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Maths Refresher Expanding and Factorising Expanding and Factorising Learning intentions …. • Recap • Expanding equations • Factorising equations • Identity: perfect pairs • Difference of two squares Introduction • Algebra requires you to manipulate algebraic expressions • We have covered simplifying expressions and solving equations • Now we look at manipulating expressions through expanding and factorising • First, we recap some mathematical ideas that will assist factorisation • Second, we revise the distributive law • Third, you will learn how to expand two or more sets of brackets Recap Multiples A multiple is a number that can be divided into a given number exactly – For example: multiples of 5 are 5, 10, 15, 20, 25… Recap • Common multiples and the LCM • A common multiple is a multiple in which two or more numbers have in common – For example: 3 and 5 have multiples in common 15, 30, 45… • The lowest common multiple LCM is the lowest multiple that two numbers have in common – For example: 15 is the LCM of 3 and 5 15 3 5 Recap • Factor – a whole number that can be multiplied a certain number of times to reach a given number – 3 is factor of 15 and 15 is a multiple of 3 – The other factors are 1and 15 – 4 is a factor of 16 and 16 is a multiple of four, other factors – 1, 2, 4, 8, 16 Recap • A common factor is a common factor that two or more numbers have in common – 3 is a common factor of 12 and 15 – 5 is a common factor of 15 and 25 – 6 is a common factor of 12 and 18 • The highest common factor – HCF – What is the HCF of 20 and 18? – the factors of 20 (1, 2, 4, 5, 10, 20) and 18 (1, 2, 3, 6, 9, 18) – the common factors are 1 and 2 and – the HCF is 2 Recap • We could say that a number is a factor of given number if it is a multiple of that number • For example, – 9 is a factor of 27 and 27 is a multiple of 9 – 7 is a factor of 35 and 35 is a multiple of 7 – 14 is a factor of 154 and 154 is a multiple of 14 YouTube clip – https://www.khanacademy.org/math/pre-algebra/factors-multiples/divisibility_and_factors/v/finding-factors-andmultiples Recap • Proper factors – All the factors apart from the number itself – 18 (1, 2, 3, 6, 9, 18) 1, 2, 3, 6, 9 are proper factors of 18 • Prime number – Any whole number greater than zero that has exactly two factors – itself and one 2, 3, 5, 7, 11, 13, 17, 19… • Composite number – Any whole number that has more than two factors – 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20… Factor tree • Prime factor – A factor that is also a prime number • Factor tree – A tree that shows the prime factors of a number Prime Numbers Click logo for link http://www.mathsisfun.com/numbers/fundamental-theorem-arithmetic.html Your turn …. 1. Create a factor tree for the numbers 72 2. List all the factors for 120 Answers 1. 2. List all the factors for 120 – – – – – – – – 120 = 1 × 120 120 = 𝟐𝟐 × 𝟔𝟔𝟔𝟔 120 = 𝟑𝟑 × 𝟒𝟒𝟒𝟒 120 = 𝟒𝟒 × 𝟑𝟑𝟑𝟑 120 = 𝟓𝟓 × 𝟐𝟐𝟐𝟐 120 = 𝟔𝟔 × 𝟐𝟐𝟐𝟐 120 = 𝟖𝟖 × 𝟏𝟏𝟏𝟏 120 = 𝟏𝟏𝟏𝟏 × 𝟏𝟏𝟐𝟐 ∴ Factors of 120 𝟏𝟏, 𝟐𝟐, 𝟑𝟑, 𝟒𝟒, 𝟓𝟓, 𝟔𝟔, 𝟖𝟖, 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏, 𝟐𝟐𝟐𝟐, 𝟐𝟐𝟐𝟐, 𝟑𝟑𝟑𝟑, 𝟒𝟒𝟒𝟒, 𝟔𝟔𝟔𝟔, 𝟏𝟏𝟏𝟏𝟏𝟏 Expand • Expanding and factorising are often used in algebra • We ‘distribute’ multiplication through addition or subtraction. • Often referred to as either expanding the brackets or removing the brackets. For example: Expand – 2 𝑛𝑛 + 3 2 × 𝑛𝑛 + 2 ×3 – 2𝑛𝑛 + 6 – 2 𝑛𝑛 + 3 = 2𝑛𝑛 + 6 http://passyworldofmathematics.com/expanding-brackets-using-distributive-rule/ Factorise • Involves working in the opposite direction (including brackets) Factorise • 2𝑛𝑛 + 6 we know that 2𝑛𝑛 is a term with 2 factors, 2 and 𝑛𝑛 the factors of 6 are – 1,2,3,6, • The common factor for both terms is 2 – 2 can be multiplied by 𝑛𝑛 and 3 2 × 𝑛𝑛 = 2𝑛𝑛 + 2 × 3=6 • So we can take 2 outside brackets 2(𝑛𝑛 + 3) Your turn …. Expand 1. 9 𝑥𝑥 + 2 = 2. 2(𝑎𝑎 + 6 + 𝑐𝑐) = Factorise 1. 9 + 27𝑐𝑐 − 3𝑎𝑎 = 2. 12 + 4𝑎𝑎 + 16 = Answers Expand 1. 9 𝑥𝑥 + 2 = 9𝑥𝑥 + 18 2. 2(𝑎𝑎 + 6 + 𝑐𝑐) = 2𝑎𝑎 + 12 + 2𝑐𝑐 Factorise 1. 9 + 27𝑐𝑐 − 3𝑎𝑎 = 3(3 + 9𝑐𝑐 − 𝑎𝑎) 2. 12 + 4𝑎𝑎 + 16 = 4 3 + 𝑎𝑎 + 4 Factorise • Factorisation requires finding the highest common factor (HCF) • For example: 5𝑥𝑥 + 15𝑥𝑥 2 − 30𝑥𝑥 3 has three terms 5𝑥𝑥 𝑎𝑎𝑎𝑎𝑎𝑎 15𝑥𝑥 2 𝑎𝑎𝑎𝑎𝑎𝑎 30𝑥𝑥 All 3 terms have the same variable (𝑥𝑥) and are a multiple of 5 If 5𝑥𝑥 is a common factor, then 5 × 𝑥𝑥 + 5 × 3 × 𝑥𝑥 × 𝑥𝑥 − (5 × 6 × 𝑥𝑥 × 𝑥𝑥 × 𝑥𝑥) 5 × 𝑥𝑥 + 5 × 3 × 𝑥𝑥 × 𝑥𝑥 − (5 × 6 × 𝑥𝑥 × 𝑥𝑥 × 𝑥𝑥) • Therefore, if we divide each term by 5𝑥𝑥 the HCF we end up with: 5𝑥𝑥(1 + 3𝑥𝑥 − 6𝑥𝑥 2 ) Remember that if we divide a number by itself it equals one. Let’s check Is 5𝑥𝑥 + 15𝑥𝑥 2 − 30𝑥𝑥 3 the same as 5𝑥𝑥(1 + 3𝑥𝑥 − 6𝑥𝑥 2 ) ? Substitute 𝑥𝑥 𝑓𝑓𝑓𝑓𝑓𝑓 2 • 5 × 2 + 15 × 4 − 30 × 8 • 10 + 60 − 240 • 70 − 240 • -170 Substitute 𝑥𝑥 𝑓𝑓𝑓𝑓𝑓𝑓 2 • 10 1 + 6 − 6 × 4 • 10 7 − 24 • 10× 17 • -170 Factorising Example Problem: • Remove the Brackets: 5𝑥𝑥 2 + 𝑦𝑦 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑡𝑡𝑡𝑡 5𝑥𝑥 × 2 + 5𝑥𝑥 × 𝑦𝑦 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑖𝑖𝑖𝑖 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡𝑡𝑡 10𝑥𝑥 + 5𝑥𝑥𝑥𝑥 • We can multiply by 2: remember the commutative law −𝑥𝑥 2𝑥𝑥 + 6 e𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 𝑡𝑡𝑡𝑡 −𝑥𝑥 × 2𝑥𝑥 + −𝑥𝑥 × 6 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑖𝑖𝑖𝑖 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡𝑡𝑡 − 2𝑥𝑥 2 + −6𝑥𝑥 ∴ −2𝑥𝑥 2 − 6𝑥𝑥 (Positive and negative make a negative) Factorising Factorise: 3𝑥𝑥 + 9𝑥𝑥 − 𝑥𝑥 2 The only thing in common is the variable 𝑥𝑥 If 𝑥𝑥 2 had a factor multiple of 3 as a coefficient, then we could factorise further So we can take the variable outside of the brackets 𝑥𝑥(3 + 9 − 𝑥𝑥) Notice that we still have one 𝑥𝑥 from 𝑥𝑥 2 inside the brackets ∴ 𝑥𝑥 12 − 𝑥𝑥 Test it 3𝑥𝑥 + 9𝑥𝑥 − 𝑥𝑥 2 let’s make 𝑥𝑥 = 2 So 6 + 18 − 4 = 20 Or 𝑥𝑥 12 − 𝑥𝑥 𝐿𝐿𝐿𝐿𝑡𝑡 ′ 𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑥𝑥 = 2; 2 × 10 = 20 Factorising Factorise 𝟏𝟏𝟏𝟏𝒙𝒙𝟑𝟑 + 𝟒𝟒𝒙𝒙𝟐𝟐 − 𝟐𝟐𝟐𝟐𝒙𝒙𝟒𝟒 What are the factors of all terms? 12𝑥𝑥𝑥𝑥𝑥𝑥 + 4𝑥𝑥𝑥𝑥 − 20𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 We can see now that 4 is a common factor as is x the HCFs: 4 and 𝑥𝑥 × 𝑥𝑥 ∴ the highest common factor is 4𝑥𝑥 2 So we can factorise to get 𝟒𝟒𝒙𝒙𝟐𝟐 (𝟑𝟑𝟑𝟑 + 𝟏𝟏 − 𝟓𝟓𝒙𝒙𝟐𝟐 ) Your turn to check: Is 12𝑥𝑥 3 + 4𝑥𝑥 2 − 20𝑥𝑥 4 the same as 4𝑥𝑥 2 (3𝑥𝑥 + 1 − 5𝑥𝑥 2 ) ? Your turn …. Is 12𝑥𝑥 3 + 4𝑥𝑥 2 − 20𝑥𝑥 4 the same as 4𝑥𝑥 2 (3𝑥𝑥 + 1 − 5𝑥𝑥 2 ) ? Answers Is 12𝑥𝑥 3 + 4𝑥𝑥 2 − 20𝑥𝑥 4 the same as 4𝑥𝑥 2 (3𝑥𝑥 + 1 − 5𝑥𝑥 2 ) ? Let’s substitute 𝑥𝑥 𝑓𝑓𝑓𝑓𝑓𝑓 2 12 × 8 + (4 × 4) − (20 × 16)= 96 + 16 − 320 = 112−320 = -208 Let’s substitute 𝑥𝑥 𝑓𝑓𝑓𝑓𝑓𝑓 2 16 6 + 1 − 5 × 4 = 16 7 − 20 = 16 × −13 = −208 Your turn … A). Factorise by grouping Example 2a +8 = 2x a +2x4 = 2(a+4) a) 6𝑡𝑡 + 3 b) 8𝑎𝑎 + 20𝑏𝑏 c) 7𝑘𝑘 − 49 B). Factorise fully Example 𝑥𝑥 2 − 7𝑥𝑥 = 𝑥𝑥 x 𝑥𝑥 – 7 x 𝑥𝑥 = 𝑥𝑥(𝑥𝑥-7) a) 𝑡𝑡 2 − 5𝑡𝑡 b) 𝑥𝑥𝑥𝑥 + 4𝑦𝑦 c) 𝑝𝑝2 + 𝑝𝑝𝑝𝑝 = d) 𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑎𝑎 Answers A). Factorise by grouping a) 6𝑡𝑡 + 3 = 3 𝑡𝑡 + 1 b) 8𝑎𝑎 + 20𝑏𝑏 = 4 2𝑎𝑎 + 5𝑏𝑏 c) 7𝑘𝑘 − 49 = 7 𝑘𝑘 − 7 B). Factorise fully a) 𝑡𝑡 2 − 5𝑡𝑡 = 𝑡𝑡(𝑡𝑡 − 5) b) 𝑥𝑥𝑥𝑥 + 4𝑦𝑦 = 𝑦𝑦(𝑥𝑥 + 4) c) 𝑝𝑝2 + 𝑝𝑝𝑝𝑝 = 𝑝𝑝(𝑝𝑝 + 𝑞𝑞) d) 𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑎𝑎 = 𝑎𝑎(𝑏𝑏 + 𝑐𝑐 + 𝑑𝑑) Common Factors • A common factor might be a combination of terms, such as a number, a term or several terms. • For example, the term 4𝑥𝑥 is one term consisting of two factors • And 3𝑎𝑎𝑎𝑎𝑐𝑐 2 is also a term consisting of several factors 3 × 𝑎𝑎 × 𝑏𝑏 × 𝑐𝑐 × 𝑐𝑐 So if we had 3𝑎𝑎𝑎𝑎𝑐𝑐 2 + 9𝑎𝑎2 𝑏𝑏𝑐𝑐 3 we could see that 3 is a common factor as is 𝑎𝑎𝑎𝑎𝑎𝑎 2 So we could factorise to 3𝑎𝑎𝑎𝑎𝑎𝑎 2 (1 + 3𝑎𝑎c) Identity: perfect squares Always true for any numerical value • From the square we can see that (𝑎𝑎 + 𝑏𝑏)2 is the same as 𝑎𝑎2 + 2𝑎𝑎𝑎𝑎 + 𝑏𝑏2 And • 𝑎𝑎 𝑎𝑎 + 𝑏𝑏 + 𝑏𝑏 𝑏𝑏 + 𝑎𝑎 • 𝑎𝑎 + 𝑏𝑏 (𝑎𝑎 + 𝑏𝑏) • 𝑎𝑎2 + ab + ba + 𝑏𝑏2 • Let’s use numerical values to check Using the FOIL method Your turn: expand & FOIL • 202 = (10 + 10)2 • 202 = (15 + 5)2 • 202 = (18 + 2)2 = (18 + 2)(18 + 2) Answers • 202 = (10 + 10)2 10 + 10 10 + 10 = 100 + 100 + 100 + 100 = 400 • 202 = (15 + 5)2 15 + 5 15 + 5 = 225 + 75 + 75 + 25 = 400 • 202 = (18 + 2)2 = (18 + 2)(18 + 2) 324+ 36 + 36 + 4 = 400 Difference of two squares (𝑎𝑎 − 𝑏𝑏)2 = 𝑎𝑎2 −2𝑎𝑎𝑎𝑎 + 𝑏𝑏 2 Difference of two squares • 𝑎𝑎 + 𝑏𝑏 𝑎𝑎 − 𝑏𝑏 = 𝑎𝑎2 − 𝑏𝑏 2 Difference of two squares Examples: Factorise: • 𝑥𝑥 2 − 64 = 𝑥𝑥 − 8 𝑥𝑥 + 8 • 9 − 𝑦𝑦 2 = (3 + 𝑦𝑦)(3 − 𝑦𝑦) Difference of two squares • A square might be the product of two or more terms • For example: 16𝑥𝑥 2 − 49𝑦𝑦 2 • Let’s factorise (4𝑥𝑥)2 − (7y)2 =(4𝑥𝑥 + 7𝑦𝑦)(4𝑥𝑥 − 7𝑦𝑦) Your turn … a) 3𝑎𝑎2 − 27 b) 2𝑥𝑥 2 − 72 c) 5𝑑𝑑 2 − 20𝑐𝑐 2 d) 16ℎ2 − 4 e) 5𝑡𝑡 2 − 180 f) 7𝑐𝑐 2 − 7𝑑𝑑 2 Answers a) 3𝑎𝑎2 − 27 = 3 𝑎𝑎2 − 9 = 3(𝑎𝑎 − 3)(𝑎𝑎 + 3) b) 2𝑥𝑥 2 − 72 = 2 𝑥𝑥 2 − 36 = 2(𝑥𝑥 − 6)(𝑥𝑥 + 6) c) 5𝑑𝑑 2 − 20𝑐𝑐 2 = 5(𝑑𝑑 2 − 4𝑐𝑐 2 )= 5(𝑑𝑑 − 2𝑐𝑐)(𝑑𝑑 + 2𝑐𝑐) d) 16ℎ2 − 4 = 4 4ℎ2 − 1 = 4(2ℎ − 1)(2ℎ + 1) e) 5𝑡𝑡 2 − 180 = 5 𝑡𝑡 2 − 36 = 5(𝑡𝑡 − 6)(𝑡𝑡 + 6) f) 7𝑐𝑐 2 − 7𝑑𝑑 2 7(𝑐𝑐 2 − 𝑑𝑑 2 ) 7(𝑐𝑐 + 𝑑𝑑)(𝑐𝑐 − 𝑑𝑑) Divisibility rules • Wouldn't it just be easier to use the 7 divisibility trick to determine if it's a multiple of 14? • All you do is double the last digit of 154, (you double the 4 to get 8) subtract that 8 from the remaining truncated number (15), giving the result of 7. • 7 is obviously a multiple of 7, meaning the original number is divisible by 7. • Since the number 154 is also even, meaning 2's a factor it's divisible by 14 (which has 7 and 2 as factors). • That sounds easier than dividing. There are a bunch more divisibility tricks for all prime numbers up to 50 here: http://www.savory.de/maths1.htm Revision 1 • The perimeter of a rectangle is 90cm • What is the value of p QSA (2011) Revision 2 • Which two expressions are equivalent? QSA (2011) Revision 3 • What is the value of k? QSA (2011) Revision 4 • Based on the information in the table, what is the value of 𝑦𝑦 when 𝑥𝑥 = −2 QSA (2011) Revision 5 • Substitute and solve: QSA (2011) Revision 6 Revision 1 answer Revision 2 answer Revision 3 answer Revision 4 answer Revision 5 answer Revision 6 answer Expanding and Factorising Reflect on the learning intentions …. • Recap • Expanding equations • Factorising equations • Identity: perfect pairs • Difference of two squares Resources • https://www.khanacademy.org/math/arithmetic/factorsmultiples/divisibility_and_factors/v/finding-factors-andmultiples?utm_medium=email&utm_content=5&utm_campaig n=khanacademy&utm_source=digest_html&utm_term=thumb nail • Baker, L. (2000). Step by step algebra 1 workbook. NSW: Pascal Press • Baker, L. (2000). Step by step algebra 2 workbook. NSW: Pascal Press • Queensland Studies Authority. (2011). 2011 NAPLAN: Year 9 numeracy. Brisbane: Queensland Government