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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.

Expanding

MQ 9 Vic - 05 43 Monday, September 17, 2001 9:13 AM

144

M a t h s Q u e s t 9 f o r V i c t o r i a

Expanding single brackets

In the previous chapter on introductoryalgebra, we learned that algebra is a type oflanguage.

In this chapter we can take the idea fur-ther and look at some of the ‘advancedfeatures’ of this language. At first we willrely on using common numerical examplesto illustrate the techniques, but after awhile, 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 reallystands for

will not

. In going from

won’t

to

will not

we have expanded the word, butthe meaning remains unchanged. It is thesame with expanding in algebra; we gofrom a more compact form (won’t) to anexpanded form (will not). Consider thefollowing 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)

=

27Consider 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

=

27You 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.

MQ 9 Vic - 05 Page 144 Monday, September 17, 2001 9:13 AM

C h a p t e r 5 E x p a n d i n g

145

Note

: It doesn’t matter what is immediately outside the brackets. It may be a number ora pronumeral or both. The following expansions are a little more complex.

Expand the following expressions.a 5(4 + 3) b 5(x + 3) c 5(x − y) d −a(x − y)THINK WRITEa Write the expression. a 5(4 + 3)

Expand the brackets. = 5(4) + 5(3)Multiply out the brackets. = 20 + 15

= 35Check that the result is valid by simplifying the brackets in the original expression first.

5(4 + 3)= 5(7) = 35, so it is indeed valid.

b Write the expression. b 5(x + 3)Expand the brackets. = 5(x) + 5(3)Multiply out the brackets. = 5x + 15

c Write the expression. c 5(x − y) Expand the brackets. = 5(x) + 5(−y) Multiply out the brackets.(Remember that a positive term multiplied by a negative term makes a negative term.)

= 5x − 5y

d Write the expression. d −a(x − y) Expand the brackets. = −a(x) − a(−y)Multiply out the brackets.(Remember that a negative term multiplied by a negative term makes a positive term.)

= −ax + ay

123

4

123123

123

1WORKEDExample

Expand each of the following.a 5x(6y − 7z) b −4y(2x + 3w) c x(2x + 3y)THINK WRITEa Write the expression. a 5x(6y − 7z)

Expand the brackets. = 5x(6y) + 5x(−7z)Multiply out the brackets.(Multiply number parts and pronumeral parts separately and write pronumerals for each term in alphabetical order.)

= 30xy − 35xz

b Write the expression. b −4y(2x + 3w) Expand the brackets. = −4y(2x) − 4y(3w) Multiply out the brackets. = −8xy − 12wy

c Write the expression. c x(2x + 3y) Expand the brackets. = x(2x) + x(3y) Multiply out the brackets.(Remember that x multiplied by itself gives x2.)

= 2x2 + 3xy

123

123123

2WORKEDExample


146 M a t h s Q u e s t 9 f o r V i c t o r i a

Expanding and collecting like termsWith more complicated expansions, like terms may need to be collected after theexpansion of the bracketed part. Remember that like terms contain the samepronumeral parts. You first expand the brackets, then collect the like terms.


1 Expand the following expressions.a 3(x + 2) b 4(x + 3) c 5(m + 4) d 2(p + 5)e 4(x + 1) f 7(x − 1) g −4(y + 6) h −5(a + 1)i −3(p − 2) j −(x − 1) k −(x + 3) l −(x − 2)m 3(2b − 4) n 8(3m − 2) o −6(5m − 4) p −3(9p − 5)

2 Expand each of the following.a x(x + 2) b y(y + 3) c a(a + 5) d c(c + 4)e x(4 + x) f y(5 + y) g m(7 − m) h q(8 − q)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)

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 WRITEa Write the expression. a 4(x − 4) + 5

Expand the brackets. = 4(x) + 4(−4) + 5Multiply out the brackets. = 4x − 16 + 5 Collect any like terms. = 4x − 11

b Write the expression. b x(y − 2) + 5xExpand the brackets. = x(y) + x(−2) + 5x Multiply out the brackets. = xy − 2x + 5xCollect any like terms. = xy + 3x

c Write the expression. c x(y − z) + 5x Expand the brackets. = x(y) + x(−z) + 5x Multiply out the brackets. = xy − xz + 5xCollect any like terms. There are no like terms.

d Write the expression. d 7x + 6(y − 2x)Expand the brackets. = 7x + 6(y) + 6(−2x)Multiply out the brackets. = 7x + 6y − 12x Collect any like terms. = −5x + 6y

1234123412341234

3WORKEDExample

remember1. 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.

remember

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WWORKEDORKEDEExample

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2

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Expanding


C h a p t e r 5 E x p a n d i n g 1473 Expand and simplify by collecting like terms.

a 2(p − 3) + 4 b 5(x − 5) + 8 c −7(p + 2) − 3d −4(3p − 1) − 1 e 6x(x − 3) − 2x f 2m(m + 5) − 3mg 3x(p + 2) − 5 h 4y(y − 1) + 7 i −4p(p − 2) + 5pj 5(x − 2y) − 3y − x k 2m(m − 5) + 2m − 4 l −3p(p − 2q) + 4pq − 1m −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 − 1b 3x + 6x = 9 + xc 4(3x) = 12x

d 8x − 3x = 5e −2(x − 7) = −2x −14f 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 gameIn 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 correctpico: one digit is correct but in the wrong positionfermi: 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 pico789 pico 387 pico fermi075 pico fermi 256 fermi087 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.

SkillSH

EET 5.1WWORKEDORKEDEExample

3


Expand the brackets in theexpressions given to find the puzzle’s

answer code.

What is a metrWhat is a metre?e?

e – 2ae e – 2ae

e – 2ae

e – 2ae 2ae – 3e6e – 3e 2

6e – 3e 2

6e – 3e 2

6e – 3e 2 5a – 10ae 6a – 2

6e – 3e 2 2ae – 3e6a – 23a + ae

3a + ae

e – 2ae5e – 108 – 4e

8 – 4e

8 – 4e 8 – 4e

2ae – 3e

2ae – 3e

2ae – 3e 3a + ae

3ae + 6e3a + ae

3a + ae

6e – 3e 2

8 – 4e –4a – 65e – 10 6a – 2 6a – 25e – 108 – 4e

8 – 4e

a 2e – ae 2

a 2e – ae 2

a 2 – ae e 2 + e e 2 + e a 2 – 4a

a 2 – 4a

a 2 – 4a

a 2 – 4a

a 2 – 4a a 2 – 4a

e 2 + e

– 4a – 6e – 2ae

a 2e – ae 2

6e – 3e 2

6e – 3e 2

6e – 3e 2 2ae – 2ae 26e – 3e 2

8 – 4e

6 – 3a 8 – 4e –4a – 6 2ae – 2ae 2

4a 2 – 4a

2ae – 2ae 2

a 2 – 4a a 2 – ae

= 2(3a – 1)

=

= 3(2 – a)

=

= 5(e – 2)

=

= –4(e – 2)

=

= –2(2a + 3)

=

= a(3 + e)

=

= –a (e – a)

=

= –a(4 – a)

=

= –e (–2a + 3)

=

= 3e(a + 2)

=

= –2ae(e – 1)

=

= –4a(–a + 1)

=

= 3e(2 – e)

=

= ae(a – e)

=

= 5a(1 – 2e)

=

= e(1 – 2a)

=

= e(e + 1)

=



C h a p t e r 5 E x p a n d i n g 149

Expanding two bracketsWhen expanding an expression that contains two (or more) brackets, the steps are thesame as before.

Step 1 Expand each bracket (working from left to right).

Step 2 Collect any like terms.

As you can see, there is really no difference between the questions in this section andthe previous section; just a little more complex ‘bookkeeping’ is required. Be carefulwhen collecting like terms which may only appear to be like. For example, 4x2y and4xy2 are not like terms at all.

Expand 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 WRITE

a Write the expression. a 5(x + 2y) + 6(x − 3y)

Expand each bracket. = 5(x) + 5(2y) + 6(x) + 6(−3y)

Multiply out the brackets. = 5x + 10y + 6x − 18y

Collect any like terms. = 11x − 8y

b Write the expression. b −5x(y − 2) + y(x + 3)

Expand each bracket. = −5x(y) − 5x(−2) + y(x) + y(3)

Multiply out the brackets. = −5xy + 10x + xy + 3y

Collect any like terms. = −4xy + 10x + 3y

c Write the expression. c 7y(x − 2y) + y2(x + 5)

Expand each bracket. = 7y(x) + 7y(−2y) + y2(x) + y2(5)

Multiply out the brackets. = 7xy − 14y2 + xy2 + 5y2

Collect any like terms. = 7xy − 9y2 + xy2

d Write the expression. d −5xy(1 + 2y) + 6x(y + 4x)

Expand each bracket. = −5xy(1) − 5xy(2y) + 6x(y) + 6x(4x)

Multiply out the brackets. = −5xy −10xy2 + 6xy + 24x2

Collect any like terms. = xy − 10xy2 + 24x2

1

2

3

4

1

2

3

4

1

2

3

4

1

2

3

4

4WORKEDExample

rememberTo expand an expression:

1. expand each bracket (working from left to right)

2. collect any like terms.

remember



Expanding two brackets

1 Expand and simplify the following expressions.a 2(x + 2y) + 3(2x − y) b 4(2p + 3q) + 2(p − 2q)c 7(2a + 3b) + 4(a + 2b) d 5(3c + 4d) + 2(2c + d)e −4(m + 2n) + 3(2m − n) f −3(2x + y) + 4(3x − 2y)g −2(3x + 2y) + 3(5x + 3y) h −5(4p + 2q) + 2(3p + q)i 6(a − 2b) − 5(2a − 3b) j 5(2x − y) − 2(3x − 2y)k 4(2p − 4q) − 3(p − 2q) l 2(c − 3d) − 5(2c − 3d)m 7(2x − 3y) − (x − 2y) n −5(p − 2q) − (2p − q)o −3(a − 2b) − (2a + 3b) p 4(3c + d) − (4c + 3d)

2 Expand and simplify the following expressions.a a(b + 2) + b(a − 3) b x(y + 4) + y(x − 2)c c(d − 2) + c(d + 5) d p(q − 5) + p(q + 3)e 3c(d − 2) + c(2d − 5) f 7a(b − 3) − b(2a + 3)g 2m(n + 3) − m(2n + 1) h 4c(d − 5) + 2c(d − 8)i 3m(2m + 4) − 2(3m + 5) j 5c(2d − 1) − (3c + cd)k −3a(5a + b) + 2b(b − 3a) l −4c(2c − 6d) + d(3d − 2c)m 6m(2m − 3) − (2m + 4) n 2p(p − 4) + 3(5p − 2)o 7x(5 − x) + 6(x − 1) p −2y(5y − 1) − 4(2y + 3)

3a What is the equivalent of 3(a + 2b) + 2(2a − b)?

b What is the equivalent of −3(x − 2y) − (x − 5y)?

c What is the equivalent of 2m(n + 4) + m(3n − 2)?

A 5a + 6b B 7a + 4b C 5(3a + b) D 7a + 8b E 12a − 12b

A −4x + 11y B −4x − 11y C 4x + 11y D 4x + 7y E 3x + 30y

A 3m + 4n − 8 B 5mn + 4m C 5mn + 10m D 5mn + 6m E 6mn − 16m

5B

SkillSH

EET 5.2 WWORKEDORKEDEExample

4a

Mathca

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Expanding two brackets

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4b, c, d

mmultiple choiceultiple choice

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1 Find the next 3 terms of this algebraic sequence.x + 3y, 2x + y, 3x + 4y, 5x + 5y, 8x + 9y, 13x + 14y, ….

2 Mind-reading tricks often use algebra as a base. Try the followingmind-reading trick. Use algebra to explain why the trick works.

Double the number of the month in which you were born. Subtract16 from your answer. Subtract 20 from your result, then multiply by 10.Finally, add the day of the month in which you were born to youranswer. The number you end up with shows the month and day youwere born. For example, if you were born on June 17, your answer willbe 617.

Try this trick on another person. ‘Read’ the person’s mind by statingthe month and day they were born from the number they tell you at theend of the calculation.



Expanding pairs of bracketsIn the previous section we expanded expressions with two brackets which were separatedby a + or − sign, such as 5(x + 2y) + 6(x − 3y). In this section we begin to look atexpressions 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 andthen 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 mayappear unnecessarily long for numeric examples (and, indeed, it is) but it works wellfor algebraic expressions.

When multiplying expressions within brackets, multiply each term in the first bracket by each term in the second bracket.

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 WRITEa Write the expression. a (6 − 5)(15 + 3)

Expand by multiplying each of the terms in the first bracket by each of the terms in the second bracket.

= 6(15 + 3) − 5(15 + 3)

Expand each of the remaining brackets. = 6(15) + 6(3) − 5(15) − 5(3)Simplify. = 90 + 18 − 75 − 15

= 18Check the results using the order of operations method.

(6 − 5)(15 + 3)= (1)(18)= 18

b Write the expression. b (x − 5)(x + 3)Expand by multiplying each of the terms in the first bracket by each of the terms in the second bracket.

= x(x + 3) − 5(x + 3)

Expand each of the remaining brackets.

= (x)(x) + x(3) − 5(x) − 5(3) = x2 + 3x − 5x − 15

Collect like terms. = x2 − 2x − 15

12

34

5

12

3

4

5WORKEDExample

Continued over page



Note: The last step in each question (collection of like terms), is most important, andoften forgotten. Without completing this step, the expansion is not fully correct.

An alternative methodPart of the problem in expanding pairs of brackets is a bookkeeping one — keepingtrack of which terms have been multiplied by which. An alternative approach uses adiagram to keep track of the various multiplication operations.

There is nothing magical or special about this method, but it forces you to keep track ofthe 4 multiplications required in the expansion.

This method is often given the name FOIL, where the letters stand for:First — multiply the first term in each bracket.Outer — multiply the 2 outer terms.Inner — multiply the 2 inner terms.Last — multiply the last term of each bracket.

THINK WRITEc Write the expression. c (x + 2)(x + 3)


= x(x + 3) + 2(x + 3)

Expand each of the remaining brackets. = x(x) + x(3) + 2(x) + 2(3)= x2 + 3x + 2x + 6

Collect like terms. = x2 + 5x + 6d Write the expression. d (2x + 2)(2x + 3)


= 2x(2x + 3) + 2(2x + 3)

Expand each of the remaining brackets. = 2x(2x) + 2x(3) + 2(2x) + 2(3)= 4x2 + 6x + 4x + 6

Collect like terms. = 4x2 + 10x + 6

12

3

412

3

4

Use a diagrammatic technique to expand (2x + 3y)(4x − 5z).THINK WRITE

Write the expression and add 4 curved lines connecting each term, according to the pattern shown.Number each of the curved lines.

Perform each of the multiplications, in order of the numbers on the lines.

1: 2x(4x) = 8x2

2: 2x(−5z) = −10xz3: 3y(4x) = 12xy4: 3y(−5z) = −15yz

Write the expression and its expansion by placing the terms on a single line.

(2x + 3y)(4x − 5z) = 8x2 − 10xz + 12xy − 15yz

Collect like terms if necessary (none in this example).

1

(2x + 3y)(4x – 5z)1

2

34

2

3

4

5

6WORKEDExample

(a + b) (c + d)F

O

IL



Expanding pairs of brackets

1 Expand and simplify each of the following expressions.a (a + 2)(a + 3) b (x + 4)(x + 3) c (y + 3)(y + 2)d (m + 4)(m + 5) e (b + 2)(b + 1) f (p + 1)(p + 4)g (a − 2)(a + 3) h (x − 4)(x + 5) i (m + 3)(m − 4)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)

2 Use a diagrammatic technique to expand the following.a (2a + 3)(a + 2) b (3m + 1)(m + 2) c (6x + 4)(x + 1)d (c − 6)(4c − 7) e (7 − 2t)(5 − t) f (1 − x)(9 − 2x)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)

4a The equivalent of (x + 7)(x − 2) is:

b What is the equivalent of (4 − y)(7 + y)?

c The equivalent of (2p + 1)(p − 5) is:

A x2 + 5x − 14 B 2x + 5 C x2 − 5x − 14 D x2 + 5x + 14 E x2 − 5x + 14

A 28 − y2 B 28 − 3y + y2 C 28 − 3y − y2 D 11 − 2y E 28 + 3y − y2

A 2p2 − 5 B 2p2 − 11p − 5 C 2p2 − 9p − 5 D 2p2 − 6p − 5 E 2p2 + 9p − 5

remember1. 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.

remember

5CWWORKEDORKEDEExample

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Mathcad

Expandingpairs ofbrackets


6

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Expanding

mmultiple choiceultiple choiceGAMEtime

Expanding— 001

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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 officeshowed 7.44 am. Halfway to school, he passed a convenience store. Thetime shown there was 7.53 am. If Xavier continues walking at the samespeed, at what time will he get to school?

2 The human heart beats about 105 times each day. Approximately howmany 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. Whatpercentage of the possible doubles teams could be mixed double teams?



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).

6If K(a − 3b) − (a + 10b) = 3a − 22b, then the missing number is:

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

What has area got to do with expanding?

1 Draw a square of side length x.2 What is the area of this square?

Consider the rectangle at right which has 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.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).

A 1 B 2 C 3 D 4 E 5

x

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mmultiple choiceultiple choice



Expansion patternsAlthough the techniques learned in the previous section are perfectly adequate for allexpansions of pairs of brackets, there are some ‘special’ cases where the expansion isparticularly simple and can be done very quickly if you recognise the pattern. Aftercomparing the result with that obtained using previous methods, perhaps you will adoptthese ‘short cuts’.

Difference of two squares ruleThe first pattern we will examine is obtained as a result of expanding a pair of bracketsto produce a ‘difference of two squares’.

That is, we produce two terms which are perfect squares (can be expressed as anumber 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 thepattern and will always happen. (Can you prove this?) Note: The terms left over are thesquares 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

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 WRITEa Write the expression. a (x + 8)(x − 8)

Check that the expression can be written as the difference of two squares by comparing it with (a + b)(a − b). It can.Write the answer as the difference of two squares using the formula(a + b)(a − b) = a2 − b2, where a = x and b = 8.

= x2 − 82 = x2 − 64

12

3

7WORKEDExample

Continued over page



Expanding identical brackets (perfect squares)The next pattern worth examining is the expansion of identical brackets. That is, eachbracket is the same, such as (x + 3)(x + 3), which can be written as (x + 3)2. Again, wecan express this using the two symbols a and b. So a perfect square could be written inthe form (a + b)(a + b).

Let us try this with a set of numbers first; using the pair of brackets (3 + 4)(3 + 4)which equals 49.

(3 + 4)(3 + 4) = 3(3 + 4) + 4(3 + 4)= 3(3) + 3(4) + 4(3) + 4(4) (Can you see a pattern here?)= 9 + 12 + 12 + 16= 49

Now let’s try it with pronumerals.(x + 3)(x + 3) = x(x + 3) + 3(x + 3)

= x(x) + x(3) + 3(x) + 3(3) = x2 + 3x + 3x + 9= x2 + 6x + 9

There is a pattern in these expansions, fairly similar to the one we just learned.

(a + b)(a + b) = a2 + 2ab + b2

There is an additional result that occurs when there is a minus sign instead of a plussign:

(a − b)(a − b) = a2 − 2ab + b2

The difference between the two patterns is quite small: the minus sign in the bracketsresults in a single minus sign in the ‘middle term’, −2ab.

This pattern can also be described in words.

Square the first term, add the square of the last term, then add (or subtract) twice their product.

THINK WRITEb Write the expression. b (6 − x)(6 + x)

Check that the difference of two squares rule can be used. It can, because (6 − x)(6 + x) means the same as (6 + x)(6 − x).Write the answer as the difference of two squares using the formula(a + b)(a − b) = a2 − b2, where a = 6 and b = x.

= 62 − x2 = 36 − x2

c Write the expression. c (2x − 3)(2x + 3) Check that the difference of two squares rule can be used. It can.Write the answer as the difference of two squares using the formula(a + b)(a − b) = a2 − b2, where a = 2x and b = 3.

= (2x)2 − 32 = 4x2 − 9

d Write the expression. d (5 + 3x)(5 − 3x) Check that the difference of two squares rule can be used. It can.Write the answer as the difference of two squares using the formula(a + b)(a − b) = a2 − b2, where a = 5 and b = 3x.

= 52 − (3x)2 = 25 − 9x2

12

3

12

3

12

3



Use the identical brackets (perfect squares) technique to expand and simplify the following.a (x + 1)(x + 1) b (x − 2)2 c (2x + 5)2 d (4x − 5y)2

THINK WRITE

a Write the expression. a (x + 1)(x + 1)Recognise the pattern of identical brackets: square the first term.

(x)(x) = x2

Square the last term. (1)(1) = 1Add (because of the + sign in the bracket) twice the product.

2(x)(1) = 2x

Apply the formula:(a + b)(a + b) = a2 + 2ab + b2.

(x + 1)(x + 1) = x2 + 2x + 1

b Write the expression and express it as a pair of brackets.

b (x − 2)2 = (x − 2)(x − 2)

Recognise the pattern of identical brackets: square the first term.

(x)(x) = x2

Square the last term. (−2)(−2) = 4Subtract (because of the − sign in the bracket) twice the product.

−2(x)(2) = −4x

Apply the formula:(a − b)(a − b) = a2 − 2ab + b2.

(x − 2)2 = x2 − 4x + 4

c Write the expression and express it as a pair of brackets.

c (2x + 5)2

= (2x + 5)(2x + 5) Recognise the pattern of identical brackets: square the first term.

(2x)(2x) = 4x2

Square the last term. (5)(5) = 25Add twice the product. 2(2x)(5) = 20xApply the formula: (a + b)(a + b) = a2 + 2ab + b2.

(2x + 5)2 = 4x2 + 20x + 25

d Write the expression and express it as a pair of brackets.

d (4x − 5y)2

= (4x − 5y)(4x − 5y)Recognise the pattern of identical brackets: square the first term.

(4x)(4x) = 16x2

Square the last term. (−5y)(−5y) = 25y2

Subtract twice the product. −2(4x)(5y) = −40xyApply the formula:(a − b)(a − b) = a2 − 2ab + b2.

(4x − 5y)2 = 16x2 − 40xy + 25y2

12

34

5

1

2

34

5

1

2

345

1

2

345

8WORKEDExample

remember1. The difference of two squares rule is: (a + b)(a − b) = a2 − b2.

2. The identical brackets (perfect squares) rules are:(a + b)(a + b) = a2 + 2ab + b2

(a − b)(a − b) = a2 − 2ab + b2.

remember

MQ 9 Vic - 05 7 Monday, September 17, 2001 9:13 AM


Expansion patterns

1 Use the difference of two squares rule, if possible, to expand and simplify each of thefollowing.a (x + 2)(x − 2) b (y + 3)(y − 3) c (m + 5)(m − 5)d (a + 7)(a − 7) e (x + 6)(x − 6) f (p − 12)(p + 12)g (a + 10)(a − 10) h (m − 11)(m + 11) i (p − q)(p + q)

2 Use the difference of two squares rule, if possible, to expand and simplify each of thefollowing.a (2x + 3)(2x − 3) b (3y − 1)(3y + 1) c (5d − 2)(5d + 2)d (7c + 3)(7c − 3) e (2 + 3p)(2 − 3p) f (1 − 9x)(1 + 9x)g (5 − 12a)(5 + 12a) h (3 + 10y)(3 − 10y) i (2b − 5c)(2b + 5c)

3 Use the identical brackets (perfect squares) rules to expand and simplify each of thefollowing.a (x + 2)(x + 2) b (a + 3)(a + 3) c (b + 7)(b + 7)d (c + 9)(c + 9) e (m + 12)2 f (n + 10)2

g (x − 6)2 h (y − 5)2 i (9 − c)2

j (8 + e)2 k (x + y)2 l (u − v)2

4 Use the identical brackets (perfect squares) rules to expand and simplify each of thefollowing.a (2a + 3)2 b (3x + 1)2 c (2m − 5)2 d (4x − 3)2

e (5a − 1)2 f (7p + 4)2 g (9x + 2)2 h (4c − 6)2

i (3 + 2a)2 j (5 + 3p)2 k (2 − 5x)2 l (7 − 3a)2

m (9x − 4y)2 n (8x − 3y)2 o (9x − 2y)2 p (7x − 4y)2

Using expanding formulas to square large numbers

Can you evaluate 9972 without a calculator and in less than 90 seconds? We would be able to evaluate this using long multiplication, but it would take a fair amount of time and effort. Mathematicians are always looking for quick and simple ways of solving problems.

What if we consider the expanding formula which produces the difference of two squares?(a + b)(a − b) = a2 − b2

Adding b2 to both sides gives (a + b)(a − b) + b2 = a2 − b2 + b2.Simplifying and swapping sides gives a2 = (a + b)(a − b) + b2.We can use this new formula and the fact that multiplying by 1000 is an easy

operation to evaluate 9972.1 If a = 997, what should we make the value of b become so that (a + b) equals 1000?2 Substitute these a and b values into the formula to evaluate 9972.3 Try this method to evaluate the following.

a 9952 b 9902 c 992 d 99912 e 99 9982

4 Can you use the expanding formulas (a + b)2 = a2 + 2ab + b2 or (a − b)2 = a2 − 2ab + b2 to evaluate 9972? Explain your method for this.

5 List three examples of your own and show how you were able to evaluate them using the method from part 4.

5D

EXCEL

Spreadsheet

Expanding(ax + b)(ax – b)


7a, b

Mathca

d

Expansionpatterns


7c, d


8a, b

EXCEL

Spreadsheet

Expanding(ax + b)2


8c, d



More complicated expansionsAlthough we have covered many expansion problems and patterns, these represent onlya small proportion of the possible algebraic expressions that can be expanded. Never-theless, the techniques we have learned so far are very useful — more complicatedexpansions just require more ‘bookkeeping’. The most important of these bookkeepingfunctions is the collection of like terms after expansion.

Expanding more than two bracketsThere are several possible combinations, such as expanding 3 brackets, 4 brackets, and so on.

Expand and simplify each of the following expressions.a (x + 3)(x + 4) + 4(x − 2) b (x − 2)(x + 3) − (x − 1)(x + 2)c (x + 2)(x − 2) − (x + 2)(x + 2) d 2(x + 3)(x − 4) + (x − 2)2

THINK WRITEa Write the expression. a (x + 3)(x + 4) + 4(x − 2)

Expand and simplify the first pair of brackets.

(x + 3)(x + 4)= x2 + 4x + 3x + 12= x2 + 7x + 12

Expand the last bracket. 4(x − 2)= 4x − 8

Add the two results. (x + 3)(x + 4) + 4(x − 2)= x2 + 7x + 12 + 4x − 8

Collect like terms. = x2 + 11x + 4b Write the expression. b (x − 2)(x + 3) − (x − 1)(x + 2)

Expand and simplify the first pair of brackets.

(x − 2)(x + 3)= x2 + 3x − 2x − 6= x2 + x − 6

Expand and simplify the second pair of brackets.

(x − 1)(x + 2)= x2 + 2x − x − 2= x2 + x − 2

Subtract all of the second result from the first result. (So, place the second result in a bracket.)Remember that −(x2 + x − 2) = −1(x2 + x − 2).

(x − 2)(x + 3) − (x − 1)(x + 2)= x2 + x − 6 − (x2 + x − 2)

Collect like terms. = x2 + x − 6 − x2 − x + 2= −4

c Write the expression. c (x + 2)(x − 2) − (x + 2)(x + 2)Expand the first pair of brackets. It is a difference of two squares expansion.

(x + 2)(x − 2)= x2 − 4

Expand the second pair of brackets. It is an identical bracket expansion.

(x + 2)(x + 2)= x2 + 4x + 4

Subtract the two results. (x + 2)(x − 2) − (x + 2)(x + 2)= x2 − 4 − (x2 + 4x + 4)

Collect like terms. = x2 − 4 − x2 − 4x − 4= −4x − 8

1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

9WORKEDExample

Continued over page



More complicated expansions

Expand and simplify each of the following expressions.

1 (x + 3)(x + 5) + (x + 2)(x + 3) 2 (x + 4)(x + 2) + (x + 3)(x + 4)

3 (x + 5)(x + 4) + (x + 3)(x + 2) 4 (x + 1)(x + 3) + (x + 2)(x + 4)

5 (p − 3)(p + 5) + (p + 1)(p − 6) 6 (a + 4)(a − 2) + (a − 3)(a − 4)

7 (p − 2)(p + 2) + (p + 4)(p − 5) 8 (x − 4)(x + 4) + (x − 1)(x + 20)

9 (y − 1)(y + 3) + (y − 2)(y + 2) 10 (d + 7)(d + 1) + (d + 3)(d − 3)

11 (x + 2)(x + 3) + (x − 4)(x − 1) 12 (y + 6)(y − 1) + (y − 2)(y − 3)

13 (x + 2)2 + (x − 5)(x − 3) 14 (y − 1)2 + (y + 2)(y − 4)

15 (p + 2)(p + 7) + (p − 3)2 16 (m − 6)(m − 1) + (m + 5)2

17 (x + 3)(x + 5) − (x + 2)(x + 5) 18 (x + 5)(x + 2) − (x + 1)(x + 2)

19 (x + 3)(x + 2) − (x + 4)(x + 3) 20 (m − 2)(m + 3) − (m + 2)(m − 4)

21 (b + 4)(b − 6) − (b − 1)(b + 2) 22 (y − 2)(y − 5) − (y + 2)(y + 6)

23 (p − 1)(p + 4) − (p − 2)(p − 3) 24 (x + 7)(x + 2) − (x − 3)(x − 4)

25 (c − 2)(c − 1) − (c + 6)(c + 7) 26 ( f − 7)( f + 2) − ( f + 4)( f + 5)

27 (m + 3)2 − (m + 4)(m − 2) 28 (a − 6)2 − (a − 2)(a − 3)

29 (p − 3)(p + 1) − (p + 2)2 30 (x + 5)(x − 4) − (x − 1)2

THINK WRITE

d Write the expression. d 2(x + 3)(x − 4) + (x − 2)2

Expand and simplify the first pair of brackets. Then multiply by the coefficient of 2 outside the pair.

2(x + 3)(x − 4)= 2(x2 − 4x + 3x − 12)= 2(x2 − x − 12)= 2x2 − 2x − 24

Expand the second pair of brackets. It is an identical bracket expansion.

(x − 2)2

= x2 − 4x + 4

Add the two results. 2(x + 3)(x − 4) + (x − 2)2

= 2x2 − 2x − 24 + (x2 − 4x + 4)

Collect like terms. = 2x2 − 2x − 24 + x2 − 4x + 4= 3x2 − 6x − 20

1

2

3

4

5

remember1. Brackets or pairs of brackets that are added or subtracted must be expanded

separately.2. Always collect any like terms following an expansion.

remember

5E

SkillSH

EET 5.3

SkillSH

EET 5.4


9

Mathca

d

More complicated expansions

GCpro

gram

Expanding



161

Expand and simplify the expressionsto find the puzzle’s answer.

WhWhy does the giraffy does the giraffe hae havve a long neck?e a long neck?

–8xx 2 + x –3x 2 + 2x + 1

–x 2 + 12x –2x 2 – 6x 2x 2 + x 2x – 9 – 95x 3x 2 – 5x – 1 2x 2 – x – 16 2x 2 – x + 1 6x 2 – 13x 6x + 10x 2 + 3x + 10 x 2 + 7x – 2x 2 + 5x – 2

–3x + 3–x 2 + 19 x 2 + 2–5x 2 + x – 3 –6x + 3 –33x + 9x 2 + 8x – 18x 2 – x 5x + 3–x – 1

= 2(x + 3) + 3(x – 1)

=

= (x – 3)(x + 2) + 6

=

= x (x – 4) + 2(2x + 1)

=

= (x + 5)(x – 2) + 2(x + 4)

=

= (x – 3)(x + 3) – x (x – 2)

=

= 5(x + 1) – 2(x – 2)

=

= (x + 2)(x – 2) + (x + 3)(x – 4)

=

= (x – 4)(x + 3) – (x + 5)(x – 3)

=

= 5(x 2 + 3x + 2) – 4(x 2 + 3x)

=

= (x + 2) 2 + 3(x – 2)

=

= (x – 1)(x + 6) + 3(x – 4)

=

= 2(x – 7)(x + 5) + 5(x + 14)

=

= –3(x + 7) – (x + 5)(x – 8)

=

= (x – 5)(x + 5) – (x + 4)(x – 4)

=

= (2 + x)(3 – x) + (x + 1)(x + 4)

=

= (x + 2)(x – 2) – (x – 1)(x + 1)

=

= –3(x 2 + x – 2) + 5(x – 1)

=

= 5(x – 7) + 2(x 2 – 3x + 18)

=

= (x – 2)2 – (x + 1)2

=

= (x – 6)(x – 7) + 2(7x – 21)

=

= (2x – 3)(3x – 2) – 6

=

= 7(2 – 3x) + 5(4x – 3)

=

= 4(2x – 6) – 3(x – 8)

== –2(x 2 + 3x – 6) – 12

== 2(x 2 + 3x – 7) – 2(x 2 + 7x – 7)

=

= 2x(x – 3) + 3x(6 – x)

=

= 5x(2 – x) – 3(3x + 1)

=

= 3(x 2 – x + 1) – 2(x + 2)

=

’

MQ 9 Vic - 05 Page 161 Thursday, September 20, 2001 2:31 PM


Higher order expansions and Pascal’s triangle

Pascal’s triangle, as shown in the figure below, is a special arrangement of numbers in a triangular shape. Any number is the sum of the two numbers immediately above it, with 1s running down the sides. The triangle was named after Blaise Pascal, a French mathematician who wrote about the triangle in 1653. However, earlier mathematicians knew about the ‘magic’ of this triangle. Chu Shih-Chieh, a Chinese mathematician, included an illustration of the triangle in a book in 1303.

There are many patterns to observe with Pascal’s triangle.

Let’s look at one of these.

1 a Expand (x + 1)2.b Which line of Pascal’s triangle links to your answer for part 1a? Describe the

pattern you have observed.2 a Expand (x + 1)3. This means expand (x + 1)(x + 1)(x + 1). Hint: First expand

(x + 1)2 then multiply this expansion by (x + 1).b Which line of Pascal’s triangle links to your answer for part 2a?

3 Use the pattern you can observe to show that (x + 1)4 = x4 + 4x3 + 6x2 + 4x + 1.4 Use Pascal’s triangle to expand each of the following.

a (x + 1)5 b (x + 1)6 c (x + 1)7 d (x + 1)8 e (x + 1)9 f (x + 1)10

(You may need to copy the diagram above and add more lines to Pascal’s triangle.)

Let’s look at what happens when we have a minus sign in the brackets.5 Expand (x − 1)2, (x − 1)3 and (x − 1)4 by multiplying terms.6 Describe what effect the minus sign has on the expansions.7 Use Pascal’s triangle and your observations from question 6 to expand each of

the following.a (x − 1)5 b (x − 1)6 c (x − 1)7 d (x − 1)8 e (x − 1)9 f (x − 1)10

ExtensionCan you work out how to use Pascal’s triangle to expand each of the following? Clearly explain how you are able to do this.a (x + 2)3, (x + 2)4, . . . (x + 2)10 b (x + y)3, (x + y)4, . . . (x + y)10

c (2x + 3)3, (2x + 3)4, . . . (2x + 3)10 d (2x − 3y)3, (2x − 3y)4, . . . (2x − 3y)10

1 16 615 1520

1 5 110 510

1 4 6 4 1

1 3 3 1

1 2 1

1

EXCEL

Spreadsheet

Expanding(x + a)n



Simplifying algebraic fractions — addition and subtraction

In chapter 3 we spent some time simplifying algebraic fractions. This section is effec-tively a continuation of that topic with more complicated fractions, requiring use of theexpansion techniques we have learned in this chapter.

Pronumerals in the numerator onlyThis type of problem is very similar to those in chapter 3.

Simplify the following.

a b

THINK WRITE

a Write the fractions. a

Find the lowest common denominator (LCD). The lowest common multiple (LCM) of 2 and 5 is 10.

=

Express as a single fraction. =

Simplify the numerator by expanding brackets and collecting like terms.

=

=

b Write the fractions. b

Find the lowest common denominator (LCD). The lowest common multiple (LCM) of 7 and 3 is 21.

=



=

=

y 3+2

------------ y 4–5

------------+ y 2+7

------------ y 3–3

------------–

1y 3+

2------------ y 4–

5-----------+

25 y 3+( )

10-------------------- 2 y 4–( )

10--------------------+

35 y 3+( ) 2 y 4–( )+

10----------------------------------------------

45y 15 2y 8–+ +

10----------------------------------------

7y 7+10

---------------

1y 2+

7------------ y 3–

3-----------–

23 y 2+( )

21-------------------- 7 y 3–( )

21--------------------–

33 y 2+( ) 7 y 3–( )–

21----------------------------------------------

43y 6 7y– 21+ +

21---------------------------------------

4y– 27+21

----------------------

10WORKEDExample



Pronumerals in the denominatorIn the case where there are pronumerals in the denominator, the common denominatoris taken to be the product of each of the denominators. Then, proceed as in previouscases.

It is customary to leave the denominator as brackets, without expanding them.

Pairs of brackets in the denominatorAlthough it is possible to have two different pairs of brackets in the denominator ofeach fraction, in this section we will consider the case where one of each pair is anidentical bracket. The common denominator will consist of each bracket that appears inthe question; the repeated bracket needs to appear only once.


a b

THINK WRITE


Find the LCD. The LCM of (a + 2) and (a − 1) is (a + 2)(a − 1).

=



=

=

b Write the fractions. b −

Find the LCD. The LCM of (2b − 5) and (4b + 1) is (2b − 5)(4b + 1).

= −



=

=

7a 2+------------ 7

a 1–------------+ 7

2b 5–--------------- 3

4b 1+---------------–

17

a 2+------------ 7

a 1–------------+

27 a 1–( )

a 2+( ) a 1–( )---------------------------------- 7 a 2+( )

a 2+( ) a 1–( )----------------------------------+

37 a 1–( ) 7 a 2+( )+

a 2+( ) a 1–( )-----------------------------------------------

47a 7– 7a 14+ +

a 2+( ) a 1–( )----------------------------------------

14a 7+a 2+( ) a 1–( )

----------------------------------

17

2b 5–--------------- 3

4b 1+---------------

2 7 4b 1+( )2b 5–( ) 4b 1+( )

----------------------------------------- 3 2b 5–( )2b 5–( ) 4b 1+( )

-----------------------------------------

37 4b 1+( ) 3 2b 5–( )–

2b 5–( ) 4b 1+( )-----------------------------------------------------

428b 7 6b– 15+ +

2b 5–( ) 4b 1+( )-------------------------------------------

22b 22+2b 5–( ) 4b 1–( )

----------------------------------------

11WORKEDExample



Simplifying algebraic fractions — addition and subtraction

1 Simplify each of the following.

a b c

d e f

g h i

j k l

m n o

Simplify.

THINK WRITE

Write the fractions.

Find the LCD. The LCM of (x + 2)(x + 3) and (x − 4)(x + 3) is (x + 2)(x + 3)(x − 4).

=



=

=

4x 2+( ) x 3+( )

----------------------------------- 1x 4–( ) x 3+( )

----------------------------------+

14

x 2+( ) x 3+( )---------------------------------- 1

x 4–( ) x 3+( )----------------------------------+

24 x 4–( )

x 2+( ) x 3+( ) x 4–( )--------------------------------------------------- 1 x 2+( )

x 2+( ) x 3+( ) x 4–( )---------------------------------------------------+

34 x 4–( ) 1 x 2+( )+x 2+( ) x 3+( ) x 4–( )

---------------------------------------------------

44x 16– x 2+ +

x 2+( ) x 3+( ) x 4–( )---------------------------------------------------

5x 14–x 2+( ) x 3+( ) x 4–( )

---------------------------------------------------

12WORKEDExample

remember1. When adding or subtracting algebraic fractions, you must first find a common

denominator.2. Pronumerals can appear in either the numerators or the denominators.3. If the pronumerals are in the denominator, the common denominator is usually

the product of the individual denominators.

remember

5FSki

llSHEET 5.5WWORKEDORKED

EExample

10 x 1+2

------------ x 3+4

------------+ m 1+6

------------- m 2+2

-------------+ x 2+4

------------ x 5+3

------------+

x 1–2

----------- x 2+3

------------+ y 3–5

----------- y 1+2

------------+ a 6+5

------------ a 2–6

------------+

p 2+10

------------ p 3–5

------------+ x 4–3

----------- x 1+5

------------+ 5x 2+5

--------------- 2x 1+15

---------------+

Mathcad

Adding andsubtractingalgebraicfractions

m 6+5

------------- 2m 1+2

-----------------+ 2 p 3–6

--------------- p 2+3

------------+ 2x 1–3

--------------- 2x 3–2

---------------+

4x 3+4

--------------- 3x 1–2

---------------+ 2x 1–2

--------------- 4x 5–3

---------------+ 2x 1–2

--------------- x 1–4

-----------+



2 Simplify each of the following.

a b c

d e f

g h i

j k l

m n o

3

a What is the equivalent of ?

b What is the equivalent of ?

c The equivalent of is:

d The equivalent of is:

4 Simplify the following.

a b c

d e f

g h i

j k l

m n o


a b c

d e f

A B C D E

A B C D E

A B C D E

A B C D E

m 2+3

------------- m 1+4

-------------–x 4+

2------------ x 5+

7------------+ p 3–

2------------ p 3+

3------------+

y 4–3

----------- y 2+5

------------+ x 5+2

------------ x 1+3

------------+ a 3+2

------------ a 6–5

------------+

m 1–8

------------- m 3+4

-------------– p 2+15

------------ p 3–5

------------– x 3–4

----------- x 2–3

-----------–

y 2–7

----------- y 5–3

-----------– x 3–2

----------- x 1–4

-----------– 4 p 1+3

---------------- 2 p 2+9

----------------–

5y 3+3

--------------- 3y 2–12

---------------– 3a 2–3

--------------- 2a 1–10

---------------– 2x 1–5

--------------- 3x 1+15

---------------–

mmultiple choiceultiple choicea 2+

3------------ a 3+

4------------+

2a 5+7

--------------- 7a 17+7

------------------ 12a 72+12

--------------------- 7a 17+12

------------------ 2a 5+12

---------------

2x 1+8

--------------- x 3–4

-----------+

4x2 6+8

-----------------4x 5–

8--------------- 5x 5–

4--------------- 4x 7+

8--------------- 3x 4–

12---------------

x 3+3

------------ x 2+6

------------–

x 4+6

------------ 3x 8+6

--------------- 3x 4–6

--------------- 53--- 2x 1+

9---------------

5m 1–2

---------------- m 2–3

-------------–

13m 7–6

------------------- 4m 1+5

----------------- 13m 1+6

-------------------- 13m 1–6

------------------- 4m 1–6

----------------


11 2m 3+------------- 1

m----+ 4

x 2+------------ 3

x---+ 7

a--- 2

a 4+------------+

9b--- 5

b 3+------------+ 3

c--- 2

c 1–-----------+ 4

m---- 5

m 3–-------------+

7p 8–------------ 2

p---+ 2

p 5–------------ 4

p---+ 4

a 6–------------ 2

a---+

6q 1–------------ 4

q---+ 1

a 1+------------ 2

a 2+------------+ 2

b 3+------------ 1

b 2+------------+

3x 2+------------ 1

x 3+------------+ 4

m 1+------------- 2

m 3+-------------+ 1

p 1–------------ 3

p 4+------------+

72m------- 2

m 1+-------------– 6

2x------ 2

x 3+------------– 9

4x------ 1

x 5+------------–

4m 5+------------- 2

m----– 7

m 3+------------- 3

m----– 5

p 3–------------ 2

p---–



g h i

j k l

m n o


a b

c d

e f

g h

i j

k l

m n

o p

7

a What is the equivalent of ?

b What is the equivalent of ?

c The equivalent of is:

d What is the equivalent of ?

A B C D E

A B C D E

A B C

D E

A B C

D E

6a 7–------------ 3

a---– 9

b--- 2

b 3–------------– 7

r--- 4

r 5–-----------–

83x------ 3

x 2+------------– 5

x 1+------------ 2

x 3+------------– 7

p 3+------------ 3

p 1+------------–

6m 7+------------- 4

m 3+-------------– 4

p 2+------------ 2

p 1+------------– 8

x 1–----------- 3

x 2+------------–


12 SkillSH

EET 5.62

x 1+( ) x 2+( )---------------------------------- 3

x 2+( ) x 3+( )----------------------------------+ 4

x 3+( ) x 1+( )---------------------------------- 2

x 1+( ) x 4+( )----------------------------------+

3x 1+( ) x 2+( )

---------------------------------- 4x 2+( ) x 3–( )

----------------------------------+ 7x 4+( ) x 3–( )

---------------------------------- 2x 1+( ) x 3–( )

----------------------------------+

5x 7+( ) x 4–( )

---------------------------------- 3x 2–( ) x 4–( )

---------------------------------+ 9x 2–( ) x 3–( )

--------------------------------- 3x 2–( ) x 1+( )

----------------------------------+

7x 6–( ) x 4–( )

--------------------------------- 2x 6–( ) x 1–( )

---------------------------------+ 4x 2+( ) x 3–( )

---------------------------------- 1x 1+( ) x 3–( )

----------------------------------+

5x 1+( ) x 3+( )

---------------------------------- 2x 3+( ) x 2+( )

----------------------------------– 6x 4+( ) x 1+( )

---------------------------------- 2x 3+( ) x 1+( )

----------------------------------–

5x 6–( ) x 2+( )

---------------------------------- 2x 1–( ) x 6–( )

---------------------------------– 4x 7+( ) x 2+( )

---------------------------------- 3x 7+( ) x 3–( )

----------------------------------–

2x 5+( ) x 6–( )

---------------------------------- 1x 1+( ) x 6–( )

----------------------------------– 11x 1+( ) x 4+( )

---------------------------------- 4x 1+( ) x 6–( )

----------------------------------–

7x 6–( ) x 2+( )

---------------------------------- 2x 2+( ) x 3–( )

----------------------------------– 9x 5+( ) x 2–( )

---------------------------------- 5x 5+( ) x 4–( )

----------------------------------–

mmultiple choiceultiple choice4

x 2–----------- 3

x---+

72x 2–--------------- 7x 6–

x x 2–( )------------------- 12x 6–

x x 2–( )------------------- 7x 6–

2x 2–--------------- 7x 6+

x x 2–( )-------------------

32m------- 2

m 5+-------------–

–m 15+2m m 5+( )-------------------------- m 15+

2m m 5+( )-------------------------- 1

2m m 5+( )-------------------------- –12m 15+

2m m 5+( )-------------------------- –m 15–

2m m 5+( )--------------------------

32x 1+--------------- 1

x 1–-----------–

2x 2+------------ –x 5+

2x 1+( ) x 1–( )------------------------------------- –6x 6+

2x 1+( ) x 1–( )-------------------------------------

x 5–2x 1+( ) x 1–( )

------------------------------------- x 4–2x 1+( ) x 1–( )

-------------------------------------

WorkS

HEET 5.2

1x 1+( ) x 2–( )

---------------------------------- 3x 1+( ) x 3+( )

----------------------------------+

GAMEtime

Expanding— 002

4x 1+( )2 x 2–( ) x 3+( )

----------------------------------------------------- 4x 18–x 1+( ) x 2–( ) x 3+( )

--------------------------------------------------- 4x 3–x 1+( ) x 2–( ) x 3+( )

---------------------------------------------------

3x2 18–x 1+( ) x 2–( ) x 3+( )

---------------------------------------------------4x 3–

x 1+( )2 x 2–( ) x 3+( )-----------------------------------------------------



1 Expand and simplify −4(x − 5).

2 Expand and simplify 3(4x − 2y) + 2(5x + y).

3 Expand and simplify (x + 5)(x + 9).

4 Find the missing term if (y + K)(y − 4) = y2 − 16.

5 Expand (6 + q)2.

6 Expand and simplify (x + 2)(x + 1) + (x + 3)(x + 1).

7 Expand and simplify (x + 6)(x − 5) − (x − 3)2.

8 True or false?

9 Simplify .

10 Simplify .

Simplifying algebraic fractions — multiplication and division

This section is an extension of the work you did in chapter 3, Introductory algebra onthe multiplication and division of algebraic fractions.

Cancelling bracketed expressions in multiplicationIf a bracketed expression, or in fact any pronumeral, appears in the numerator of oneterm and the denominator of another, then they may be cancelled.

2

x 2+4

------------ x 5+3

------------+ 7x 26+12

------------------=

63z 2–-------------- 4

4z 1+---------------+

4x 4+( ) x 3–( )

---------------------------------- 4x 6+( ) x 3–( )

----------------------------------–

Simplify each of the following.

a b c

THINK WRITE


Cancel (x + 3) and replace with a 1, since it appears in the first denominator and second numerator.

= ×

Simplify by multiplying the remaining numerators and denominators.

=

4x 3+------------ x 3+

5------------× y 2–

2 y------------ 4 y–

3 y 2–( )--------------------× z 3–( ) z 4–( )

z 5–( ) z 1+( )--------------------------------- z 1+( ) z 1–( )

z 3+( ) z 3–( )---------------------------------×

14

x 3+------------ x 3+

5------------×

241--- 1

5---

345---

13WORKEDExample



169

As before, it is customary to leave bracketed expressions intact without expanding anyfurther.

Cancelling bracketed expressions in division

When dividing fractions, we first turn the second fraction ‘upside down’ and change thedivision sign to a multiplication sign (times and tip). We then proceed as for multi-plication.

THINK WRITE


Cancel (y − 2) and replace with a 1, since it appears in the first numerator and second denominator. Cancel y and replace with a 1, since it appears in the first denominator and second numerator.

=

= × −

Multiply the remaining numerators and denominators and simplify the resultant fraction.

= −

= −

c Write down the expression. c

Cancel (z − 3) and replace with a 1, since it appears in the first numerator and second denominator. Cancel (z + 1) and replace with a 1, since it appears in the first denominator and second numerator.

=

Simplify the resultant fraction. =

1y 2–2y

----------- 4y–3 y 2–( )--------------------×

21

2 1×------------ 4 1×–

3 1×---------------×

12--- 4

3---

346---

23---

1z 3–( ) z 4–( )z 5–( ) z 1+( )

--------------------------------- z 1+( ) z 1–( )z 3+( ) z 3–( )

---------------------------------×

21 z 4–( )×z 5–( ) 1×

------------------------- 1 z 1–( )×z 3+( ) 1×

-------------------------×

3z 4–( ) z 1–( )z 5–( ) z 3+( )

---------------------------------


a b c

THINK WRITE


Convert to a multiplication problem by turning the second fraction upside down.

=

Cancel (x − 3) and replace with a 1. = ×


1x 3–------------ 3

x 3–------------÷ 3 y–

y 2–------------ 5 y

y 2–------------÷ z 2+( ) z 3–( )

z 4–( )--------------------------------- z 3–

z 4–-----------÷

11

x 3–----------- 3

x 3–-----------÷

21

x 3–----------- x 3–

3-----------×

311--- 1

3---

413---

14WORKEDExample

Continued over page

MQ 9 Vic - 05 Page 169 Wednesday, September 19, 2001 10:45 AM


So, division of fractions is the same as multiplication with an additional first step.

Simplifying algebraic fractions — multiplication and division


a b c

d e f

THINK WRITE



=

Cancel (y − 2) and replace with a 1. Cancel y and replace with a 1.

=

= − ×

Simplify the resultant fraction. = −

c Write the fractions. c


=

Cancel (z − 3) and replace with a 1. Cancel (z − 4) and replace with a 1.

=


= z + 2

13y–

y 2–----------- 5y

y 2–-----------÷

23y–

y 2–----------- y 2–

5y-----------×

33 1×–

1--------------- 1

5 1×------------×

31--- 1

5---

435---

1z 2+( ) z 3–( )

z 4–( )--------------------------------- z 3–

z 4–-----------÷

2z 2+( ) z 3–( )

z 4–( )--------------------------------- z 4–( )

z 3–( )----------------×

3z 2+( ) 1×

1------------------------- 1

1---×

4z 2+

1-----------

remember1. If a bracketed expression appears in the numerator of one term and the

denominator of another, they can be cancelled when multiplying.2. When dividing fractions, take the second fraction, turn it ‘upside down’ and

change the division sign to a multiplication sign (times and tip).

remember

5G

SkillSH

EET 5.7 WWORKEDORKEDEExample

13 x 1+2

------------ 1x 1+------------× y 1–

3----------- 2

y 1–-----------× p 2–

4------------ 4

p 2–------------×

b 5+3

------------ 9b 5+------------× m 4+

m------------- m

m 4+-------------× 3 p

p 6–------------ p 6–

p------------×



g h i

j k l


a b c

d e f

g h i

j k l


a b

c d

e f

4

a Which is the equivalent of ?

b The equivalent of is:

c What is the equivalent of ?

d Which is the equivalent of ?

A 2 B 4 C 2(m + 3)2 D E 32

A B 9 C D E

A B C 3m D E

A a − 1 B C

D E

y 7–4y

----------- yy 7–-----------× 4a

a 2–------------ a 2–

2a------------× p 3–

p 3+------------ 5 p 3+( )

p 2–--------------------×

x 2+( ) x 7–( )x 1+

---------------------------------- x 1+x 7–------------× b 2–( ) b 3+( )

b 5+( ) b 4–( )---------------------------------- b 4–

b 2–------------× x 3–

x 2+------------ x 2+( ) x 1+( )

x 6–( ) x 3–( )----------------------------------×


14 SkillSH

EET 5.8x 2+

3------------ x 2+

x------------÷ m 3–

4------------- m 3–

8-------------÷ 5

x 3–----------- 10

x 3–-----------÷

12a 2+------------ 4

a 2+------------÷ m

m 6–------------- m

m 6–-------------÷ p 2+

p 1–------------ p 2–

p 1–------------÷

s 6–3 s 1+( )------------------- s

s 1+-----------÷ a 1+( ) a 3–( )

a 2+---------------------------------- a 1+

a 2+------------÷ 8m

m 2+( ) m 5–( )------------------------------------- 2m

m 5–-------------÷

a 2+( ) a 3–( )a 5–( ) a 4+( )

---------------------------------- a 2+a 4+------------÷ m m 2–( )

m 1+( ) m 5+( )------------------------------------- m

m 1+-------------÷ p

p 7+------------ p p 4–( )

p 2+( ) p 7+( )-----------------------------------÷

Mathcad

Multiplyingand dividing

algebraicfractions

y 3–( ) y 3+( )y y 1+( )

---------------------------------- yy 3–----------- y 3+

3------------÷× x 6+( ) x 2–( )

2x x 3+( )---------------------------------- 6x

x 6+------------ x 2–

2-----------÷×

4mm 7–------------- 12m m 3–( )

m 2+( ) m 7–( )------------------------------------- 9

m 1+-------------×÷ b 7+

2b------------ b 6–( ) b 7+( )

10 b 2+( )----------------------------------÷ 2b

b 2–------------×

d 8–5d

------------ 15d d 6–( )d 7–( ) d 8–( )

----------------------------------× 12d 7–------------÷ 12s

s 11–-------------- s 2–( ) s 11–( )

2s s 6+( )-----------------------------------× 30

s s 6+( )-------------------÷

mmultiple choiceultiple choicem 3+

4------------- 8

m 3+-------------×

12---

1x 3–----------- x 3–

9-----------×

13--- 1

9---

x 3–( )2

9-------------------

9x 3–-----------

m 8–3

------------- m 8–m

-------------÷

3m---- m 8–( )2

3m--------------------

m3---- 1

3m-------

a 2+( ) a 1–( )a 3+

---------------------------------- a 2+a 3+------------÷

1a 1–------------ a 1–

a 3+------------

a 2+( )2 a 1–( )a 3+( )2

------------------------------------ a 3+a 1–------------



Applications There are many problems in algebra where the expansion of brackets is a key com-ponent in finding a solution. A most important skill is to be able to convert a word, or‘real-life’ problem into an algebra ‘sentence’ or expression.

The algebraic expression found in part c of worked example 15 allows us to calculatethe area of the path for any given width, x.

A rectangular swimming pool measures 30 m by 20 m. A path around the edge of the pool is x m wide on each side.a Find the area of the pool.b Find an expression for the area of the pool plus the path.c Find an expression for the area of the path.d If the path is 1.5 m wide, calculate the area of the path.

THINK WRITE

a Construct a drawing of the pool. a

Calculate the area of the pool. Area = 20 × 30= 600 m2

b Write expressions for the total length and width.

b Length = 30 + x + x= 30 + 2x

Width = 20 + x + x= 20 + 2x

Find an expression for the area using the formula:Area = length × width.

Area = length × width= (30 + 2x)(20 + 2x)= 30(20) + 30(2x) + 2x(20) + 2x(2x)= 600 + 60x + 40x + 4x2

= (600 + 100x + 4x2) m2

c Find an expression for the area of the path by subtracting the area of the pool from the total area.

c Area of path = total area − area of pool= 600 + 100x + 4x2 − 600= (100x + 4x2) m2

d Substitute 1.5 for x in the expression found for the area of the path.

d When x = 1.5,Area of path = 100(1.5) + 4(1.5)2

= 159 m2

1

x

x

x30 m

20 m x

2

1

2

15WORKEDExample

20 m

30 m



Suppose that the page of a typical textbook is 24 cm high by 16 cmwide.

The page has margins at the top and bottom of x cm and on the left and right of y cm.a Write an expression for the height of the page that is ‘inside’

the margins.b Write an expression for the width of the page that is ‘inside’

the margins.c Write an expression for the area of the page that is ‘inside’

the margins.d Show that if the margins on the left and right are doubled, then the area available to be

printed is reduced by (48y − 4xy) cm.

THINK WRITE

a Construct a drawing, showing the key dimensions.

a

The real height (24 cm) is effectively reduced by x cm at the top and x cm at the bottom.

Height = 24 − x − x= 24 − 2x

b Similarly the width (16 cm) is reduced by y on the left and y on the right.

b Width = 16 − y − y= 16 − 2y

c The ‘effective’ area of the page is the width times the height.

c Area = (24 − 2x)(16 − 2y)= 24(16) + 24(−2y) − 2x(16) − 2x(−2y)= 384 − 48y − 32x + 4xy

d If the left and right margins are doubled, they become 2y and 2y respectively. Determine the new expression for the width of the page.

d Width = 16 − 2y − 2y= 16 − 4y

Determine the new expression for the area.

Area = (24 − 2x)(16 − 4y)= 24(16) + 24(−4y) − 2x(16) − 2x(−4y)= 384 − 96y − 32x + 8xy

Determine the difference in area by subtracting this from the first result for the ‘effective’ area.

Difference in area= 384 − 48y − 32x + 4xy − (384 − 96y − 32x + 8xy)= 384 − 48y − 32x + 4xy − 384 + 96y + 32x − 8xy= 48y − 4xySo the amount by which the area is reduced is (48y − 4xy) cm.

ANTHONY GRAY • SANDRA KENMAN

1

x

y

16 cm

24 cmy

x

2

1

2

3

16WORKEDExample



Applications

1 Answer the following for each shape.i Find an expression for the perimeter.ii Find the perimeter when x = 5.iii Find an expression for the area and simplify by expanding if necessary.iv Find the area when x = 5.

a b c d

e f g

2 A rectangular swimming pool measures 50 m by 25 m.A path around the edge of the pool is x m wide on each side.a Find the area of the pool.b Find an expression for the area of the pool plus the path.c Find an expression for the area of the path.d If the path is 2.3 m wide, calculate the area of the path.

remember1. When working on worded problems try to convert English ‘sentences’ into

algebraic ones.2. Drawing a diagram is an excellent aid in problem solving.

remember

5H

3x x + 2 4x – 1

4x

x

3x + 1

2x x – 3

5x + 2 3x + 5

x + 4

6x

3xWWORKEDORKEDEExample

15


C h a p t e r 5 E x p a n d i n g 1753 The page of a book is 20 cm high by 15 cm wide.

The page has margins at the top and bottom of the pageof x cm, and on the left and right of the page of y cm.a Write an expression for the height of the page that

is ‘inside’ the margins.b Write an expression for the width of the page that

is ‘inside’ the margins.c Write an expression for the area of the page that is

‘inside’ the margins.d Show that if the margins on the left and right are

doubled, then the area available to be printed on isreduced by (40y − 4xy) cm.

4 A rectangular book cover is 8 cm long and 5 cm wide.a Find the area of the book cover.b i If the length of the book cover is increased by v cm, write an expression for its

new length.ii If the width of the book cover is increased by v cm, write an expression for its

new width.iii Write an expression for the new area of the book cover and expand.iv Find the area of the book cover if v = 2 cm.

c i If the length of the book cover is decreased by d cm, write an expression for itsnew length.

ii If the width of the book cover is decreased by d cm, write an expression for itsnew width.

iii Write an expression for the new area of the rectangle and expand.iv Find the area of the book cover if d = 2 cm.

d i If the length of the book cover is made x times as long, write an expression forits new length.

ii If the width of the book cover is increased by x cm, write an expression for itsnew width.

iii Write an expression for the new area of the book cover and expand.iv Find the area of the book cover if x = 5 cm.

5 A square has dimensions of 5x metres.a Write an expression for its perimeter. b Write an expression for its area. c i If its length is decreased by 2 m, write an expression for its new length.

ii If its width is decreased by 3 m, write an expression for its new width.iii Write an expression for its new area and expand.iv Find its area when x = 6 m.

6 A rectangular sign has a length of 2x cm and a width of x cm.a Write an expression for its perimeter.b Write an expression for its area.c i If its length is increased by y cm,

find an expression for its new length.ii If its width is decreased by y cm,

find an expression for its new width.iii Write an expression for its new area

and expand.iv Find its area when x = 4 cm and

y = 3 cm.


16

2x

x



7 A square has a side length of x cm.a Write an expression for its perimeter.b Write an expression for its area.c i If its side length is increased by y cm, write an expression for its new side

length.ii Write an expression for its new perimeter and expand.iii Find the perimeter when x = 5 cm and y = 9 cm.iv Write an expression for its new area and expand.

d Find the area when x = 3.2 cm and y = 4.6 cm.

8 A swimming pool with length 4p + 2 metres and width 3p metres is surrounded by a path of width p metres.

Find the following in expanded form.a An expression for the perimeter of the pool.b An expression for the area of the pool.c An expression for the length of the pool and path.d An expression for the width of the pool and path.e An expression for the perimeter of the pool and path.f An expression for the area of the pool and path.g An expression for the area of the path.h The area of the path when p = 2 m.

Billboard costsAt the beginning of this chapter we were looking at the increased cost of a billboard if the length and height are increased by 2 metres and 3 metres respectively. Does the increase in cost depend on the initial size of the billboard?

1 Let the length of the billboard be l and the height be h. (Both l and h are in metres.) Write an expression for the area of the billboard.

2 What is the cost of this billboard at a rate of $50 per square metre?

3 If the length of the billboard is increased by 2 metres, write an expression for the new length.

4 If the height of the billboard is increased by 3 metres, write an expression for the new height.

5 Use your answers to parts 3 and 4 to find an expression for the new area.

6 What is the new cost of the billboard?

7 Subtract the original cost of the billboard to find the increase in the cost. Is this cost constant or does it depend on the original length and height?

4p + 2

3p

p

WorkS

HEET 5.3



Copy the sentences below. Fill in the gaps by choosing the correct word or expression from the word list that follows.

1 Expansion, in algebra, means to multiply everything thebrackets by what is outside.

2 After expanding an expression you should always look to .3 The expansion of 6(x + 3) − 4(3 − x) is .4 When multiplying two expressions within brackets each

term in the first bracket by each term in the second bracket.5 The expansion of (4 + x)(4 − x) is an example of a pattern called

.6 The expansion of (7 − x)(7 + x) is . This follows the rule

(a + b)(a − b) = .7 The identical bracket expansions or rules are:

(a + b)(a + b) = a2 + 2ab + b2

(a − b)(a − b) = a2 − 2ab + b2.8 When adding algebraic fractions, the first step is to find the .

9 The common denominator used when subtracting is

.10 Division of fractions is done by turning the problem into a

problem.11 The fraction is turned when dividing fractions.12 Brackets can be left in the of the answer when working with

algebraic fractions.13 When working on , you are trying to convert English

‘sentences’ into algebraic ones.14 Drawing a diagram is an excellent aid in solving.

summary

4x 2–------------ 1

x---–

W O R D L I S Tdenominatormultiplycommon

denominatorupside down

insideproblemcollect like termsword problems49 − x2

10x + 6perfect squaresdifference of two

squaressecond

multiplicationx(x − 2)a2 − b2


178

M a t h s Q u e s t 9 f o r V i c t o r i a

1

Expand these expressions.

a

5(

x

+

3)

b

−

(

y

+

5)

c

−

x

(3

−

2

x

)

d

−

4

m

(2

m

+

1)

2

Expand and simplify by collecting like terms.

a

3(

x

−

2)

+

9

b

−

2(5

m

−

1)

−

3

c

4

m

(

m

−

3)

+

3

m

−

5

d

7

p

−

2

−

(3

p

+

4)

3

Expand and simplify the following expressions.

a

3(

a

+

2

b

)

+

2(3

a

+

b

)

b

−

4(2

x

+

3

y

)

+

3(

x

−

2

y

)

c

2

m

(

n

+

6)

−

m

(3

n

+

1)

d

−

2

x

(3

−

2

x

)

−

(4

x

−

3)

4

Expand and simplify these expressions.

a

(

x

+

4)(

x

+

5)

b

(

m

−

2)(

m

+

1)

c

(3

m

−

2)(

m

−

5)

d

(2

a

+

b

)(

a

−

3

b

)

5

Expand and simplify these expressions.

a

(

x

+

4)(

x

−

4) b (9 − m)(9 + m) c (x + y)(x − y) d (1 − 2a)(1 + 2a)

6 Expand and simplify these expressions.a (x + 5)2 b (m − 3)2 c (4x + 1)2 d (2 − 3y)2

7 Expand and simplify these expressions.a (x + 2)(x + 1) + (x + 3)(x + 2) b (m + 7)(m − 2) + (m + 3)2

c (x + 6)(x + 2) − (x + 3)(x − 1) d (b − 7)2 − (b − 3)(b − 4)


a b c d


a b

c d

e f


a b c

d e f

11 A rectangular table top has a length of 3x cm and a width of x cm.a Write an expression for its perimeter.b Write an expression for its area.c i If its side length is increased by y cm, write an

expression for its new side length.ii Write an expression for its new perimeter and expand.iii Find the perimeter when x = 90 cm and y = 30 cm.iv Write an expression for its new area and expand.v Find the area when x = 90 cm and y = 30 cm.

5A

CHAPTERreview

5A5B

5C5D5D5E

5F x 2+5

------------ x 1+2

------------+ 3m 1+8

----------------- 2m 3–4

----------------+ a 6+2

------------ a 3+7

------------– 2y 5+4

--------------- 3y 4–12

---------------–

5F 1m 4+------------- 2

m----+ 5

m 1+------------- 3

m 2–-------------+

73x------ 1

x 2+------------– 4

2y 1+--------------- 2

3y 2–---------------–

3x 1+( ) x 2+( )

---------------------------------- 2x 1+( ) x 4+( )

----------------------------------+ 43x 1–( ) x 2+( )

------------------------------------- 1x 2+( ) 2x 3+( )

-------------------------------------–

5G x 3+4

------------ 1x 3+------------× p 3+

p------------ 3 p

p 3+------------× a 7+

a 3–------------ a 7+

a 3–------------÷

b 6–( ) b 2+( )b 5+( ) b 3–( )

---------------------------------- b 2+b 5+------------÷ 3m

m 2+( ) m 3+( )------------------------------------- m 2+

9-------------× p 3–

10------------ p 3–( ) p 4+( )

2 p-----------------------------------÷

5H

testtest

CHAPTERyyourselfourself

testyyourselfourself

5

MQ 9 Vic - 05 Page 178 Monday, September 17, 2001 1:12 PM

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