Chapter 9 Algebraic Techniques PDF

Chapter 9 Algebraic techniques Essential mathematics: why skills with algebra are important Algebra skills are essential when applying formulas in business, the professions and the trades, such as...

Chapter 9 Algebraic techniques Essential mathematics: why skills with algebra are important Algebra skills are essential when applying formulas in business, the professions and the trades, such as the air-conditioning, aviation, construction, electrical, electronic, manufacturing, mechanic, mechanical, metal working, plumbing, retail and welding trades. Apprenticeship training for the electrical trades involves using many formulas requiring algebraic techniques including fractions. Nurses use algebraic techniques including fractions to determine medical dosage amounts and intravenous fluid quantities. Financial mathematical analysis is a key to success for businesses. Algebraic skills are required to apply formulas that calculate the financial aspects of a business, including expenses, revenue, losses, profits, GST, wages, tax amounts, insurance and any loan repayments. Examples of small businesses that require financial analysis include hairdressers, bakers, cake makers, café or food truck owners, dog groomers, fashion designers, florists, food delivery services, personal fitness trainers and numerous technology start-ups. Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  In this chapter Online 3 9A Reviewing algebra resources (Consolidating) 9B Expanding binomial products 9C Expanding perfect squares 9D Forming a difference of perfect squares 9E Factorising algebraic expressions 9F Simplifying algebraic fractions: multiplication and division 9G Simplifying algebraic fractions: addition and subtraction Victorian Curriculum NUMBER AND ALGEBRA Patterns and algebra Apply the distributive law to the expansion of algebraic expressions, including binomials, and collect like terms where appropriate (VCMNA306) © Victorian Curriculum and Assessment Authority (VCAA) Online resources A host of additional online resources are included as part of your Interactive Textbook, including HOTmaths content, video demonstrations of all worked examples, auto-marked quizzes and much more. Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party. 482 Chapter 9 Algebraic techniques 1 Write down the coefficient of x in these expressions. Warm-up quiz a 4x − 1 b 3y + 6x c 3xy − 2x d −4x2 − 7x 2 What is the constant term in these expressions? a 5 − 2x b −16a + 2 c 5xy − 2x + 1 d 101x − 6 3 Evaluate the following if a = 2 and b = −5. a ab b −3b + 1 c 4−b+a d 5a2 b —2 e a f 2b2 − a g 5 − b2 h a2 b − 2b2 Ë 4 Expand the following using a(b + c) = ab + ac. a 2(x + 3) b 3(a − 5) c 4x(3 − 2y) d −3(2b − 1) 5 Write down the highest common factor of: a 4 and 6 b 12 and 18 c 2x and 4x d 3xy and 9y e 10x and 15x2 f 3a2 b and 4ab2 6 Factorise by taking out the highest common factor. a 2a + 6 b 3x + 12y c 5x2 − 15x d 4m − 6mn 7 Add or subtract these fractions. You will need a common denominator for parts c and d. a 3+2 b 4−5 c 1+2 d 3−1 7 7 9 9 2 3 8 2 8 Multiply or divide these fractions. a 2×4 b 3×2 c 7 ×2 d 6 × 22 3 5 4 3 14 3 11 12 e 2÷1 f 7 ÷ 14 g 3÷4 h 7÷4 3 3 8 24 2 3 9 3 9 Expand and simplify. a 3(x − 1) + 5 b 4(1 − x) + 5x c −2(5 + x) − x 10 Write two expressions for the area of this rectangle, one with brackets and one without. x 2 Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  9A Reviewing algebra 483 9A 9A Reviewing algebra CONSOLIDATING Learning intentions To know the names of the parts of an algebraic expression To be able to form algebraic expressions from simple word phrases To know that only like terms can be combined under addition and subtraction To be able to simplify algebraic expressions using the four operations: +, −, × and ÷ To be able to expand expressions involving brackets To be able to evaluate expressions by substituting given values Key vocabulary: expression, pronumeral, variable, term, like terms, constant term, coefficient, distributive law, evaluate A high level of skill in algebra is required to solve more complex mathematical problems. Skills include adding, subtracting, multiplying and dividing algebraic expressions, as well as expanding and factorising expressions. In this section we will review some basic concepts in algebra. Lesson starter: Vocab review Consider the expression 3x2 + 4xy − 2 − xy. How many terms are given in the expression? What letters are used as pronumerals? What is the coefficient of x2 ? What is the constant term? Are there any like terms? Can the expression be simplified? If so, how? What would be the value of the expression if x = 2 and y = −1? Key ideas An expression is a combination of numbers and pronumerals connected by mathematical operations. A term is part of an expression with numbers and pronumerals connected only by multiplication and division A coefficient is the number part in front of a term A constant term is a term that does not contain any pronumerals This is an example of a 3-term expression 7y2 + 2xy − 4 coefficient of y2 (7) constant term (−4) Like terms have the same pronumeral part. They can be collected using addition and subtraction. For example: 5a − 7a = −2a and 3xy2 + 2y2 x = 5xy2 For multiplication and division, the symbols × and ÷ are not usually shown. For example: (−3 × a × b) × (2 × b) = −3ab × 2b = −6ab2 2 14a2 b ÷ (7ab) = 14a b = 2a 7ab The distributive law is used to expand brackets. For example: −3(x − 2) = −3x + 6 4x + 2(1 − x) = 4x + 2 − 2x = 2x + 2 Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party. 484 Chapter 9 Algebraic techniques To evaluate an expression, substitute a value for each pronumeral and simplify. For example: if a = 2 and b = −3 then b2 − a = (−3)2 − 2 =9−2 =7 Exercise 9A Understanding 1–3 3 1 Write expressions for each of the following. a The cost of: i x movie passes at $11 each ii n apples at 50 cents each b The number of tickets purchased for: i x adults and y children ii b boys and g girls c The cost of hiring an electrician for n hours if the callout fee is $50 and the cost of labour is $60 per hour 2 Consider the expression 5x − 2xy − y2 − 7. Hint: Remember to include a What is the coefficient of x? the negative sign in the b What is the coefficient of xy? coefficient of xy. c What is the constant term? d Are there any like terms? e Evaluate the expression if x = 1 and y = 2. f Evaluate the expression if x = 2 and y = −1. 3 Write expressions for the following. a The sum of 5 and 2x. Hint: ‘Product’ means multiply. b 7 less than 4a. c 1 less than the square of y. d The product of x and z. e The sum of the squares of a and b. f The square root of 8 more than x. Fluency 4–6(½) 4–6(½) Example 1 Collecting like terms Collect like terms to simplify the following. a 5y − x + 2y b 4a2 b + ab − ba2 Solution Explanation a 5y − x + 2y = 7y − x 5y and 2y are like terms. b 4a2 b + ab − ba2 = 3a2 b + ab 4a2 b and ba2 (or a2 b) are like terms and 4 − 1 = 3. Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  9A Reviewing algebra 485 Now you try Collect like terms to simplify the following. a 7m + 2n − 5m − 3n b 2x2 y − xy − 4yx2 4 Simplify by collecting like terms. a 3a − 4a b 4x + 7x c 6ab + 2ab Hint: Only collect ‘like’ terms. d 5xy − 9xy e 3a − 1 − 2a f 7y + 2x − 11y Note that ab = ba. g 5x − y − 3y + x h ab + ba i a2 b − 5a2 b − 3 j 11xy − 14yx k 10r2 a + 2ar2 l 9st2 − st − 8t2 s Example 2 Multiplying and dividing terms Simplify these expressions. a −4a × 2ab b 21a2 b ÷ (3abc) Solution Explanation a −4a × 2ab = −8a2 b −4 × 2 = −8 and a × a = a2 2 b 21a2 b ÷ (3abc) = 21a b Write as a fraction then cancel where possible. 3abc 2 a × a
1 Note: a = =a = 7a a a
1 c Now you try Simplify these expressions. a 5xy × (−3y) b 6a2 b ÷ (12ab) 5 Simplify the following. a 5 × 3x b 7a × 2 c −4x × 2y d 6a × (−7a) e −3ab × b f −6a2 b × 2b g 4x ÷ 2 h 7a ÷ a i 12a ÷ (4a) j 22a2 ÷ (11a) k 40a ÷ (4a2 ) l 100x2 y ÷ (25xy) 2 48a2 bc 12xy m 7ab n o 7b 16abc2 36xy2 Example 3 Expanding brackets Use the distributive law to expand and simplify: a −x(2 + x) b 4x − 2(x − 1) Solution Explanation a −x(2 + x) = −2x − x2 −x × 2 = −2x and −x × x = −x2 b 4x − 2(x − 1) = 4x − 2x + 2 First expand the brackets then collect like terms. = 2x + 2 Note: −2 × (−1) = 2, not −2. Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party. 486 Chapter 9 Algebraic techniques 9A Now you try Use the distributive law to expand and simplify: a −3(x + 2) b −4x(1 − x) + 2x2 6 Use the distributive law to expand and simplify: a 3(x + 2) b 2(4 + x) Hint: Remember: a negative c −3(x + 4) d −6(x + 1) times a negative is a positive. e −2(x − 3) f −x(x + 1) g −3y(2 − 3y) h −7a(a − b) i 3 + 2(x − 1) j 3x + 4(1 − x) k 3(x + 1) − 7x l 4(x + 2) − 2(x + 1) m 7(x − 3) − 3(x − 4) n −4(1 − x) − 21x + 1 o −6(2 − x) − 7(4 − x) p −2(3 − x) − (5 − x) Problem-solving and reasoning 7, 8(½), 9 8(½), 9–11 7 For these shapes, write expressions for i perimeter and ii area. a b b c a x a b c 8 Evaluate the following if a = 3, b = −2 and c = −4. a 2a + b b abc c 2a ÷ (3b) d −3c + b 2 e a +b 2 f c2 ÷ b Hint: Substitute pronumeral values and work out the answer. 2 2 g a +c h a−c i 1 (b + c) 2 7 a 3 j a −b 3 3 k 2b − c l Ëa2 + b2 9 You can hire a sports car for an upfront fee of $100 plus $80 per hour after that. a What is the cost of hiring a sports car for: i 2 hours? ii 9 hours? b Write an expression for the cost of hiring a sports car for n hours. Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  9A Reviewing algebra 487 10 Find the area of these shapes in expanded form. All angles are 90°. a 2x − 1 b c x 3 x x 2 x+3 1 11 Identify the errors, then correct to find the answer. 4a a −3(x − 1) b 3ab = 2a c = 2a d 3(x + 2) − 3(x − 1) 6b 2a2 = −3x − 3 = 3x + 6 − 3x − 3 =3 Jake vs Lucas — 12 12 Jake and Lucas operate separate computer consulting services. Jake charges $60 per hour. Lucas charges a $40 callout fee plus $50 per hour after that. a What is the cost of hiring Jake for: i 2 hours? ii 10 hours? b What is the cost of hiring Lucas for: i 2 hours? ii 10 hours? c Write an expression for the cost of hiring Jake for n hours. d Write an expression for the cost of hiring Lucas for n hours. e After what time is the cost of hiring Jake or Lucas the same? Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  488 Chapter 9 Algebraic techniques 9B 9B Expanding binomial products Learning intentions To understand the distributive law for expanding binomial products To be able to expand and simplify binomial products Key vocabulary: binomial product, distributive law, expand A binomial is an expression with two terms such as x + 5 or x2 + 3. You will recall from the previous section that we looked at the product of a single term with a binomial expression; e.g. 2(x − 3) or x(3x − 1). The product of two binomial expressions can also be expanded using the distributive law. This involves multiplying every term in one expression by every term in the other expression. Expanding the product of two expressions can be applied to problems involving the expansion of rectangular areas, such as a farmer’s paddocks. Lesson starter: Rectangular expansions If (x + 1) and (x + 2) are the side lengths of a rectangle as shown, the total area can be x 2 found as an expression in two different ways. Write an expression for the total area of the rectangle using length = (x + 2) and x width = (x + 1). Now find the area of each of the four parts of the rectangle and combine to give an 1 expression for the total area. Compare your two expressions above and complete this equation: (x + 2)( ) = x2 + +. Can you explain a method for expanding the left-hand side to give the right-hand side? Key ideas A binomial is an expression with two terms. Expanding binomial products uses the a b distributive law. (a + b)(c + d) = a(c + d) + b(c + d) c ac bc = ac + ad + bc + bd d ad bd Diagrammatically (a + b)(c + d) = ac + ad + bc + bd x 5 For example: (x + 3)(x + 5) = x2 + 5x + x + 5 x x2 5x = x2 + 6x + 5 1 1x 5 Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  9B Expanding binomial products 489 Exercise 9B Understanding 1–3 2, 3 1 The given diagram shows the area (x + 2)(x + 3). x 3 a Write down an expression for the area of each of the four regions x inside the rectangle. b Copy and complete: (x + 2)(x + 3) = + 3x + +6 2 = + 5x + 2 The given diagram shows the area (2x + 3)(x + 1). 2x 3 a Write down an expression for the area of each of the four regions inside the rectangle. x b Copy and complete: (2x + 3)( ) = 2x2 + + 3x + 1 = + + 3 Copy and complete these expansions a (x + 1)(x + 5) = + 5x + +5 Hint: The product is negative if = + 6x + there are opposite signs (+, −) or (−, +) and positive if they are b (x − 3)(x + 2) = + − 3x − the same sign (+, +) or (−, −). = −x− c (3x − 2)(7x + 2) = + 6x − − = − − d (4x − 1)(3x − 4) = − − 3x + = − 19x + Fluency 4–5(½) 4–5(½) Example 4 Expanding binomial products Expand (x + 3)(x + 5). Solution Explanation (x + 3)(x + 5) = x2 + 5x + 3x + 15 Use the distributive law to expand the brackets 2 = x + 8x + 15 and then collect the like terms 5x and 3x. Now you try Expand (x + 2)(x + 9). 4 Expand the following. a (x + 2)(x + 5) b (b + 3)(b + 4) c (t + 8)(t + 7) Hint: First expand with the distributive law then collect the d (p + 6)(p + 6) e (x + 9)(x + 6) f (d + 15)(d + 4) two like terms. g (a + 1)(a + 7) h (y + 10)(y + 2) i (m + 4)(m + 12) Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party. 490 Chapter 9 Algebraic techniques 9B Example 5 Expanding binomial products involving subtraction signs Expand the following. a (x − 4)(x + 7) b (2x − 1)(x − 6) c (5x − 2)(3x + 7) Solution Explanation a (x − 4)(x + 7) = x2 + 7x − 4x − 28 After expanding to get the four terms, collect 2 = x + 3x − 28 the like terms 7x and −4x. Note: x × 7 = 7x and −4 × x = −4x. b (2x − 1)(x − 6) = 2x2 − 12x − x + 6 Remember: 2x × x = 2x2 and −1 × (−6) = 6. = 2x2 − 13x + 6 c (5x − 2)(3x + 7) = 15x2 + 35x − 6x − 14 Recall: 5x × 3x = 5 × 3 × x × x = 15x2. = 15x2 + 29x − 14 Now you try Expand the following. a (x + 3)(x − 6) b (3x − 2)(x − 4) c (6x + 5)(3x − 4) 5 Expand the following. Hint: First expand with a (x + 3)(x − 4) b (x + 5)(x − 2) c (x + 4)(x − 8) the distributive law then d (x − 6)(x + 2) e (x − 1)(x + 10) f (x − 7)(x + 9) collect the two like g (x − 2)(x + 7) h (x − 1)(x − 2) i (x − 4)(x − 5) terms. j (4x + 3)(2x + 5) k (3x + 2)(2x + 1) l (3x + 1)(5x + 4) m (2x − 3)(3x + 5) n (8x − 3)(3x + 4) o (3x − 2)(2x + 1) p (5x + 2)(2x − 7) q (2x + 3)(3x − 2) r (4x + 1)(4x − 5) s (3x − 2)(6x − 5) t (5x − 2)(3x − 1) u (7x − 3)(3x − 4) Problem-solving and reasoning 6–8, 9(½) 7–11 6 A 10 m by 7 m rectangular factory shed is expanded by x metres on two sides. 10 m a Write expressions for: 7m i the new length of the shed (horizontal length) x m ii the new width of the shed (vertical length) xm b Using your results from part a, expand brackets to find an expression for the new total area. c What would be the new area if x = 2? 7 A rectangular room in a house with dimensions 4 m by 5 m is to be extended. Both xm the length and width are to be increased by x m. 5m a Find an expanded expression for the area of Hint: First label the length and x 4m the new room. width of the new room. m b i If x = 3, find the area of the new room. ii By how much has the area increased? Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  9B Expanding binomial products 491 8 A rectangular trampoline of length 3 m and width 2 m is to be surrounded by padding of width x metres. xm 2m x 3m m a Write expressions for: Hint: Don’t forget to count the x i the total length of the trampoline and padding on both sides. ii the total width of the trampoline and padding b Find an expression for the total area by expanding brackets. c What would be the total area if x = 1? 9 Write the missing terms in these expansions. a (x + 2)(x + ) = x2 + 5x + 6 b (x + )(x + 5) = x2 + 7x + 10 c (x + 1)(x + ) = x2 + 7x + d (x + )(x + 9) = x2 + 11x + e (x + 3)(x − ) = x2 + x − f (x − 5)(x + ) = x2 − 2x − 10 Expand these binomial products. a (a + b)(a + c) b (a − b)(a + c) c (b − a)(a + c) d (2x + y)(x − 2y) e (2a + b)(a − b) f (3x − y)(2x + y) 11 A picture frame 5 cm wide has a length that is twice the width, x cm. Hint: Length of picture = a Find an expression for the total area of the frame and picture. 2x − 5 − 5 = 2x − 10 b Find an expression in expanded form for the area of the picture only. Picture x cm 5 cm 2x cm Paving the pool — 12 12 The outside edge of a path around a rectangular swimming pool is xm 15 m long and 10 m wide. The path is x metres wide. a Write expressions for: Pool 10 m i the length of the pool ii the width of the pool 15 m b Find an expression for the area of the pool in expanded form. c Find the area of the pool if x = 2. d What value of x makes the pool area 50 m2 ? Use trial and error. Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  492 Chapter 9 Algebraic techniques 9C 9C Expanding perfect squares Learning intentions To be able to recognise a perfect square To understand that expressions that are perfect squares can be expanded To be able to expand and simplify perfect squares Key vocabulary: perfect square, distributive law, expand A special type of binomial product involves perfect squares. Examples of perfect squares are 22 = 4, 152 = 225, x2 and (a + b)2. To expand (a + b)2 we multiply (a + b) by (a + b) and use the distributive law: (a + b)2 = (a + b)(a + b) = a(a + b) + b(a + b) = a2 + ab + ba + b2 = a2 + 2ab + b2 A similar result is obtained for the square of (a − b): (a − b)2 = (a − b)(a − b) = a(a − b) − b(a − b) = a2 − ab − ba + b2 = a2 − 2ab + b2 Lesson starter: Seeing the pattern Using (a + b)(c + d) = ac + ad + bc + bd, expand and simplify the binomial products below. (x + 1)(x + 1) = x2 + x + x + 1 (x − 5)(x − 5) = = = (x + 3)(x + 3) = (x − 7)(x − 7) = = = Describe the patterns you see in the expansions above. Generalise your observations by completing the following expansions. (a + b)(a + b) = a2 + + + (a − b)(a − b) = 2 = =a + + Key ideas 32 = 9, a2 , (2y)2 , (x − 1)2 and (3 − 2y)2 are all examples of perfect squares. They are expressions that can be written as a single square. Expanding perfect squares: (a + b)2 = (a + b)(a + b) (a − b)2 = (a − b)(a − b) = a(a + b) + b(a + b) = a(a − b) − b(a − b) 2 2 = a + ab + ba + b = a2 − ab − ba + b2 = a2 + 2ab + b2 = a2 − 2ab + b2 Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  9C Expanding perfect squares 493 Exercise 9C Understanding 1–3 3 1 The side lengths of this square are (x + 3) units. x 3 a What are the areas of each of the four regions? Write expressions each time. x b Add up all the area expressions to find an expression for the total area. 3 c Complete the following: (x + 3)(x + 3) = x2 + 3x + — + — = x2 + 6x + — 2 Complete these expansions. Hint: Expand using the a (x + 4)(x + 4) = distributive law: x2 + 4x + — + — b (x + 5)(x + 5) = x2 + 5x + — + — (a + b)(c + d) = = —————— = —————— ac + ad + bc + bd c (x − 2)(x − 2) = x2 − 2x − — + — d (x − 7)(x − 7) = x2 − 7x − — + — = —————— = —————— 3 a Substitute the given value of b into x2 + 2bx + b2 and simplify. i b=3 ii b = 11 b Substitute the given value of b into x2 − 2bx + b2 and simplify. i b=2 ii b = 9 Fluency 4–5(½) 4–6(½) Example 6 Expanding perfect squares Expand each of the following. a (x + 3)2 b (x − 2)2 Solution Explanation a (x + 3)2 = (x + 3)(x + 3) Write in expanded form. = x2 + 3x + 3x + 9 Use the distributive law. 2 = x + 6x + 9 Collect like terms. Alternative solution: (x + 3)2 = x2 + 2 × x × 3 + 32 Expand using (a + b)2 = a2 + 2ab + b2 , = x2 + 6x + 9 where a = x and b = 3. b (x − 2)2 = (x − 2)(x − 2) Write in expanded form. = x2 − 2x − 2x + 4 Use the distributive law. 2 = x − 4x + 4 Collect like terms. Alternative solution: (x − 2)2 = x2 − 2 × x × 2 + 22 Expand using (a − b)2 = a2 − 2ab + b2 = x2 − 4x + 4 where a = x and b = 2. Now you try Expand each of the following. a (x + 8)2 b (x − 5)2 Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party. 494 Chapter 9 Algebraic techniques 9C 4 Expand each of the following perfect squares. a (x + 1)2 b (x + 3)2 c (x + 2)2 d (x + 5)2 Hint: Recall: (a + b)2 = (a + b)(a + b) e (x + 4)2 f (x + 9)2 g (x + 7)2 h (x + 10)2 (a − b)2 = (a − b)(a − b) i (x − 2)2 j (x − 6)2 k (x − 1)2 l (x − 3)2 m (x − 9) 2 n (x − 7) 2 o (x − 4)2 p (x − 12)2 Example 7 Expanding more perfect squares Expand (2x + 3)2. Solution Explanation (2x + 3)2 = (2x + 3)(2x + 3) Write in expanded form. = 4x2 + 6x + 6x + 9 Use the distributive law. 2 = 4x + 12x + 9 Collect like terms. Alternative solution: (2x + 3)2 = (2x)2 + 2 × 2x × 3 + 32 Expand using (a + b)2 = a2 + 2ab + b2 = 4x2 + 12x + 9 where a = 2x and b = 3. Recall (2x)2 = 2x × 2x = 4x2. Now you try Expand (5x − 3)2. 5 Expand each of the following perfect squares. Hint: a (2x + 1)2 b (2x + 5)2 c (3x + 2)2 (2x)2 = 2x × 2x d (3x + 1)2 e (5x + 2)2 f (4x + 3)2 = 4x2 g (7 + 2x)2 h (5 + 3x)2 i (2x − 3)2 j (3x − 1) 2 k (4x − 5)2 l (2x − 9)2 6 Expand each of the following perfect squares. a (3 − x)2 b (5 − x)2 c (1 − x)2 d (6 − x) 2 e (11 − x)2 f (4 − x)2 Hint: −x × (−x) = x2 g (7 − x)2 h (12 − x)2 i (8 − 2x)2 j (2 − 3x)2 k (9 − 2x)2 l (10 − 4x)2 Problem-solving and reasoning 7, 8 8–10 7 A farmer extends a 12 m square sheep pen on two sides by x metres. 12 m xm 12 m xm Hint: An example of an a Write the expression for the side length of the new pen. expression is 10 − x b Write an expression for the area of the new pen and expand or x2 − 20x + 100. this expression. c Use your result to find the new area if: i x=2 ii x = 5 Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  9C Expanding perfect squares 495 8 A child at pre-school cuts off a strip of width x cm from two sides of a 10 cm x cm square piece of paper of side length 10 cm. a Write an expression for the new side length of the remaining paper. b Find the area of the new piece of paper in expanded form. 10 cm c Use your result to find the new area if: i x=2 ii x = 6 x cm d What value of x makes the new area one quarter of the original? 9 A square piece of tin of side length 20 cm has four squares of side x cm length x cm removed from each corner. The sides are folded up to form a tray. The centre square forms the tray base. x cm a Write an expression for the side length of the base of the tray. Tray 20 cm base b Write an expression for the area of the base of the tray. Expand your answer. c Find the area of the tray base if x = 3. Hint: For part d, what is the d Find the volume of the tray if x = 3. height of the tray? 20 cm 10 A square of side length b is removed from a square of side length a. a Using subtraction, write down an expression for the C b remaining area. b Write expressions for the area of the regions in expanded form: i A ii B iii C a c Add all the expressions from part b to see if you get your answer A B from part a. a Arranging tennis courts — 11 11 Four tennis courts are arranged as shown, with a square storage space in the b centre. Each court area has the same dimensions, a × b. a Write an expression for the side length of the total area. b Write an expression for the area of the total area. a c Write an expression for the side length of the inside storage space. d Write an expression for the area of the storage space in expanded form. e Subtract your answer to part d from your answer to part b to find the area of the four courts. f Find the area of one court. Does your answer confirm that your answer to part e is correct? Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  496 Chapter 9 Algebraic techniques 9D 9D Forming a difference of perfect squares Learning intentions To understand how a difference of perfect squares is formed To be able to expand and simplify to form a difference of perfect squares Key vocabulary: difference of perfect squares, distributive law, expand Another type of expansion deals with the product of the sum and difference of the same two terms. The result is the difference of two perfect squares: (a + b)(a − b) = a(a − b) + b(a − b) or (a + b)(a − b) = a2 − ab + ba − b2 = a2 − ab + ba − b2 = a2 − b2 = a2 − b2 Lesson starter: How is 16 × 14 a difference of perfect squares? Using the fact that 152 = 225, follow these steps to show how 16 × 14 can be calculated mentally using a difference of perfect squares. 16 × 14 = (15 + ) × (15 − ) Rewrite. 2 = 15 − + − Expand. = −1 Simplify. = Evaluate. Now try this technique on 17 × 13 and 19 × 21. Key ideas A difference of perfect squares (DOPS) is formed when one square is subtracted from another. This is formed when (a + b)(a − b) is expanded and simplified. (a + b)(a − b) = a2 − ab + ba − b2 = a2 − b2 (a − b)(a + b) also expands to a2 − b2 The result is a difference of two perfect squares. Exercise 9D Understanding 1–3 2, 3 1 Why do the two middle terms in an expansion of (x + a)(x − a) (i.e. x2 + ax − xa − a2 ) cancel out? 2 Decide whether each of the following shows a single perfect square or a difference of perfect squares. a 42 b 7 2 − 32 c a2 − b2 d x2 3 Complete these expansions. Hint: (a + b)(a − b) = a2 − ab + ab − b2 a (x + 4)(x − 4) = x2 − 4x + − = a2 − b2 = b (2x − 1)(2x + 1) = 4x2 + − − = Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  9D Forming a difference of perfect squares 497 Fluency 4-5(½) 4-5(½) Example 8 Forming a difference of perfect squares Expand and simplify the following. a (x + 2)(x − 2) b (x − 7)(x + 7) Solution Explanation a (x + 2)(x − 2) = x2 −  + 2x −4 2x Expand using the distributive law. = x2 − 4 −2x + 2x = 0. Alternate solution: (x + 2)(x − 2) = (x)2 − (2)2 (a + b)(a − b) = a2 − b2. Here a = x and b = 2. = x2 − 4 b (x − 7)(x + 7) = x2 + 7x − 7x − 49 Expand, then note that 7x − 7x = 0. 2 = x − 49 Alternate solution: (x − 7)(x + 7) = (x)2 − (7)2 (a − b)(a + b) = a2 − b2 , with a = x and b = 7. = x2 − 49 Now you try Expand and simplify the following. a (x + 5)(x − 5) b (x − 15)(x + 15) 4 Expand and simplify the following to form a difference of perfect squares. a (x + 1)(x − 1) b (x + 3)(x − 3) c (x + 8)(x − 8) d (x + 4)(x − 4) e (x + 12)(x − 12) f (x + 11)(x − 11) Hint: Recall: g (x − 9)(x + 9) h (x − 5)(x + 5) (a + b)(a − b) = a2 − b2 i (x − 6)(x + 6) j (5 − x)(5 + x) k (2 − x)(2 + x) l (7 − x)(7 + x) Example 9 Forming more differences of perfect squares Expand and simplify (3x − 2y)(3x + 2y). Solution Explanation (3x − 2y)(3x + 2y) = 9x2 + − 6xy  − 4y2 6xy Expand using the distributive law. 2 2 = 9x − 4y 6xy − 6xy = 0. Alternate solution: (3x − 2y)(3x + 2y) = (3x)2 − (2y)2 (a + b)(a − b) = a2 − b2 , with a = 3x and b = 2y here. Recall that (3x)2 = 3x × 3x. = 9x2 − 4y2 Now you try Expand and simplify (12x − 5y)(12x + 5y). Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party. 498 Chapter 9 Algebraic techniques 9D 5 Expand and simplify the following. a (3x − 2)(3x + 2) b (5x − 4)(5x + 4) Hint: c (4x − 3)(4x + 3) d (7x − 3y)(7x + 3y) (3x)2 = 3x × 3x e (9x − 5y)(9x + 5y) f (11x − y)(11x + y) = 9x2 g (8x + 2y)(8x − 2y) h (10x − 9y)(10x + 9y) i (7x − 5y)(7x + 5y) j (6x − 11y)(6x + 11y) k (8x − 3y)(8x + 3y) l (9x − 4y)(9x + 4y) Problem-solving and reasoning 6(½), 7 6(½), 8 6 To calculate 21 × 19, here is a method using the difference of perfect squares. 21 × 19 = (20 + 1) × (20 − 1) = 202 − 12 = 400 − 1 = 399 Use this technique to evaluate the following (mentally if you can). a 31 × 29 (Note: 302 = 900) b 41 × 39 (Note: 402 = 1600) c 26 × 24 (Note: 252 = 625) d 51 × 49 (Note: 502 = 2500) 2 e 22 × 18 (Use 20 = 400) f 23 × 17 (Use 202 = 400) g 35 × 25 (Use 302 = 900) h 54 × 46 (Use 502 = 2500) 7 Lara is x years old and her two best friends are (x − 2) and (x + 2) years old. a Write an expression for: i the square of Lara’s age Hint: Expand in part a ii. ii the product of the ages of Lara’s best friends b Are the answers from parts a i and ii equal? If not, by how much do they differ? 8 A square of side length x has one side reduced by 1 unit and the other increased by 1 unit. a Find an expanded expression for the area of the resulting Hint: Determine expressions for the length and width rectangle. of the rectangle first. b Is the area of the original square the same as the area of the resulting rectangle? Explain why/why not. Classroom renovation — 9 9 A square classroom is to be shortened on the south side by 3 m xm 3m and extended on the east side by 3 m. The original side length of the classroom was x m. a Write expressions for: xm i the original area of the classroom ii the new length of the classroom 3m iii the new width of the classroom b Expand to find an expression for the new area of the classroom. c Is the new area the same as the original area? If not, by how much do they differ? Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  9E Factorising algebraic expressions 499 9E 9E Factorising algebraic expressions Learning intentions To understand that factorising and expanding are reverse processes To be able to identify the highest common factor To know the form of a factorised expression To be able to factorise algebraic expressions involving a common factor Key vocabulary: highest common factor, factorise The process of factorisation is a key step in the simplification of many algebraic expressions and in the solution of equations. It is the reverse process of expansion and involves writing an expression as a product of its factors. Expanding 5x(x + 2) 5x2 + 10x Factorising Lesson starter: Factorised areas Here is a rectangle of length (x + 3) and width x. Write an expression for the total area using the given length and width. x 3 Write an expression for the total area by adding up the area of the two smaller regions. Are your two expressions equivalent? How could you work from your x second expression (expanded) to the first expression (factorised)? Key ideas To factorise means to write an expression as a product of its factors. When factorising expressions with common factors, take out HCF of 2x and 10 the highest common factor (HCF). The HCF could be: a number 2x + 10 = 2(x + 5) For example: 2x + 10 = 2(x + 5) a pronumeral For example: x2 + 5x = x(x + 5) expanded form factorised form the product of numbers and pronumerals For example: 2x2 + 10x = 2x(x + 5) A factorised expression can be checked by using expansion. For example: 2x(x + 5) = 2x2 + 10x. Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party. 500 Chapter 9 Algebraic techniques Exercise 9E Understanding 1–3 3 1 Write down the highest common factor (HCF) of these pairs of numbers. a 8, 12 b 10, 20 c 5, 60 d 24, 30 2 Write down the missing factor. a 5× = 5x b 7× = 7x c 2a × = 2a2 d 5a × = 10a2 e × 3y = −6y2 f × 12x = −36x2 3 a Write down the missing factor in each part. i (x2 + 2x) = 6x2 + 12x ii (2x + 4) = 6x2 + 12x iii (x + 2) = 6x2 + 12x b Which equation above uses the HCF of 6x2 and 12x? Fluency 4–6(½) 4–6(½) Example 10 Finding the highest common factor (HCF) Determine the HCF of the following. a 6a and 8ab b 3x2 and 6xy Solution Explanation a HCF of 6a and 8ab is 2a HCF of 6 and 8 is 2. HCF of a and ab is a. b HCF of 3x2 and 6xy is 3x HCF of 3 and 6 is 3. HCF of x2 and xy is x. Now you try Determine the HCF of the following. a 10ab and 5b b 7xy2 and 21xy 4 Determine the HCF of the following. a 6x and 14xy b 12a and 18a c 10m and 4 Hint: HCF stands for d 12y and 8 e 15t and 6s f 15 and p highest common factor. g 9x and 24xy h 6n and 21mn i 10y and 2y 2 j 8x and 14x k 4x2 y and 18xy l 5ab2 and 15a2 b Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  9E Factorising algebraic expressions 501 Example 11 Factorising expressions Factorise the following. a 4x + 12 b 10y − 25y2 Solution Explanation a 4x + 12 = 4(x + 3) 4 is the HCF of 4x and 12. Place 4 in front of the brackets and divide each term by 4: 4x ÷ 4 = x and 12 ÷ 4 = 3. Check your answer using expansion. b 10y − 25y2 = 5y(2 − 5y) The HCF of 10y and 25y2 is 5y. Place 5y in front of the brackets and divide each term by 5y. Now you try Factorise the following. a 15a + 20 b 2x2 − 6x 5 Factorise the following. a 7x + 7 b 3x + 3 c 4x − 4 d 5x − 5 Hint: Always take out the highest e 4 + 8y f 10 + 5a g 3 − 9b h 6 − 2x common factor (HCF). i 12a + 3b j 6m + 6n k 10x − 8y l 4a − 20b m x2 + 2x n a2 − 4a o y2 − 7y p x − x2 q 3p2 + 3p r 8x − 8x2 s 4b2 + 12b t 6y − 10y2 Example 12 Taking out the common negative sign Factorise −8x2 − 12x. Solution Explanation −8x2 − 12x = −4x(2x + 3) The HCF of the terms is −4x, including the common negative. Place the factor in front of the brackets and divide each term by −4x. Note: −8x2 ÷ (−4x) = 2x and −12x ÷ (−4x) = 3. Now you try Factorise −11a − 22a2. 6 Factorise the following by including the negative sign in the common factor. a −8x − 4 b −4x − 2 c −10x − 5y Hint: Take a negative factor out d −7a − 14b e −9x − 12 f −6y − 8 of both terms. g −10x − 15y h −4m − 20n i −3x2 − 18x j −8x2 − 12x k −16y2 − 6y l −5a2 − 10a Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party. 502 Chapter 9 Algebraic techniques 9E Problem-solving and reasoning 7, 8(½) 8(½), 9–11 7 Write the missing number or expression. a 3x + 9 = (x + 3) b xy + x = x( + 1) c a2 − a = (a − 1) d 5xy + 10x = (y + 2) e −7a − 14 = (a + 2) 2 f −24a − 36a = (2a + 3) 8 Write down the perimeter of these shapes in factorised form. Hint: First find an a b c expression for the x perimeter, then 5x factorise it. 4 6 10 2x d 5 e f 3 2 4 x x x+1 1 9 The expression for the area of a rectangle is (4x2 + 8x). Find an expression for ? its width if the length is (x + 2). 4x2 + 8x x+2 10 The height, in metres, of a ball thrown in the air is given by 5t − t2 , where t is the time in seconds. a Write an expression for the ball’s height in factorised form. b Find the ball’s height at these times: i t=0 ii t = 2 iii t = 4 c How long does it take for the ball’s height to return to 0 metres? Use trial and error if required. 11 7 × 9 + 7 × 3 can be evaluated by first factorising to 7(9 + 3). This gives 7 × 12 = 84. Use a similar technique to evaluate the following. a 9×2+9×5 b 6×3+6×9 c −2 × 4 − 2 × 6 d −5 × 8 − 5 × 6 e 23 × 5 − 23 × 2 f 63 × 11 − 63 × 8 Further factorisation — 12, 13 12 Common factors can also be removed from expressions with more than two terms. For example: 2x2 + 6x + 10xy = 2x(x + 3 + 5y) Factorise these expressions by taking out the HCF. a 3a2 + 9a + 12 b 5z2 − 10z + zy c x2 − 2xy + x2 y d 4by − 2b + 6b2 e −12xy − 8yz − 20xyz f 3ab + 4ab2 + 6a2 b 13 You can factorise some expressions by taking out a binomial factor. Hint: x − 2 is the common For example: 3(x − 2) + x(x − 2) = (x − 2)(3 + x) factor. Factorise the following by taking out a binomial common factor. a 4(x + 3) + x(x + 3) b 3(x + 1) + x(x + 1) c 7(m − 3) + m(m − 3) d x(x − 7) + 2(x − 7) e 8(a + 4) − a(a + 4) f 5(x + 1) − x(x + 1) g y(y + 3) − 2(y + 3) h a(x + 2) − x(x + 2) i t(2t + 5) + 3(2t + 5) j m(5m − 2) + 4(5m − 2) k y(4y − 1) − (4y − 1) l (7 − 3x) + x(7 − 3x) Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  Progress quiz 503 Progress quiz 9A 1 Simplify the following. a 3x − 2y + 5x − 4y b 3ab2 − ab + 4ab2 2 c 4a × 5ab d 3xy 9xy 9A 2 Expand and simplify the following. a 3(x − 4) b 2y(3y + 5) c 5 − 2(3x + 1) 9B 3 Expand the following binomial products. a (x + 4)(x + 6) b (x + 3)(x − 2) c (2x − 3)(x + 4) d (3x − 2)(2x − 5) 9C 4 Expand the following perfect squares. a (x + 6)2 b (x − 8)2 c (3x + 7)2 d (5 − x)2 9B/9C 5 Find the area of these figures in expanded form. a b (x − 3) (2x + 1) (3x − 1) 9D 6 Expand and simplify the following. a (x + 10)(x − 10) b (x − 1)(x + 1) c (3x + 2)(3x − 2) d (4x − 5y)(4x + 5y) 9C/9D 7 Expand and simplify (x + 1)2 − (x + 1)(x − 1). 9E 8 Determine the HCF of the following. a 8x and 12 b 6a and 15a c 5ab and 10ab 2 d 12x2 and 9x 9E 9 Factorise the following. Take a negative factor out in parts e and f. a 6x + 12 b 15y − 20 c 8ab + 20a 2 d 4x − 2x e −10x − 5 f −4x2 − 10x 9E 10 Give the perimeter of these shapes in factorised form a b x+3 2x + 4 3x Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  504 Chapter 9 Algebraic techniques 9F 9F Simplifying algebraic fractions: multiplication and division Learning intentions To know that expressions must be factorised before common factors can be cancelled To be able to simplify algebraic fractions by cancelling common factors To be able to multiply and divide algebraic fractions Key vocabulary: algebraic fraction, common factor, factorise, numerator, denominator, reciprocal Fractions such as 4 and 24 can be simplified by cancelling common factors. This is also true of 6 16 2 algebraic fractions such as 3x, 5a and 2x + 4. 6 10a 2 2x+ 6
3 Lesson starter: Does = 2x + 3? 2
1 For the expression 2x + 6, a student attempts to cancel the 6 with the 2 in the denominator. 2 2x + 6
3 They write = 2x + 3. 2
1 Is this a correct method? Substitute x = 1 into the left hand side and then the right hand side. Are they equal? Can you show a correct method? Does it involve factorisation? Key ideas Simplify algebraic fractions by cancelling common factors 3
1 x x 5a
2 5a 2x + 4 = 2
1 (x + 2) = x + 2 = = 6
2 2 3a
3 2 2
1 2x + 4
2 This is an incorrect cancellation: = 2x + 2 2
1 To multiply algebraic fractions, first cancel any numerator with any denominator. 2
1  x1 3 15 ×  = 1×3 = 3 5
1 x 1 y 1 × 2y 2y 4
2  To divide algebraic fractions, multiply by the reciprocal of the fraction following the division sign. The reciprocal of a = b. b a 3x ÷ 9 = 3
1 x × 8
2 y = 2xy 4 8y 4
1 9
3 3 Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  9F Simplifying algebraic fractions: multiplication and division 505 Exercise 9F Understanding 1–3 3 1 Cancel to simplify these fractions. a 4 b 4x c 4(x + 2) d 4(x − 1) 2 2 2 2 2 Write down the reciprocal of these fractions. Hint: a 2 b 4 c 7x Recall: 7 = 7 for part e. 1 3 3 2 5 d e 7 f 6x 4a2 Hint: 3 Divide these simple fractions. Copy the first a 2÷3 b 7÷3 c 6÷4 fraction. 3 4 8 4 7 7 Change the ÷ to ×. Flip over the second d 5 ÷ 10 e 21 ÷ 7 f 4÷4 fraction. 14 7 4 8 9 9 Fluency 4–7(½) 4–7(½) Example 13 Cancelling common factors Simplify by cancelling common factors. a 7x b 5(x + 2)(x − 1) 21 x+2 Solution Explanation 71 x a
=x The HCF of 7 and 21 is 7. 3 3 21  + (x 5 1 (x − 1) 2) b   = 5(x − 1) Treat (x + 2) as a common factor to both the x+  21 numerator and denominator. Now you try Simplify by cancelling common factors. 2 a −2ab b −3(x − 2)(x + 7) 4b 3(x − 2) 4 Simplify by cancelling common factors. a 4x b 8x c −4ab d 7a 8 16 12 14b 2 e 4(x − 1) f −6(2x + 1) g −7x h 4a 4 3 21x 40a i 3(x + 2) j −2(x − 7) k 5(x + 3) l −16(x − 14) x+2 x−7 10(x + 3) 24(x − 14) m (x + 2)(x − 4) n 2(x − 1)(x + 3) o −4(x + 2)(x − 5) p −7(x + 1)(x − 9) x+2 x+3 2(x − 5) 7(x + 1)(x − 9) Essential Mathematics for the Victorian Curriculum ISBN 978-1-108-87854-8 © Greenwood et al. 2021 Cambridge University Press CORE Year 9 Photocopying is restricted under law and this material must not be transferred to another party. 506 Chapter 9 Algebraic techniques 9F Example 14 Simplifying by factorising Simplify these fractions by factorising first. a 2x + 6 b −3x − 9 2 x+3 Solution Explanation a 2x + 6 = 2
1 (x + 3) First factorise the numerator, then cancel the 2. 2 2
1 =x+3 (x 1 + b −3x − 9 = −3 3) −3 is common to both −3x and −9. (x + 3) can x+3 x+ 3 1 now be cancelled. = −3 Now you try Simplify these fractions by factorising first. a 5a − 30 b −7x − 14 5 x+2 5 Simplify these fractions by factorising first. a 4x + 8 b 2x − 6