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 1111xx 33 + 44xx 22 2222xx 44 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 44xx 22 (3333 + 11 55xx 22 ) 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) pp 2 + 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) pp 2 + 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 + 9aa 2 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 aa 2 + 2aaaa + bb 2 And aa aa + bb + bb bb + aa aa + bb (aa + bb) aa 2 + ab + ba + bb 2 Let s use numerical values to check
Using the FOIL method
Your turn: expand & FOIL 20 2 = (10 + 10) 2 20 2 = (15 + 5) 2 20 2 = (18 + 2) 2 = (18 + 2)(18 + 2)
Answers 20 2 = (10 + 10) 2 10 + 10 10 + 10 = 100 + 100 + 100 + 100 = 400 20 2 = (15 + 5) 2 15 + 5 15 + 5 = 225 + 75 + 75 + 25 = 400 20 2 = (18 + 2) 2 = (18 + 2)(18 + 2) 324+ 36 + 36 + 4 = 400
Difference of two squares (aa bb) 2 = aa 2 2aaaa + bb 2
Difference of two squares aa + bb aa bb = aa 2 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) 3aa 2 27 b) 2xx 2 72 c) 5dd 2 20cc 2 d) 16w 2 4 e) 5tt 2 180 f) 7cc 2 7dd 2
Answers a) 3aa 2 27 = 3 aa 2 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) 16w 2 4 = 4 4w 2 1 = 4(2w 1)(2w + 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