# Expanding brackets and factorising

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1 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 x x y y y 2 gh g h More complicated multiplications can also be simplified. EXAMPLE 1 Simplify f 4g. f 4g f 4 g 4 f g 12 fg 12fg To multiply algebraic terms, multiply the numbers then multiply the letters. Multiplying a bracket You can work out 6 4 by thinking of 4 as (0 1 4) Algebra 101

2 Brackets are often used in algebra. 6(x 1 4) means 6 (x 1 4) As in the 6 4 example, you have to multiply each term inside the brackets by 6. 6(x 1 4) 6 x x 1 24 It is like working out the area of a rectangle that has length x 1 4 and width 6 x 4 6 area = 6 x = 6x area = 6 4 = 24 Total area 6(x 1 4) 6 x x 1 24 When you do this it is called expanding the brackets. It is also known as removing the brackets or multiplying out the brackets. You find the total area by adding the area of the two smaller rectangles. 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) (a 1 6) (b) 2(x 2 8) (c) (2c 2 d) (a) (a 1 6) a 1 6 a 1 0 (b) 2(x 2 8) 2 x x 2 16 (c) (2c 2 d) 2c 2 d 6c 2 d You must multiply each term inside the bracket by the term outside the bracket. 2c 2 c 6 c 6c 102 Algebra

3 Expanding brackets and factorising EXERCISE 7A 1 Simplify these expressions. (a) 2 k (b) 6b (c) 4a (d) a 2b (e) 4c d (f) x y 2 Expand the brackets to find the value of these expressions. Check your answers by working out the brackets first. (a) 2(0 1 7) (b) (40 1 6) (c) 6(70 1 ) (d) (40 2 2) (e) 7(0 2 4) (f) 8(40 2 ) Remove the brackets from these. (a) (p 1 6) (b) (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. (a) (2c 1 6) (b) (4t 1 ) (c) 2(p 1 q) (d) (2a 2 b) (e) 6(c 2 2d) (f) 7(2x 1 y 2 ) (g) 6(a 2 4b 1 c) (h) 2(x 2 1 x 1 2) (i) 4(y 2 2 y 2 10) Write down the 6 pairs of cards which show equivalent expressions. Remember, you must multiply each term inside the brackets by the term outside the bracket. A common mistake is to forget to multiply the second term. Remember 2c 2 c 6c 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 Expand the brackets in these expressions. (a) a(a 1 4) (b) x(2x 2 y) (c) t(t 2 1 1) (a) a(a 1 4) a a 1 a 4 a 2 1 4a (b) x(2x 2 y) x 2x 2 x y 2x 2 2 xy (c) t(t 2 1 1) t t 2 1 t 1 t 1 t Remember a a a 2 x 2x x 2 x 2 x x 2x 2 Remember x y xy t t 2 t t t t Algebra 10

4 EXERCISE 7B Expand the brackets in these expressions. 1 b(b 1 4) 2 a( 1 a) k(k 2 6) 4 m(9 2 m) a(2a 1 ) 6 g(4g 1 1) 7 p(2p 1 q) 8 t(t 1 w) 9 m(m 1 n) 10 x(2x 2 y) 11 r(4r 2 t) 12 a(a 2 4b) 1 2t(t 1 ) 14 x(x 2 8) 1 k(k 1 l) 16 a(2a 1 4) 17 2g(4g 1 h) 18 p(p 2 2q) 19 x(2y 1 z) 20 r(r 2 1 1) 21 a(a 2 1 ) 22 t(t 2 7) 2 2p(p 2 1 q) 24 4x(x 2 1 x) Remember x 4x 4 x x 12x 2 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) (a 1 4) 1 2a 1 10 (b) (2x 1 ) 1 2(x 2 4) (a) (a 1 4) 1 2a 1 10 a a 1 10 a 1 2a a 1 22 (b) (2x 1 ) 1 2(x 2 4) 6x x 2 8 6x 1 2x x 1 7 Expand the brackets first. Then collect like terms. Expand both sets of brackets first. Subtracting If you have an expression like 2(2x 2 ), multiply both terms in the brackets by x 26x and So 2(2x 2 ) 2 2x x 1 1 Multiplying 104 Algebra

5 Expanding brackets and factorising EXAMPLE Expand these expressions. (a) 22(t 1 4) (b) 2(4x 2 1) (a) 22(t 1 4) 22 t t t 2 8 (b) 2(4x 2 1) 2 4x x EXAMPLE 6 Expand then simplify these expressions. (a) (2t 1 1) 2 2(2t 1 4) (b) 8(x 1 1) 2 (2x 2 ) (a) (2t 1 1) 2 2(2t 1 4) 6t 1 2 4t 2 8 6t 2 4t t 2 (b) 8(x 1 1) 2 (2x 2 ) 8x x 1 1 8x 2 6x x 1 2 Remember to multiply both terms in the second bracket by 22. Expand the brackets first. Remember that Then collect like terms. EXERCISE 7C Expand these expressions. 1 22(2k 1 4) 2 2(2x 1 6) 2(n 2 1) 4 24(t 1 ) 2(4p 2 1) 6 22(x 2 7) Expand then simplify these expressions. 7 (y 1 4) 1 2y (k 1 6) 1 k (a 1 ) 2 2a (t 2 2) 1 4t (2y 1 ) 1 2(y 1 ) 12 4(x 1 7) 1 (x 1 4) 1 (2x 1 ) 1 2(x 2 4) 14 2(4n 1 ) 1 (n 2 ) 1 (x 2 ) 1 2(x 2 ) 16 4(2x 2 1) 1 2(x 2 2) 17 (2b 1 1) 2 2(2b 1 4) 18 4(2m 1 ) 2 2(2m 1 ) 19 2(t 1 ) 2 2(t 1 1) 20 (2k 1 2) 2 4(2k 1 6) 21 8(a 1 1) 2 (2a 2 ) 22 2(4p 1 1) 2 4(p 2 ) 2 (2g 2 4) 2 2(4g 2 6) 24 2(w 2 4) 2 (2w 2 1) 2 x(x 1 ) 1 4(x 1 2) 26 x(2x 1 1) 2 (x 2 4) Algebra 10

6 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 ) 20. Method A 4(c 1 ) 20 4c c c 8 4c c 2 Method B 4(c 1 ) 20 c c 1 c 2 c 2 Expand the bracket by multiplying both terms inside the bracket by the term outside the bracket. You must subtract 12 from both sides before dividing both sides by 4. Since 4 divides exactly into 20 you can divide both sides by 4 first. EXAMPLE 8 Solve 2(p 2 4) 7. 2(p 4) 7 6p 8 7 6p p 1 6p p 2. Expand the bracket. You must add 8 to both sides before dividing both sides by Algebra

7 Expanding brackets and factorising EXERCISE 7D 1 Solve these equations. (a) 4(g 1 6) 2 (b) 7(k 1 1) 21 (c) (s 1 10) 6 (d) 2(n 2 4) 6 (e) (f 2 2) 24 (f) 6(v 2 ) 42 (g) 4(m 2 ) 14 (h) 2(w 1 7) 19 2 Solve these equations. (a) 4(t 1 2) 48 (b) (2r 1 4) 0 (c) 2(2b 1 2) 22 (d) 2(w 2 6) 27 (e) (4x 2 2) 24 (f) (2y 1 11) 40 (g) 6(2k 2 1) 6 (h) (2a 2 1) 18 When two brackets are involved, expand both brackets then collect like terms before solving. Like terms are terms of the same kind. In Example 9 there are only terms in m and number terms. EXAMPLE 9 Solve 2(2m 1 10) 12(m 2 1). 2(2m 1 10) 12(m 2 1) 4m m m 2 4m 2 8m m 4 Expand the brackets on both sides of the equation and collect like terms. Collect terms in m on the RHS because 12m on the RHS is greater than 4m on the LHS. This keeps the m term positive. When an equation involves fractions, it can be transformed into an equation without fractions by multiplying all terms by the LCM of the numbers in the denominators. LCM means Lowest Common Multiple. EXAMPLE 10 Solve the equation x x 1 2. x 1 17 x (x 1 17) 4(x 1 2) 4 x x x 2 x 9 x x Multiply both sides by 4, collect like terms and then finally divide by. Note the use of brackets. Collect the terms in x on the RHS and the numbers on the LHS. 4x on the RHS is greater than x on the LHS. Algebra 107

8 EXAMPLE 11 Solve the equation x 2 6 x 2 6 x 1 4 1(x 2 6) 1(x 1 4) (x 6) (x 1 4) x 2 0 x 1 12 x 2 x x 42 x 21 x 1 4. Look at the denominators. and have a LCM of 1 so multiply both sides of the equation by 1. Note the use of brackets. Always put them in when you multiply in this way. Then solve using the method shown in Example 9. EXAMPLE 12 Solve the equation 2x x (2x 1 ) 1 6(x 2 2) 6() 6 2 2x 1 1 2(x 2 2) 1 2x 1 1 2x x x 16 x 4 The LCM here is 6. Note the use of brackets. The most common mistake is to forget to multiply all terms by the LCM. 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 (a 2 1) (b) (d 2 2) 2d 2 1 (c) (x 1 ) 11x 1 (d) 12p 1 (p 1 7) (e) 4t 1 (2t 2 ) (f) b 2 4 2(2b 2 7) (g) 8(g 2 1) 1g 1 10 (h) (2k 1 6) 17k 1 7 (i) 2(y 1 ) y 1 12 (j) r 1 4(2r 1 ) 2 Solve the following equations by expanding both brackets. (a) 2(b 1 1) 8(2b 2 ) (b) (4a 1 7) (8a 1 9) (c) 6(x 2 2) (x 2 8) (d) (2p 1 2) 6(p 1 ) (e) 9(s 2 4) (4s 2 ) (f) 4(10t 2 7) (6t 2 2) (g) 4(2w 1 2) 2(w 1 7) (h) (y 2 2) 7(y 2 2) Use Example 9 to help. 108 Algebra

9 Expanding brackets and factorising Solve these equations. (a) d 1 1 (c) x (e) c Solve these equations. (a) x 1 1 x (c) x 1 1 (e) x d (b) 6y 2 x 2 2 (d) 6 1 a 2 c 1 1 (f) 10 2 b 2x x 1 6 Solve these equations. (a) x x (c) x 1 2 (e) x x x 1 (b) 2x 2 1 (d) x 1 2 (f) 8 2 x 2 (b) x (d) x (f) 2x 1 4 y 1 a b x 2 x 2 2x x x x Use Example 10 to help. Use Example 11 to help. Use Example 12 to help. 7. 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 1 Solve these inequalities. (a) 9 < (y 2 1) (b) (2x 2 ). 2(x 1 4) (a) 9 < (y 2 1) (b) (2x 2 ). 2(x 1 4) 9 < y 2 6x x < y 6x 2 2x < y 4x. 2 4 < y x. 2 4 x. 4 You must remember to keep the inequality sign in your answer. For example, if you leave (a) as 4 y you will lose a mark because you have not included all possible values of y. If you are asked for integer solutions to (b) the final answer will be x > 6. Algebra 109

10 EXAMPLE 14 n is an integer. List the values of n such that 211, 2(n 2 ), , 2(n ), 1 211, 2n 6, , 2n, , 2n, 7 22., n,. Values of n are 22, 21, 0, 1, 2, This is a double inequality. Expand the bracket. Add 6 throughout. Remember to list the integer solutions as you were asked to in the question. Remember to include 0. EXERCISE 7F 1 Solve these inequalities. (a) 2(x 2 7) < 8 (b) 7, 2(m 1 ) (c) 4(w 2 1). 20 (d) (2y 1 1) < 21 (e) 2(p 2 ). 4 1 p (f) 1 2 k, 2( 1 2k) (g) (x 2 1) > (x 1 2) (h) 2(n 1 ) < (2n 2 2) 2 Solve these inequalities then list the integer solutions. (a) 24 < 2x < 8 (b) 26, y, 1 (c) 28 < 4n, 17 (d) 212, 6m < 0 (e) 2, 2(t 1 1), 7 (f) 2, (x 2 4), 6 (g) 26 < (y 1 1) < 11 (h) 217, 2(2x 2 ) < 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 that is, a number that divides into all the terms in the expression. To factorise completely, use the HCF of the terms. For example, 6x 1 10 can be written as 2(x 1 ) because 6x 2 x and 10 2 HCF means highest common factor. 2 is a factor of 6x. 2 is also a factor of 10. So 2 is a common factor of 6x and 10. Notice that the common factor is the term outside the bracket. 110 Algebra

11 Expanding brackets and factorising EXAMPLE 1 Copy and complete these. (a) t 1 1 (h 1 ) (b) 4n 1 12 h(n 1 ) (a) t 1 1 (t 1 ) (b) 4n (n 1 ) Because t t (and 1) 4 n 4n and 4 12 EXAMPLE 16 Factorise these expressions. (a) a 1 20 (b) 4x 2 12 (c) x 2 1 7x (d) 6p 2 q 2 2 9pq (a) a 1 20 a 1 4 (a 1 4) (a 1 4) Check (a 1 4) a 1 4 a 1 20 (b) 4x x 2 4 4(x 2 ) 4(x 2 ) (c) x 2 1 7x x x 1 x 7 x(x 1 7) x(x 1 7) (d) 6p 2 q 2 2 9pq 2 p p q q 2 p q q q pq 2 (2p 2 q) pq 2 (2p 2 q) is a factor of a a a is also a factor of So is a common factor of a and 20 and is the term outside the bracket. Check your answer by removing the brackets. 2 is a factor of 4x and is also a common factor of 4x and 12. Use 4 because it is the highest common factor (HCF) of 4x and 12. Always look for the HCF. x is a common factor of x 2 and 7x. The HCF is pq 2. EXERCISE 7G 1 Copy and complete these. (a) x 1 1 (h 1 ) (b) a 1 10 (h 1 2) (c) 2x (x 2 h) (d) 4m (m 2 h) (e) 4t 1 12 h(t 1 ) (f) n 1 18 h(n 1 6) (g) 2b 2 14 h(b 2 7) (h) 4t 2 20 h(t 2 ) Use Example 1 to help. Don t forget to check your answers by removing the brackets. Algebra 111

12 2 Factorise these expressions. (a) p 1 20 (b) 2a 1 12 (c) y 1 1 (d) 7b 1 21 (e) 4q 1 12p (f) 6k 1 24l Factorise these expressions. (a) 4t 2 12 (b) x 2 9 (c) n 2 20 (d) 2b 2 8 (e) 6a 2 18b (f) 7k Factorise these expressions. (a) y 2 1 7y (b) x 2 1 x (c) n 2 1 n (d) x 2 2 7x (e) p 2 2 8p (f) a 2 2 ab Factorise these expressions. (a) 6p 1 4 (b) 4a 1 10 (c) 6 2 4t (d) 12m 2 8n (e) 2x 1 1y (f) 12y 2 9z 6 Factorise completely. (a) x 2 2 6x (b) 8x 2 2 xy (c) 8a 1 4ab (d) p 2 p 2 (e) t 1 6t 2 (f) 10yz 2 1y 2 (g) 18a ab (h) 16p pq 7 Factorise these expressions. (a) 4ab 2 1 6ab (b) 10xy 2 x 2 (c) p 2 q 2 6p q 2 (d) 8mn 1 4n 2 2 6m 2 n (e) 6h 2 k 2 12hk 2 18h 2 k 2 Use Example 16(a) to help. Remember 1 Use Example 16(c) to help. Remember n n 1 Remember 6p 2 p Use Example 16(a) to help. Look for the common factors in the terms. Expanding two brackets You can use a grid method to multiply two numbers. For example, (0 1 4) (0 1 7) 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 ) x x x 2 x 2 2x 10 (x 1 2)(x 1 ) x x 1 x 1 2 x 1 2 x 2 1 x 1 2x 1 10 x 2 1 7x 1 10 You simplify the final expression by collecting the like terms. x 1 2x 7x. 112 Algebra

13 Expanding brackets and factorising It is like working out the area of a rectangle of length x 1 and width x 1 2. Total area (x 1 2)(x 1 ) x 2 1 x 1 2x 1 10 x 2 1 7x 1 10 x 2 x area = x x = x 2 area = x = x area = 2 x = 2x area = 2 = 10 EXAMPLE 17 Expand and simplify these. (a) (a 1 4)(a 1 10) (b) (t 1 6)(t 2 2) (a) a 10 a a 2 10a 4 4a 40 (a 1 4)(a 1 10) a a 1 a a a a 1 4a 1 40 a a 1 40 (b) t 22 t t 2 22t 6 6t 212 (t 1 6)(t 2 2) t t 1 t (22) 1 6 t 1 6 (22) t 2 2 2t 1 6t 2 12 t 2 1 4t 2 12 Remember you can use a grid to help. Remember to multiply each term in the first bracket by each term in the second bracket. Remember you are multiplying by ve 2 ve 2 ve. Look again at the last example. (t 6) (t 2) t 2 2 2t 1 6t 2 12 t 2 1 4t 2 12 The First terms in each bracket multiply to give t 2. The Outside pair of terms multiply to give 22t. The Inside pair of terms multiply to give 16t. The Last terms in each bracket multiply to give 212. This method is often known as FOIL and is another way of expanding brackets. Algebra 11

14 EXERCISE 7H Expand and simplify. 1 (a 1 2)(a 1 7) 2 (x 1 )(x 1 1) (x 1 )(x 1 ) 4 (t 1 )(t 2 2) (x 1 7)(x 2 4) 6 (n 2 )(n 1 8) 7 (x 2 4)(x 1 ) 8 (p 2 4)(p 1 4) 9 (x 2 9)(x 2 4) 10 (h 2 )(h 2 8) 11 (y 2 )(y 2 ) 12 (4 1 a)(a 1 7) 1 (m 2 7)(8 1 m) 14 (6 1 q)(7 1 q) 1 (d 1 )(4 2 d) 16 (8 2 x)( 2 x) 17 (x 2 12)(x 2 7) 18 (y 2 16)(y 1 6) Be careful when there are negative signs 2 this is where a lot of mistakes are made. 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 Expand and simplify (x 1 4) 2. (x 4) (x 4) (x 1 4) 2 (x 1 4)(x 1 4) x 2 1 4x 1 4x 1 16 x 2 1 8x 1 16 You need to multiply the expression (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. Notice that you do not just square the x and the 4, there are two other terms in the expansion. EXERCISE 7I 1 Expand and simplify. (a) (x 1 ) 2 (b) (x 1 6) 2 (c) (x 2 ) 2 (d) (x 1 1) 2 (e) (x 2 4) 2 (f) (x 2 ) 2 (g) (x 1 7) 2 (h) (x 2 8) 2 (i) ( 1 x) 2 (j) (2 1 x) 2 (k) ( 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 x 2 1 hx 1 6 (b) (x 2 h) 2 x 2 2 hx 1 49 (c) (x 1 h) 2 x x 1 h (d) (x 2 h) 2 x x 1 h In question 1, see if you can spot the pattern between the terms in the brackets and the final expression. 114 Algebra

15 Expanding brackets and factorising Expand and simplify. (a) (x 1 4)(x 2 4) (b) (x 1 )(x 2 ) (c) (x 1 2)(x 2 2) (d) (x 2 11)(x 1 11) (e) (x 2 )(x 1 ) (f) (x 2 1)(x 1 1) (g) (x 1 9)(x 2 9) (h) (x 1 a)(x 2 a) (i) (t 1 x)(t 2 x) What happens to the x term when you multiply brackets of the form (x 1 a)(x 2 a)? EXAMPLE 19 Expand and simplify (x 2 y)(x 2 2y). x 22y x x 2 26xy 2y 2xy 2y 2 (x 2 y)(x 2 2y) x 2 2 6xy 2 xy 1 2y x 2 2 7xy 1 2y 2 Remember to multiply each term in the first bracket by each term in the second bracket. Be careful when there are negative signs. This is where a lot of mistakes are made. 1 ve 2 ve 2 ve. 2 ve 2 ve 1 ve. EXERCISE 7J Expand and simplify. 1 (a 1 2)(a 1 4) 2 (x 1 )(x 1 2) (2t 1 )(t 1 ) 4 (4y 1 1)(2y 1 7) (6x 1 )(2x 1 ) 6 (4x 1 )(x 2 1) 7 (2z 1 )(z 2 2) 8 (y 1 1)(7y 2 8) 9 (n 2 )(n 1 8) 10 (b 2 )(2b 1 1) 11 (p 2 4)(7p 1 ) 12 (2z 2 )(z 2 4) 1 (x 2 9)(2x 2 1) 14 (2y 2 )(2y 2 ) 1 (2 1 a)(4a 1 ) 16 (x 1 4) 2 17 (2x 2 7) 2 18 ( 2 4x) 2 19 (2x 1 1)(2x 2 1) 20 (y 1 2)(y 2 2) 21 (n 1 4)(n 2 4) 22 (x 1 )(x 2 ) 2 (1 1 2x)(1 2 2x) 24 (t 1 2x)(t 2 2x) Can you see the connection between questions and question in Exercise 7I? Algebra 11

16 EXAMINATION QUESTIONS 1 Solve the inequality 7 2 x > 217, given that x is a positive integer. [] (CIE Paper 2, Nov 2000) 2 Solve the inequality 2 2 x, 7. [2] (CIE Paper 2, Jun 2001) Solve the inequality (x 1 7), x 2 9. [2] (CIE Paper 2, Jun 2002) 4 (a) Solve the inequality 2 2x > x 4. [] (b) List the positive integers which satisfy the inequality 2 2x > x 4. [1] (CIE Paper 2, Nov 2002) Solve the equation x [2] 4 (CIE Paper 2, Jun 2004) 6 (a) Factorise completely 12x 2 2 y 2. [2] (b) (i) Expand (x 2 ) 2. [2] (ii) x 2 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) 7 Solve the equation x 2 2 (CIE Paper 2, Nov 2004) 8. [2] 116 Algebra

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