will lose a mark because you have not included all possible 9 < 3y 2 3 6x 2 15 . 2x 1 8 values of y. 9 1 3 < 3y 6x 2 2x . 8 1 15 12 < 3y 4x . 23 If you are asked for integer solutions to the final answer will 23 (b) 4 < y x . 4 be x > 6. 3 x . 54

Algebra 109

M07_CME_SB_IGCSE_6867_U07.indd 109 28/8/09 11:35:34 EXAMPLE 14

n is an integer. List the values of n such that 211 , 2(n 2 3) , 1.

This is a double inequality. 211 , 2(n – 3) , 1 Expand the bracket. 211 , 2n – 6 , 1 Add 6 throughout. 2 11 1 6 , 2n , 1 1 6 2 , , 5 2n 7 Remember to list the integer 22.5 , n , 3.5 solutions as you were asked to in the question. Values of n are 22, 21, 0, 1, 2, 3 Remember to include 0.

EXERCISE 7F

1 Solve these inequalities. (a) 2(x 2 7) < 8 (b) 7 , 2(m 1 5) (c) 4(3w 2 1) . 20 (d) 3(2y 1 1) < 215 (e) 2(p 2 3) . 4 1 3p (f) 1 2 5k , 2(5 1 2k) (g) 5(x 2 1) > 3(x 1 2) (h) 2(n 1 5) < 3(2n 2 2) 2 Solve these inequalities then list the integer solutions. (a) 24 < 2x < 8 (b) 26 , 3y , 15 (c) 28 < 4n , 17 (d) 212 , 6m < 30 (e) 25 , 2(t 1 1) , 7 (f) 23 , 3(x 2 4) , 6 (g) 26 < 5(y 1 1) < 11 (h) 217 , 2(2x 2 3) < 10

7.4 Factorising by removing a common factor

Factorising an algebraic expression is the opposite of expanding brackets. To factorise an expression, look for a common factor – HCF means highest common factor. that is, a number that divides into all the terms in the expression. To factorise completely, use the HCF of the terms. 2 is a factor of 6x. 2 is also a factor of 10. So 2 is a common factor of 6x For example, and 10. 6x 1 10 can be written as 2(3x 1 5) Notice that the common factor is because 6x 5 2 × 3x the term outside the bracket. and 10 5 2 × 5

110 Algebra

M07_CME_SB_IGCSE_6867_U07.indd 110 28/8/09 11:35:35 Expanding brackets and factorising

EXAMPLE 15

Copy and complete these. (a) 3t 1 15 5 3(h 1 5) (b) 4n 1 12 5 h(n 1 3)

Because 3 × t 5 3t (and 3 × 5 5 15) (a) 3t 1 15 5 3(t 1 5) (b) 4n 1 12 5 4(n 1 3) 4 × n 5 4n and 4 × 3 5 12

EXAMPLE 16

Factorise these expressions. (a) 5a 1 20 (b) 4x 2 12 (c) x2 1 7x (d) 6p2q2 2 9pq3 5 is a factor of 5a 5a 5 5 × a 5 is also a factor of 20 20 5 5 × 4 (a) 5a 1 20 5 5 × a 1 5 × 4 So 5 is a common factor of 5a and 5 5(a 1 4) 20 and is the term outside the bracket. 5 5(a 1 4) Check your answer by removing the Check 5(a 1 4) 5 5 × a 1 5 × 4 brackets. 5 5a 1 20 3 (b) 4x 2 12 5 4 × x 2 4 × 3 2 is a factor of 4x and 12. 4 is also a common factor of 4x and 5 4(x 2 3) 12. Use 4 because it is the highest 5 4(x 2 3) common factor (HCF) of 4x and 12. Always look for the HCF. (c) x2 1 7x 5 x × x 1 x × 7 5 x(x 1 7) x is a common factor of x2 and 7x. 5 x(x 1 7) (d) 6p2q2 2 9pq3 The HCF is 3pq2. 5 3 × 2 × p × p × q × q 2 3 × 3 × p × q × q × q 5 3pq2(2p 2 3q) 5 3pq2(2p 2 3q)

EXERCISE 7G

1 Copy and complete these. Use Example 15 to help. (a) 3x 1 15 5 3(h 1 5) (b) 5a 1 10 5 5(h 1 2) (c) 2x 2 12 5 2(x 2 h) (d) 4m 2 16 5 4(m 2 h) Don’t forget to check your answers by removing the brackets. (e) 4t 1 12 5 h(t 1 3) (f) 3n 1 18 5 h(n 1 6) (g) 2b 2 14 5 h(b 2 7) (h) 4t 2 20 5 h(t 2 5)

Algebra 111

M07_CME_SB_IGCSE_6867_U07.indd 111 28/8/09 11:35:37 2 Factorise these expressions. Use Example 16(a) to help. (a) 5p 1 20 (b) 2a 1 12 (c) 3y 1 15 Remember 5 5 5 × 1 (d) 7b 1 21 (e) 4q 1 12p (f) 6k 1 24l 3 Factorise these expressions. (a) 4t 2 12 (b) 3x 2 9 (c) 5n 2 20 (d) 2b 2 8 (e) 6a 2 18b (f) 7k 2 7

4 Factorise these expressions. Use Example 16(c) to help. (a) y2 1 7y (b) x2 1 5x (c) n2 1 n Remember n 5 n × 1 (d) x2 2 7x (e) p2 2 8p (f) a2 2 ab 5 Factorise these expressions. Remember 6p 5 2 × 3p (a) 6p 1 4 (b) 4a 1 10 (c) 6 2 4t (d) 12m 2 8n (e) 25x 1 15y (f) 12y 2 9z 6 Factorise completely. (a) 3x2 2 6x (b) 8x2 2 xy (c) 8a 1 4ab (d) p3 2 5p2 (e) 3t3 1 6t2 (f) 10yz 2 15y2 (g) 18a2 1 12ab (h) 16p2 2 12pq 7 Factorise these expressions. 2 3 2 (a) 4ab 1 6ab (b) 10xy 2 5x Use Example 16(a) to help. Look for (c) 3p2q 2 6p3q2 (d) 8mn3 1 4n2 2 6m2n the common factors in the terms. (e) 6h2k 2 12hk3 2 18h2k2

Expanding two brackets You can use a grid method to multiply two numbers. For example, 34 × 57 34 × 57 5 (30 1 4) × (50 1 7) × 50 7 5 30 × 50 1 30 × 7 1 4 × 50 1 4 × 7 30 1500 210 5 1500 1 210 1 200 1 28 4 200 28 5 1938

You can also use a grid method when you multiply two brackets together. You have to multiply each term in one bracket by each term in the other bracket. For example, To expand and simplify (x 1 2)(x 1 5) × x 5 x x2 5x 2 2x 10 (x 1 2)(x 1 5) 5 x × x 1 x × 5 1 2 × x 1 2 × 5 You simplify the final expression by 5 x2 1 5x 1 2x 1 10 collecting the like terms. 5 x2 1 7x 1 10 5x 1 2x 5 7x.

112 Algebra

M07_CME_SB_IGCSE_6867_U07.indd 112 28/8/09 11:35:37 Expanding brackets and factorising

It is like working out the area of x 5 a rectangle of length x 1 5 and 1 area = x × x width x 2. x area = x × 5 = x2 = 5x Total area 5 (x 1 2)(x 1 5) 5 x2 1 5x 1 2x 1 10 2 area = 2 × x area = 2 × 5 5 x2 1 7x 1 10 = 2x = 10

EXAMPLE 17

Expand and simplify these. (a) (a 1 4)(a 1 10) (b) (t 1 6)(t 2 2)

(a) × a 10 Remember you can use a grid to 2 a a 10a help. 4 4a 40 (a 1 4)(a 1 10) 5 a × a 1 a × 10 1 4 × a 1 4 × 10 5 a2 1 10a 1 4a 1 40 Remember to multiply each term in the first bracket by each term in the 2 5 a 1 14a 1 40 second bracket.

(b) × t 22 Remember you are multiplying t t2 22t by 22. 1 2 2 6 6t 212 ve × ve 5 ve. (t 1 6)(t 2 2) 5 t × t 1 t × (22) 1 6 × t 1 6 × (22) 5 t2 2 2t 1 6t 2 12 5 t2 1 4t 2 12

Look again at the last example.

5 t2 2 2t 1 6t 2 12 � � (t 6) (t 2) 5 t2 1 4t 2 12

The First terms in each bracket multiply to give t2. The Outside pair of terms multiply to give 22t. This method is often known as FOIL 1 and is another way of expanding The Inside pair of terms multiply to give 6t. brackets. The Last terms in each bracket multiply to give 212.

Algebra 113

M07_CME_SB_IGCSE_6867_U07.indd 113 28/8/09 11:35:38 EXERCISE 7H

Expand and simplify. 1 (a 1 2)(a 1 7) 2 (x 1 3)(x 1 1) 3 (x 1 5)(x 1 5) 4 (t 1 5)(t 2 2) 5 (x 1 7)(x 2 4) 6 (n 2 5)(n 1 8) Be careful when there are negative signs 2 this is where a lot of 7 (x 2 4)(x 1 5) 8 (p 2 4)(p 1 4) 9 (x 2 9)(x 2 4) mistakes are made. 10 (h 2 3)(h 2 8) 11 (y 2 3)(y 2 3) 12 (4 1 a)(a 1 7) 13 (m 2 7)(8 1 m) 14 (6 1 q)(7 1 q) 15 (d 1 5)(4 2 d) 16 (8 2 x)(3 2 x) 17 (x 2 12)(x 2 7) 18 (y 2 16)(y 1 6)

Squaring an expression You can use the same method of expanding two brackets for examples involving the square of an expression.

To square an expression, write out the bracket twice and expand.

EXAMPLE 18

You need to multiply the expression 1 2 Expand and simplify (x 4) . (x 1 4) by itself so write down the bracket twice and expand as you did in Example 17 or use FOIL as in this example. (x � 4) (x � 4) Notice that you do not just square the and the 4, there are two other (x 1 4)2 5 (x 1 4)(x 1 4) x terms in the expansion. 5 x2 1 4x 1 4x 1 16 5 x2 1 8x 1 16

EXERCISE 7I

1 Expand and simplify. 2 2 2 (a) (x 1 5) (b) (x 1 6) (c) (x 2 3) In question 1, see if you can spot (d) (x 1 1)2 (e) (x 2 4)2 (f) (x 2 5)2 the pattern between the terms in 2 2 2 the brackets and the final (g) (x 1 7) (h) (x 2 8) (i) (3 1 x) expression. (j) (2 1 x)2 (k) (5 2 x)2 (l) (x 1 a)2 2 Copy and complete these by finding the correct number to go in each box. (a) (x 1 h)2 5 x2 1 hx 1 36 (b) (x 2 h)2 5 x2 2 hx 1 49 (c) (x 1 h)2 5 x2 1 18x 1 h (d) (x 2 h)2 5 x2 2 20x 1 h

114 Algebra

M07_CME_SB_IGCSE_6867_U07.indd 114 28/8/09 11:35:40 Expanding brackets and factorising

3 Expand and simplify. (a) (x 1 4)(x 2 4) (b) (x 1 5)(x 2 5) (c) (x 1 2)(x 2 2) What happens to the x term when (d) (x 2 11)(x 1 11) (e) (x 2 3)(x 1 3) (f) (x 2 1)(x 1 1) you multiply brackets of the form (x 1 a)(x 2 a)? (g) (x 1 9)(x 2 9) (h) (x 1 a)(x 2 a) (i) (t 1 x)(t 2 x)

EXAMPLE 19

Expand and simplify (3x 2 y)(x 2 2y). Remember to multiply each term in the first bracket by each term in the second bracket. × x 22y 3x 3x2 26xy Be careful when there are negative 2y 2xy 2y2 signs. This is where a lot of mistakes are made. (3x 2 y)(x 2 2y) 5 3x2 2 6xy 2 xy 1 2y 1ve × 2ve 5 2ve. 2 2 1 5 3x2 2 7xy 1 2y2 ve × ve 5 ve.

EXERCISE 7J

Expand and simplify. 1 (3a 1 2)(a 1 4) 2 (5x 1 3)(x 1 2) 3 (2t 1 3)(3t 1 5) 4 (4y 1 1)(2y 1 7) 5 (6x 1 5)(2x 1 3) 6 (4x 1 3)(x 2 1) 7 (2z 1 5)(3z 2 2) 8 (y 1 1)(7y 2 8) 9 (3n 2 5)(n 1 8) 10 (3b 2 5)(2b 1 1) 11 (p 2 4)(7p 1 3) 12 (2z 2 3)(3z 2 4) 13 (5x 2 9)(2x 2 1) 14 (2y 2 3)(2y 2 3) 15 (2 1 3a)(4a 1 5) 1 2 2 2 2 2 16 (3x 4) 17 (2x 7) 18 (5 4x) Can you see the connection 19 (2x 1 1)(2x 2 1) 20 (3y 1 2)(3y 2 2) 21 (5n 1 4)(5n 2 4) between questions 19–24 and question 3 in Exercise 7I ? 22 (3x 1 5)(3x 2 5) 23 (1 1 2x)(1 2 2x) 24 (3t 1 2x)(3t 2 2x)

Algebra 115

M07_CME_SB_IGCSE_6867_U07.indd 115 28/8/09 11:35:41 EXAMINATION QUESTIONS

1 Solve the inequality 7 2 5x > 217, given that x is a positive integer. [3]

(CIE Paper 2, Nov 2000)

2 Solve the inequality 25 2 3x , 7. [2]

(CIE Paper 2, Jun 2001)

3 Solve the inequality 3(x 1 7) , 5x 2 9. [2]

(CIE Paper 2, Jun 2002)

2x 1 x 4 (a) Solve the inequality 5 2 > + . [3] 3 2 4 2x 1 x (b) List the positive integers which satisfy the inequality 5 2 > + . [1] 3 2 4 (CIE Paper 2, Nov 2002)

x 5 Solve the equation 2 8 5 22. [2] 4 (CIE Paper 2, Jun 2004)

6 (a) Factorise completely 12x2 2 3y2. [2] (b) (i) Expand (x 2 3)2. [2] (ii) x2 2 6x 1 10 is to be written in the form (x 2 p)2 1 q. Find the values of p and q. [2]

(CIE Paper 2, Jun 2004)

3x 2 2 7 Solve the equation 5 8. [2] 5 (CIE Paper 2, Nov 2004)

116 Algebra

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