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# Ch 05

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### Ch 05

1. 1. Expanding 5 Belinda works for an advertising company that produces billboard advertising. The cost of a billboard is based on the area of the sign and is \$50 per square metre. If we increase the length of the sign by 2 m and the height of the sign by 3 m, can you write a rule for the increase in the cost of the billboard? This chapter shows you how to manipulate algebraic terms and expressions to place them in the most useful form.
2. 2. 144 Maths Quest 9 for VictoriaExpanding singlebrackets In the previous chapter on introductory algebra, we learned that algebra is a type of language. In this chapter we can take the idea fur- ther and look at some of the ‘advanced features’ of this language. At ﬁrst we will rely on using common numerical examples to illustrate the techniques, but after a while, when your conﬁdence has increased, the techniques will become easier to under- stand using algebra only. What is expanding? Consider the English word: won’t. It really stands for will not. In going from won’t to will not we have expanded the word, but the meaning remains unchanged. It is the same with expanding in algebra; we go from a more compact form (won’t) to an expanded form (will not). Consider the following example with numbers. 3(4 + 5) How can we ﬁnd its value? We know from our work on order of operations in chapter 1 that we do brackets ﬁrst, so that: 3(4 + 5) = 3(9) Now the brackets mean multiply, so: 3(4 + 5) = 3(9) = 27 Consider now an alternative way of expanding the original expression, temporarily ‘ignoring’ order of operations. 3(4 + 5) = 3(4) + 3(5) This may seem unusual, or even incorrect, but it isn’t. (It is not!) 3(4 + 5) = 3(4) + 3(5) = 12 + 15 = 27 You can try this with any numbers you like, but the result is the same. This expan- sion is valid and correct. Expansion means to multiply everything inside the brackets by what is directly outside the brackets.
3. 3. Chapter 5 Expanding 145 WORKED Example 1Expand the following expressions.a 5(4 + 3) b 5(x + 3) c 5(x − y) d −a(x − y)THINK WRITEa 1 Write the expression. a 5(4 + 3) 2 Expand the brackets. = 5(4) + 5(3) 3 Multiply out the brackets. = 20 + 15 = 35 4 Check that the result is valid by simplifying the 5(4 + 3) brackets in the original expression ﬁrst. = 5(7) = 35, so it is indeed valid.b 1 Write the expression. b 5(x + 3) 2 Expand the brackets. = 5(x) + 5(3) 3 Multiply out the brackets. = 5x + 15c 1 Write the expression. c 5(x − y) 2 Expand the brackets. = 5(x) + 5(−y) 3 Multiply out the brackets. = 5x − 5y (Remember that a positive term multiplied by a negative term makes a negative term.)d 1 Write the expression. d −a(x − y) 2 Expand the brackets. = −a(x) − a(−y) 3 Multiply out the brackets. = −ax + ay (Remember that a negative term multiplied by a negative term makes a positive term.) Note: It doesn’t matter what is immediately outside the brackets. It may be a number or a pronumeral or both. The following expansions are a little more complex. WORKED Example 2Expand each of the following.a 5x(6y − 7z) b −4y(2x + 3w) c x(2x + 3y)THINK WRITEa 1 Write the expression. a 5x(6y − 7z) 2 Expand the brackets. = 5x(6y) + 5x(−7z) 3 Multiply out the brackets. = 30xy − 35xz (Multiply number parts and pronumeral parts separately and write pronumerals for each term in alphabetical order.)b 1 Write the expression. b −4y(2x + 3w) 2 Expand the brackets. = −4y(2x) − 4y(3w) 3 Multiply out the brackets. = −8xy − 12wyc 1 Write the expression. c x(2x + 3y) 2 Expand the brackets. = x(2x) + x(3y) 3 Multiply out the brackets. = 2x2 + 3xy (Remember that x multiplied by itself gives x2.)
4. 4. 146 Maths Quest 9 for Victoria Expanding and collecting like terms With more complicated expansions, like terms may need to be collected after the expansion of the bracketed part. Remember that like terms contain the same pronumeral parts. You ﬁrst expand the brackets, then collect the like terms. WORKED Example 3 Expand and simplify by collecting like terms. a 4(x − 4) + 5 b x(y − 2) + 5x c x(y − z) + 5x d 7x + 6(y − 2x) THINK WRITE a 1 Write the expression. a 4(x − 4) + 5 2 Expand the brackets. = 4(x) + 4(−4) + 5 3 Multiply out the brackets. = 4x − 16 + 5 4 Collect any like terms. = 4x − 11 b 1 Write the expression. b x(y − 2) + 5x 2 Expand the brackets. = x(y) + x(−2) + 5x 3 Multiply out the brackets. = xy − 2x + 5x 4 Collect any like terms. = xy + 3x c 1 Write the expression. c x(y − z) + 5x 2 Expand the brackets. = x(y) + x(−z) + 5x 3 Multiply out the brackets. = xy − xz + 5x 4 Collect any like terms. There are no like terms. d 1 Write the expression. d 7x + 6(y − 2x) 2 Expand the brackets. = 7x + 6(y) + 6(−2x) 3 Multiply out the brackets. = 7x + 6y − 12x 4 Collect any like terms. = −5x + 6y remember remember 1. Expansion means to multiply everything inside the brackets by what is directly outside the brackets. 2. After expanding brackets, simplify by collecting any like terms. reads L Sp he 5A Expanding single brackets etEXCE Expanding single WORKED 1 Expand the following expressions. brackets Example 1 a 3(x + 2) b 4(x + 3) c 5(m + 4) d 2(p + 5) d hca e 4(x + 1) f 7(x − 1) g −4(y + 6) h −5(a + 1)Mat Expanding i −3(p − 2) j −(x − 1) k −(x + 3) l −(x − 2) single m 3(2b − 4) n 8(3m − 2) o −6(5m − 4) p −3(9p − 5) brackets WORKED 2 Expand each of the following. Example ogram a x(x + 2) b y(y + 3) c a(a + 5) d c(c + 4) 2GC pr e x(4 + x) f y(5 + y) g m(7 − m) h q(8 − q) Expanding i 2x(y + 2) j 5p(q + 4) k −3y(x + 4) l −10p(q + 9) m −b(3 − a) n −7m(5 − n) o −6a(5 − 3a) p −4x(7 − 4x)
5. 5. Chapter 5 Expanding 147WORKED 3 Expand and simplify by collecting like terms. 5.1Example a 2(p − 3) + 4 b 5(x − 5) + 8 c −7(p + 2) − 3 HEET SkillS 3 d −4(3p − 1) − 1 e 6x(x − 3) − 2x f 2m(m + 5) − 3m g 3x(p + 2) − 5 h 4y(y − 1) + 7 i −4p(p − 2) + 5p j 5(x − 2y) − 3y − x k 2m(m − 5) + 2m − 4 l −3p(p − 2q) + 4pq − 1 m −7a(5 − 2b) + 5a − 4ab n 4c(2d − 3c) − cd − 5c o 6p + 3 − 4(2p + 5) p 5 − 9m + 2(3m − 1) Oops! Any errors? Here are 6 expressions that someone has simpliﬁed. Have any errors been made? a 5(x − 1) = 5x − 1 b 3x + 6x = 9 + x c 4(3x) = 12x d 8x − 3x = 5 e −2(x − 7) = −2x −14 f x(x + 5) = 2x + 5x 1 Which expressions have been simpliﬁed correctly? 2 Explain why someone might make the errors you have found. 3 Correct any errors you ﬁnd by rewriting the expression on the right-hand side of the equals sign. 4 Choose 2 values for x and evaluate the left- and right-hand sides to check whether they are now equivalent. The Bagels game In the game of Bagels, a player is to determine a 3-digit number (no digit repeated) by making educated guesses. After each guess, a clue is given about the guess. Here are the clues. bagels: no digit correct pico: one digit is correct but in the wrong position fermi: one digit is correct and in the correct position 1 In each of the problems below, a number of guesses have been made with the clue for each guess shown to its right. From the given set of guesses and clues, determine the 3-digit number. a 123 bagels b 908 bagels 456 pico 134 pico 789 pico 387 pico fermi 075 pico fermi 256 fermi 087 pico 237 pico pico ??? ??? 2 Now try this game with a partner. One person is to decide on a 3-digit number and provide clues to the other person who is guessing what the 3-digit number is.
6. 6. 148 Maths Quest 9 for VictoriaWhat is a metre? metre? Expand the brackets in the expressions given to find the puzzle’s answer code. e – 2ae 6e – 3 e 2 e – 2ae a 2e – ae 2 5e – 10 8 – 4e –4a – 6 e – 2ae 2ae – 3e 8 – 4e 5e – 10 6a – 2 a 2e – ae 2 6a – 2 6e – 3e 2 8 – 4e 2ae – 3e a 2 – ae e – 2ae e 2 + e e 2 + e e – 2ae a 2 – 4a 2ae – 3e 6e – 3e 2 3a + ae a 2 – 4a – 4a – 6 2 2 6e – 3e 2 3a + ae 8 – 4e 5e – 10 e – 2ae a 2e – ae 2 6e – 3e 2 6a – 2 2ae – 3e 6 – 3a 8 – 4e –4a – 6 2ae – 2ae 2 a – 4a a – ae 6e – 3 e 2 3a + ae 8 – 4e 2ae – 3e a 2 – 4a 2ae – 2ae 2 6e – 3 e 2 3a + ae 3ae + 6e a 2 – 4a e2 + e 8 – 4e 6e – 3 e 2 a 2 – 4a 6e – 3 e 2 3a + ae 8 – 4e 8 – 4e 5a – 10ae 4 a 2 – 4a 6a – 2 6e – 3 e 2 a 2 – 4 a 2ae – 2ae 2 = – a (e – a ) = = 2(3 a – 1) = = – e (–2 a + 3) = = 3(2 – a ) = = – a (4 – a ) = = 5(e – 2) = = 3 e (a + 2) = = –4(e – 2) = = 5 a (1 – 2 e ) = = –2(2 a + 3) = = –2 ae (e – 1) = = a (3 + e ) = = ae (a – e ) = = e (1 – 2 a ) = = 3 e (2 – e ) = = e (e + 1) = = –4 a (– a + 1) =
7. 7. Chapter 5 Expanding 149Expanding two brackets When expanding an expression that contains two (or more) brackets, the steps are the same as before. Step 1 Expand each bracket (working from left to right). Step 2 Collect any like terms. WORKED Example 4Expand and simplify the following expressions.a 5(x + 2y) + 6(x − 3y) b −5x(y − 2) + y(x + 3)c 7y(x − 2y) + y2(x + 5) d −5xy(1 + 2y) + 6x(y + 4x)THINK WRITEa 1 Write the expression. a 5(x + 2y) + 6(x − 3y) 2 Expand each bracket. = 5(x) + 5(2y) + 6(x) + 6(−3y) 3 Multiply out the brackets. = 5x + 10y + 6x − 18y 4 Collect any like terms. = 11x − 8yb 1 Write the expression. b −5x(y − 2) + y(x + 3) 2 Expand each bracket. = −5x(y) − 5x(−2) + y(x) + y(3) 3 Multiply out the brackets. = −5xy + 10x + xy + 3y 4 Collect any like terms. = −4xy + 10x + 3yc 1 Write the expression. c 7y(x − 2y) + y2(x + 5) 2 Expand each bracket. = 7y(x) + 7y(−2y) + y2(x) + y2(5) 3 Multiply out the brackets. = 7xy − 14y2 + xy2 + 5y2 4 Collect any like terms. = 7xy − 9y2 + xy2d 1 Write the expression. d −5xy(1 + 2y) + 6x(y + 4x) 2 Expand each bracket. = −5xy(1) − 5xy(2y) + 6x(y) + 6x(4x) 3 Multiply out the brackets. = −5xy −10xy2 + 6xy + 24x2 4 Collect any like terms. = xy − 10xy2 + 24x2 As you can see, there is really no difference between the questions in this section and the previous section; just a little more complex ‘bookkeeping’ is required. Be careful when collecting like terms which may only appear to be like. For example, 4x2y and 4xy2 are not like terms at all. remember remember To expand an expression: 1. expand each bracket (working from left to right) 2. collect any like terms.
9. 9. Chapter 5 Expanding 151Expanding pairs of brackets In the previous section we expanded expressions with two brackets which were separated by a + or − sign, such as 5(x + 2y) + 6(x − 3y). In this section we begin to look at expressions where there are two brackets being multiplied together, such as (x + 2y) (x − 3y). These require a more careful analysis and technique. Let us again refer to a numerical example: (2 + 6)(7 − 3) The ‘traditional’ approach, using order of operations from chapter 1, results in: (2 + 6)(7 − 3) = (8)(4) = 32 Consider the ‘alternative’ approach. First multiply the 2 by the second bracket and then the 6 by the second bracket. (2 + 6)(7 − 3) = 2(7 − 3) + 6(7 − 3) = 2(7) + 2(−3) + 6(7) + 6(−3) = 14 − 6 + 42 − 18 = 32 Again, we end up with identical results, no matter which method is used. It may appear unnecessarily long for numeric examples (and, indeed, it is) but it works well for algebraic expressions. When multiplying expressions within brackets, multiply each term in the ﬁrst bracket by each term in the second bracket. WORKED Example 5 Expand and simplify each of the following expressions. a (6 − 5)(15 + 3) b (x − 5)(x + 3) c (x + 2)(x + 3) d (2x + 2)(2x + 3) THINK WRITE a 1 Write the expression. a (6 − 5)(15 + 3) 2 Expand by multiplying each of the = 6(15 + 3) − 5(15 + 3) terms in the ﬁrst bracket by each of the terms in the second bracket. 3 Expand each of the remaining brackets. = 6(15) + 6(3) − 5(15) − 5(3) 4 Simplify. = 90 + 18 − 75 − 15 = 18 5 Check the results using the order of (6 − 5)(15 + 3) operations method. = (1)(18) = 18 b 1 Write the expression. b (x − 5)(x + 3) 2 Expand by multiplying each of the = x(x + 3) − 5(x + 3) terms in the ﬁrst bracket by each of the terms in the second bracket. 3 Expand each of the remaining = (x)(x) + x(3) − 5(x) − 5(3) brackets. = x2 + 3x − 5x − 15 4 Collect like terms. = x2 − 2x − 15 Continued over page
10. 10. 152 Maths Quest 9 for VictoriaTHINK WRITEc 1 Write the expression. c (x + 2)(x + 3) 2 Expand by multiplying each of the terms in the ﬁrst = x(x + 3) + 2(x + 3) bracket by each of the terms in the second bracket. 3 Expand each of the remaining brackets. = x(x) + x(3) + 2(x) + 2(3) = x2 + 3x + 2x + 6 4 Collect like terms. = x2 + 5x + 6d 1 Write the expression. d (2x + 2)(2x + 3) 2 Expand by multiplying each of the terms in the ﬁrst = 2x(2x + 3) + 2(2x + 3) bracket by each of the terms in the second bracket. 3 Expand each of the remaining brackets. = 2x(2x) + 2x(3) + 2(2x) + 2(3) = 4x2 + 6x + 4x + 6 4 Collect like terms. = 4x2 + 10x + 6 Note: The last step in each question (collection of like terms), is most important, and often forgotten. Without completing this step, the expansion is not fully correct. An alternative method Part of the problem in expanding pairs of brackets is a bookkeeping one — keeping track of which terms have been multiplied by which. An alternative approach uses a diagram to keep track of the various multiplication operations. WORKED Example 6 Use a diagrammatic technique to expand (2x + 3y)(4x − 5z). THINK WRITE 1 Write the expression and add 4 curved lines 2 connecting each term, according to the pattern shown. 1 2 Number each of the curved lines. (2x + 3y)(4x – 5z) 3 4 3 Perform each of the multiplications, in order of the 1: 2x(4x) = 8x2 numbers on the lines. 2: 2x(−5z) = −10xz 3: 3y(4x) = 12xy 4: 3y(−5z) = −15yz 4 Write the expression and its expansion by placing the (2x + 3y)(4x − 5z) terms on a single line. = 8x2 − 10xz + 12xy − 15yz 5 Collect like terms if necessary (none in this example). There is nothing magical or special about this method, but it forces you to keep track of the 4 multiplications required in the expansion. This method is often given the name FOIL, where the letters stand for: O First — multiply the ﬁrst term in each bracket. F Outer — multiply the 2 outer terms. (a + b) (c + d) Inner — multiply the 2 inner terms. I L Last — multiply the last term of each bracket.
11. 11. Chapter 5 Expanding 153 remember remember 1. When multiplying expressions within pairs of brackets, multiply each term in the ﬁrst bracket by each term in the second bracket, then collect the like terms. 2. You can use a ‘diagrammatic method’ (or FOIL) to help you keep track of which terms are to be multiplied together. 5C Expanding pairs of bracketsWORKED 1 Expand and simplify each of the following expressions. MathExample a (a + 2)(a + 3) b (x + 4)(x + 3) c (y + 3)(y + 2) cad 5 d (m + 4)(m + 5) e (b + 2)(b + 1) f (p + 1)(p + 4) Expanding pairs of g (a − 2)(a + 3) h (x − 4)(x + 5) i (m + 3)(m − 4) brackets j (y + 5)(y − 3) k (y − 6)(y + 2) l (x − 3)(x + 1) m (x − 3)(x − 4) n (p − 2)(p − 3) o (x − 3)(x − 1)WORKED 2 Use a diagrammatic technique to expand the following. GC proExample a (2a + 3)(a + 2) b (3m + 1)(m + 2) (6x + 4)(x + 1) gram 6 c d (c − 6)(4c − 7) e (7 − 2t)(5 − t) f (1 − x)(9 − 2x) Expanding g (2 + 3t)(5 − 2t) h (7 − 5x)(2 − 3x) i (5x − 2)(5x − 2) 3 Expand and simplify each of the following. a (x + y)(z + 1) b (p + q)(r + 3) c (2x + y)(z + 4) d (3p + q)(r + 1) e (a + 2b)(a + b) f (2c + d)(c − 3d) g (x + y)(2x − 3y) h (4p − 3q)(p + q) i (3y + z)(x + z) j (a + 2b)(b + c) k (3p − 2q)(1 − 3r) l (7c − 2d)(d − 5) m (4x − y)(3x − y) n (p − q)(2p − r) o (5 − 2j)(3k + 1) GAME 4 multiple choice time a The equivalent of (x + 7)(x − 2) is: Expanding A x2 + 5x − 14 B 2x + 5 C x2 − 5x − 14 D x2 + 5x + 14 E x2 − 5x + 14 — 001 b What is the equivalent of (4 − y)(7 + y)? A 28 − y2 B 28 − 3y + y2 C 28 − 3y − y2 D 11 − 2y E 28 + 3y − y2 SHE 5.1 ET Work c The equivalent of (2p + 1)(p − 5) is: A 2p2 − 5 B 2p2 − 11p − 5 C 2p2 − 9p − 5 D 2p2 − 6p − 5 E 2p2 + 9p − 5 QUEST SM AT H GE 1 Xavier left for school in the morning. One quarter of the way to school, he passed a post ofﬁce. The clock on the outside of the post ofﬁce EN showed 7.44 am. Halfway to school, he passed a convenience store. The CH L time shown there was 7.53 am. If Xavier continues walking at the same AL speed, at what time will he get to school? 2 The human heart beats about 105 times each day. Approximately how many times does the heart beat in an 80-year lifetime? 3 There are 12 people trying out for a tennis team. Five of them are girls. What percentage of the possible doubles teams could be mixed double teams?
12. 12. 154 Maths Quest 9 for Victoria What has area got to do with expanding? 1 Draw a square of side length x. x 3 2 What is the area of this square? Consider the rectangle at right which has x a length 3 units longer and a width 2 units wider than the square you have just drawn. Notice that it has been divided up into 4 regions. 3 Find the area of each of the 4 regions. 2 4 What is the total area of the rectangle? 5 Write an expression for the length of the rectangle. 6 Write an expression for the width of the rectangle. 7 Using the relationship, area = length × width and your answers to parts 5 and 6, write an expression for the area of the rectangle. 8 Relate your answers to parts 4 and 7. What have you noticed? 9 Show, using a diagram and areas of 4 regions, how to obtain the expanded expression for (x + 5)(x + 7). 10 Show, using a diagram and areas of 4 regions, how to obtain the expanded expression for (a + b)(c + d). 11 Challenge: Show, using a diagram and areas of 4 regions, how to obtain the expanded expression for (x − 2)(x + 3). 1 1 Expand 5(x + 3). 2 Expand z(3 − 7z). 3 Expand 2(p − 7q) + 3p − 5q and simplify by collecting like terms. 4 Expand and simplify 5(a + 2b) + 2(3a + b). 5 Expand and simplify m(n + 1) + n(m − 4). 6 multiple choice If K(a − 3b) − (a + 10b) = 3a − 22b, then the missing number is: A 1 B 2 C 3 D 4 E 5 7 Expand and simplify (a + 1)(a + 4). 8 Expand and simplify (p − 7)(6p − 3). 9 Expand and simplify (4 − 3j)(4k + 2). 10 True or false? (x + 4)(x + 10) = x2 + 14x + 40
13. 13. Chapter 5 Expanding 155Expansion patterns Although the techniques learned in the previous section are perfectly adequate for all expansions of pairs of brackets, there are some ‘special’ cases where the expansion is particularly simple and can be done very quickly if you recognise the pattern. After comparing the result with that obtained using previous methods, perhaps you will adopt these ‘short cuts’. Difference of two squares rule The ﬁrst pattern we will examine is obtained as a result of expanding a pair of brackets to produce a ‘difference of two squares’. That is, we produce two terms which are perfect squares (can be expressed as a number and/or a pronumeral squared) where one term is subtracted from the other. Consider expanding (x + 3)(x − 3). (x + 3)(x − 3) = x(x − 3) + 3(x − 3) = x(x) + x(−3) + 3(x) + 3(−3) = x2 − 3x + 3x − 9 = x2 − 9 Notice how the ‘middle terms’, −3x + 3x cancel each other out. This is the key to the pattern and will always happen. (Can you prove this?) Note: The terms left over are the squares of each of the original terms. In other words, (x + 3)(x − 3) = x2 − 32. Notice the pattern of terms in the pair of brackets which produce the difference of two squares. Here are some more examples. (x + 5)(x − 5) (x + 4)(x − 4) (x + h)(x − h) (2x + 7)(2x − 7) = x2 − 52 = x2 − 42 = x2 − h2 = (2x)2 − 72 = x2 − 25 = x2 − 16 = 4x2 − 49 Therefore, when we recognise this pattern we merely have to write the squares of each of the two terms and place a minus sign between them. (a + b)(a − b) = a2 − b2 WORKED Example 7 Use the difference of two squares rule, if possible, to expand and simplify each of the following. a (x + 8)(x − 8) b (6 − x)(6 + x) c (2x − 3)(2x + 3) d (5 + 3x)(5 − 3x) THINK WRITE a 1 Write the expression. a (x + 8)(x − 8) 2 Check that the expression can be written as the difference of two squares by comparing it with (a + b)(a − b). It can. 3 Write the answer as the difference of = x2 − 82 two squares using the formula = x2 − 64 (a + b)(a − b) = a − b , where a = x 2 2 Continued over page and b = 8.