mathematics for australia 10a year 10 advanced

Specialists in mathematics publishing

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HAESE MATHEMATICS

Michael HaeseSandra HaeseMark Humphries

Austral ian Curriculum

10A10AMathematicsfor Australia

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MATHEMATICS FOR AUSTRALIA 10A

Michael Haese B.Sc.(Hons.), Ph.D.Sandra Haese B.Sc.Mark Humphries B.Sc.(Hons.)

152 Richmond Road, Marleston, SA 5033, AUSTRALIA

Support material: Katie Richer.

set in Australia by Charlotte Frost and Deanne Gallasch. Typeset in Times Roman 10 /11

Published by:Haese Mathematics152 Richmond Road, Marleston, SA 5033, AUSTRALIAEmail:

National Library of Australia Card Number & ISBN 978-1-921972-24-9

Haese Mathematics 2013

Published by Haese Mathematics

First Edition 2013

Cartoon artwork by John Martin.

Artwork by Benjamin Fitzgerald, Gregory Olesinski and Brian Houston.

Cover design by Piotr Poturaj.

Computer software by Troy Cruickshank, Adrian Blackburn, Ashvin Narayanan, Edward Ross andTim Lee.

Type .

Printed in Malaysia through Bookpac Production Services, Singapore.

. Except as permitted by the Copyright Act (any fair dealing for the purposes ofprivate study, research, criticism or review), no part of this publication may be reproduced, stored in aretrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying,recording or otherwise, without the prior permission of the publisher. Enquiries to be made to HaeseMathematics.

: Where copies of part or the whole of the book are made under PartVB of the Copyright Act, the law requires that the educational institution or the body that administers ithas given a remuneration notice to Copyright Agency Limited (CAL). For information, contact theCopyright Agency Limited.

: While every attempt has been made to trace and acknowledge copyright, the authorsand publishers apologise for any accidental infringement where copyright has proved untraceable. Theywould be pleased to come to a suitable agreement with the rightful owner.

: All the internet addresses (URLs) given in this book were valid at the time of printing.While the authors and publisher regret any inconvenience that changes of address may cause readers,no responsibility for any such changes can be accepted by either the authors or the publisher.

\Qw_ \Qw_

This book is copyright

Copying for educational purposes

Acknowledgements

Disclaimer

[email protected]:

Cover photography by iStockphoto.com (Spiral Fountain, Sydney)


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FOREWORD

Mathematics for Australia 10A has been designed and written for the AustralianCurriculum. The textbook covers all of the content outlined in the Year 10 and Year 10Acurricula. Students who have an interest in mathematics, and are intending to study morerigorous mathematics courses in Years 11 and 12, will benefit from completing the morechallenging material provided in this textbook. The textbook is best used in conjunctionwith the related year levels in our 'Mathematics for Australia' series.

The textbook has been structured to give an abbreviated coverage of the more basic Year10 curriculum content, allowing students to complete both the Year 10 and the Year 10Arequirements within the school year. There is a lot of work in the book; students shouldnot feel obliged to complete all of the problems in a section if they understand the topicwell.

The material ispresented in a clear, easy-to-follow style, free from unnecessary distractions, while efforthas been made to contextualise questions so that students can relate concepts to everydayuse.

Each chapter begins with an Opening Problem, offering an insight into the application ofthe mathematics that will be studied in the chapter. Important information and key notesare highlighted, while worked examples provide step-by-step instructions with conciseand relevant explanations. Discussions, Activities, Investigations, Puzzles, and Researchexercises are used throughout the chapters to develop understanding, problem solving,and reasoning, within an interactive environment.

Extensive Review Sets are located at the end of each chapter, comprising a range ofquestion types including short answer, extended response, and multiple choice.

Graphics calculator instructions are provided throughout the book to help students buildan understanding of the technology. Instructions are provided for the Casio fx-9860GPlus, TI-84 Plus, and TI- spire calculator models.

The accompanying CD contains specially designed SELF TUTOR software. Click on anyworked example throughout the book to activate a teacher's voice which will explain eachstep in the worked example. SELF TUTOR is an excellent tool for students who havebeen absent from class and for those who need extra revision and practice.

In addition to SELF TUTOR, the interactive CD contains links to geometry software,statistics packages, demonstrations, calculator instructions, and a range of printableworksheets, tables, spreadsheets and diagrams, allowing teachers to demonstrate conceptsand students to experiment for themselves.

The textbook and interactive student CD provide an engaging and structured package,allowing students to explore and develop their confidence in mathematics.

n

We welcome your feedback.

The authors and publishers would like to thank all those teachers who offered advice andsupport during the writing of this book.

Acknowledgements

Email: [email protected]

Web: www.haesemathematics.com.au


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Simply click on the (or anywhere in the example box) to access the workedexample, with a teachers voice explaining each step necessary to reach the answer.

Play any line as often as you like. See how the basic processes come alive usingmovement and colour on the screen.

Ideal for students who have missed lessons or need extra help.

SELF TUTOR is an exciting feature of this book.

The icon on each worked example denotes an active link on the CD.

The interactive CD is ideal for independent study.

Students can revisit concepts taught in class and undertake their ownrevision and practice. The CD also has the text of the book, allowingstudents to leave the textbook at school and keep the CD at home.

By clicking on the relevant icon, a range of features can be accessed:

Self Tutor

Interactive Links to spreadsheets,calculator instructions, graphing andgeometry software, computer demonstrationsand simulations

USING THE INTERACTIVE CD

See , , p. 415Chapter 19 Quadratic Functions

INTERACTIVELINK

Self Tutor

Sketch each of the following functions on the same set of axes as y = x2. In each casestate the coordinates of the vertex.

a y = (x 2)2 + 3 b y = (x+ 2)2 5

a We draw y = x2, then translateit 2 units to the right and 3 unitsupwards.

The vertex is at (2, 3).

b We draw y = x2, then translateit 2 units to the left and 5 unitsdownwards.

The vertex is at (2, 5).

Example 6

+2

+3

y

x

y = (x - 2) + 3\ \ \ \ \ \2

y = x\ \ 2 -2

y

x

y = (x + 2) - 5\ \ \ \ \ \2y = x\ \ 2

-5

INTERACTIVE STUDENT CDINTERACTIVE STUDENT CD

spre

adsh

eets

gr

aphi

ngan

dgeo

metry so

ftware worksheets demonstrations

sim

ulationsincludesSelf Tutor

Mathematicsfor AustraliaMathematicsfor Australia

Haese MathematicsHaese Mathem

aticsHaese Mathem

aticsHaese Mathem

atics

2013

10A10A


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TABLE OF CONTENTS 5

TABLE OF CONTENTS

GRAPHICS CALCULATOR

INSTRUCTIONS

1 INDICES

2 ALGEBRAIC EXPANSION AND

FACTORISATION

3 RADICALS

4 ALGEBRAIC FRACTIONS

8

9

25

47

63

Casio fx-9860G PLUS CD

Texas Instruments TI-84 Plus CD

Texas Instruments TI- spire CD

A Index laws 10

B Rational (fractional) indices 13

C Scientific notation (standard form) 17

Review set 1 21

Practice test 1A Multiple choice CD

Practice test 1B Short response 22

Practice test 1C Extended response 23

A Expansion laws 26

B Further expansion 31

C The binomial expansion 32

D Revision of factorisation 34

E Factorising expressions with four terms 38

F Factorising quadratic trinomials 39

G Factorising 40

H Miscellaneous factorisation 43

Review set 2 44




A Radicals and surds 48

B Simplifying radicals 49

C Adding and subtracting radicals 53

D Multiplications involving radicals 54

E Division by radicals 56

Review set 3 60




A Evaluating algebraic fractions 64

B Simplifying algebraic fractions 65

C Multiplying and dividing algebraic

fractions 70

D Adding and subtracting algebraic fractions 72

Review set 4 76

n

ax + bx + c a 12 , =




A Solving linear equations 80

B Linear equation problems 85

C Linear inequalities 87

D Solving linear inequalities 89

E Linear inequality problems 93

Review set 5 94




A Pythagoras theorem 98

B The converse of Pythagoras theorem 103

C Pythagorean triples 104

D Problem solving using Pythagoras 106

E Circle problems 110

F Three-dimensional problems 114

Review set 6 116




A Length and perimeter 122

B Area 125

C Surface area 130

D Volume 137

E Capacity 146

Review set 7 148




A Formula construction 154

B Formula substitution 158

C Formula rearrangement 160

D Rearrangement and substitution 164

E Formulae by induction 167

Review set 8 169



5 LINEAR EQUATIONS AND

INEQUALITIES

6 PYTHAGORAS THEOREM

7 MEASUREMENT

8 FORMULAE

79

97

121

153

6


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6 TABLE OF CONTENTS


A Congruence of figures 174

B Congruent triangles 175

C Proof using congruence 179

D Similarity 181

E Similar triangles 183

F Areas and volumes 188

Review set 9 192




A Equations of the form 198

B The Null Factor law 200

C Solution by factorisation 201

D Completing the square 203

E The quadratic formula 207

F Problem solving 209

Review set 10 213

Practice test 10A Multiple choice 214



A Distance between two points 219

B Midpoints 222

C Gradient 224

D Parallel and perpendicular lines 228

E The equation of a line 232

F Graphing lines from equations 234

G Finding the equation of a line 237

Review set 11 242




A Labelling right angled triangles 246

B The trigonometric ratios 247

C Finding side lengths 250

D Finding angles 252

E Problem solving with trigonometry 254

F True bearings 258

G 3-dimensional problem solving 261

Review set 12 263




x = k2

9 CONGRUENCE AND

SIMILARITY

10 QUADRATIC EQUATIONS

11 COORDINATE GEOMETRY

12 TRIGONOMETRY

173

197

217

245

13 STATISTICS

14 FINANCIAL MATHEMATICS

15 NON-RIGHT ANGLED

TRIANGLE TRIGONOMETRY

16 SIMULTANEOUS EQUATIONS

17 PROBABILITY

267

303

321

343

359

A Discrete data 268

B Continuous data 271

C Measuring the centre 273

D Cumulative data 280

E Measuring the spread 282

F Box plots 286

G Standard deviation 292

H Evaluating reports 295

Review set 13 298




A Business calculations 304

B Appreciation and depreciation 309

C Simple interest 311

D Compound interest 315

Review set 14 319




A The unit circle 322

B The area of a triangle 326

C The sine rule 328

D The cosine rule 333

E Problem solving using the sine and

cosine rules 337

Review set 15 339




A Graphical solution 344

B Solution by substitution 347

C Solution by elimination 348

D Problem solving 350

E Non-linear simultaneous equations

(Extension) 354

Review set 16 355




A Theoretical probability 361

B Compound events 365

C Expectation 372


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TABLE OF CONTENTS 7

D Conditional probability 373

Review set 17 380




A Relations 386

B Functions 389

C Function notation 390

D Transforming 393

Review set 18 402




A Quadratic functions 408

B Graphs of quadratic functions 410

C Axes intercepts 417

D Axis of symmetry 422

E Vertex 424

F Quadratic optimisation 426

Review set 19 428




A Exponential functions 432

B Graphs of exponential functions 433

C Growth and decay 436

D Exponential equations 440

E Logarithms 441

Review set 20 445




A Circle theorems 451

B Further circle theorems 455

C Geometric proof 459

Review set 21 462




A Polynomials 466

B Polynomial operations 467

C The Remainder theorem 471

D The Factor theorem 473

y = f(x)

18 RELATIONS AND FUNCTIONS

19 QUADRATIC FUNCTIONS

20 EXPONENTIAL FUNCTIONS

AND LOGARITHMS

21 GEOMETRY OF CIRCLES

22 POLYNOMIALS

385

407

431

449

465

E Graphs of polynomials 475

Review set 22 480




A The unit circle 484

B The relationship between 487

C The multiples of 30 and 45 489

D Trigonometric functions 491

E Trigonometric equations 498

Review set 23 500




A Circles 504

B Ellipses 511

C Hyperbolae 514

Review set 24 521




A Line graphs 526

B Scatter plots 529

C Correlation 530

D Measuring correlation 533

E Line of best fit 537

Review set 25 543




sin cos_ _and

23 ADVANCED TRIGONOMETRY

24 CONIC SECTIONS

25 BIVARIATE STATISTICS

ANSWERS

INDEX

483

503

525

548

614


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TI-84 PlusCASIO

fx-9860G Plus

GRAPHICSCALCULATOR

INSTRUCTIONS

When additional calculator help may be needed, specific instructions

can be printed from icons within the text.

GRAPHICS CALCULATOR INSTRUCTIONS

Printable graphics calculator instruction booklets are available for the ,

, and the . Click on the relevant icon below.

Casio fx-9860G Plus

TI-84 Plus TI- spiren

TI- spiren


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1Chapter1

Indices

Contents:

A Index lawsB Rational (fractional) indicesC Scientific notation (standard form)

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10

Opening problem

INDICES (Chapter 1)

INDEX NOTATION

We often deal with numbers that are repeatedly multiplied together, such as 5 5 5. We canuse indices or exponents to conveniently represent such expressions.

Using index notation, we represent 5 5 5 as 53, which reads 5 to the power 3. We saythat 5 is the base, and 3 is the index or power or exponent.

If n is a positive integer, then an is the product of n factors of a.

an = a a a a :::: a| {z }n factors

The definition of an given above is only meaningful if n is a positive integer.

Things to think about:

a What does it mean if a number is raised to an index which is a negative integer? Can you

write a3 without a negative index?b What does it mean if a number is raised to an index which is a fraction?

c i Can you use the index laws to show that (91

2 )2 = 9?

ii What is the value of 91

2 ?

In previous years we have seen the following index laws:

If the bases a and b are both positive and the indices m and n are integers then:

am an = am+n To multiply numbers with the same base, keep the base and addthe indices.

am

an= amn To divide numbers with the same base, keep the base and subtract

the indices.

(am )n = amn When raising a power to a power, keep the base and multiplythe indices.

(ab)n = anbn The power of a product is the product of the powers.a

b

n=an

bnThe power of a quotient is the quotient of the powers.

a0 = 1, a 6= 0 Any non-zero number raised to the power of zero is 1.

an =1

anand in particular, a1 =

1

a.

INDEX LAWSA


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INDICES (Chapter 1) 11

Self Tutor

Express in simplest form with a prime number base:

a 94 b 4 2p c 3x

9yd 25x1

a 94

= (32)4

= 324

= 38

b 4 2p= 22 2p= 22+p

c3x

9y

=3x

(32)y

=3x

32y

= 3x2y

d 25x1

= (52)x1

= 52(x1)

= 52x2

EXERCISE 1A

1 Simplify using the index laws:

a 32 35 b x6 x3 c x5 xn d t3 t4 t5

e79

75f

x7

x3g

t6

txh t3m t

i (53)2 j (t4)3 k (y3)m l (a3m)4

2 Express in simplest form with a prime number base:

a 121 b 32 c 81 d 42

e 252 f 7t 49 g 3a 9 h 8p 4i

7n

7n2j

9

3xk (25t)2 l 16k3 2k

m4a

2bn

8x

16yo

125x+1

5x1p

27a+2

3a 9a

Self Tutor

Remove the brackets of: a (2x)3 b

3c

b

4a (2x)3

= 23 x3= 8x3

b

3c

b

4=

34 c4b4

=81c4

b4

3 Remove the brackets of:

a (xy)2 b (ab)3 c (xyz)2 d (3b)3 e (5a)4 f (10xy)5

g

p

q

2h

m

n

3i

x

3

4j

5

z

3k

2a

b

4l

3x

4y

3

Example 2

Example 1

Each factor within the

brackets is raised to the

power outside them.


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12 INDICES (Chapter 1)

4 Simplify the following expressions using one or more of the index laws:

a 4b2 2b3 b a6b3

a4bc 3ab2 2a3 d 5x

3y2

15xy

e

a2

5b

3f

24t6r4

15t6r2g

(4c3d2)2

c2dh

10k7

(2k)5

Self Tutor

a 70 b 32 c 30 31 d5

3

2a 70 = 1 b 32 = 1

32=

1

9

c 30 31 = 1 13=

2

3d

5

3

2=

3

5

2=

9

25

5

a 30 b 61 c 41 d 50

e 42 f 42 g 53 h 53

i 72 j 72 k 103 l 103

6 Simplify, giving your answers in simplest rational form:

a

1

2

0b

54

54c 2t0 d (2t)0

e 70 f 3 40 g 53

55h

26

210

ix4

x9j

3

8

1k

2

3

1l

1

5

1m 20 + 21 n 50 51 o 30 + 31 31 p

1

3

2q

2

3

3r

11

2

3s

4

5

2t

21

2

27 Write the following without brackets or negative indices:

a (3b)1 b 3b1 c 7a1 d (7a)1

e

1

t

2f

3x

y

1g (5t)2 h (5t2)1

i xy1 j (xy)1 k xy3 l (xy)3

m (3pq)1 n 3(pq)1 o 3pq1 p (xy)3

y2

q (5x2y3)3 r

c

2d3

2s

3r3t

2t

2p

5q2

3

Example 3

Notice thata

b

2=

b

a

2Simplify, giving your answers in simplest rational form:

Simplify, giving your answers in simplest rational form:


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INDICES

Investigation Rational (fractional) indices

(Chapter 1) 13

8 Use the index laws to show that, for positive a and b, and integer n:

a1

an= an b

a

b

n=

bn

an

9 The units for speed kilometres per hour can be written as km/h or km h1.Write these units in index form:

a m/s b cubic metres/hour c square centimetres per second

d cubic centimetres per minute e grams per second

f kilogram metres per second g metres per second per second.

10 Find the smaller of 2125 and 375 without a calculator. Hint: 2125 = (25)25

11 Order the following numbers from smallest to largest: 290, 360, 536, 1024.

A rational number is a number which can be written in the formp

qwhere p and q are integers.

The integers themselves are rational numbers, since for example 2 =2

1.

The index laws can be applied not just to integer indices, but to rational indices in general. This

helps to give meaning to values such as 51

2 and 71

3 .

What to do:

1 Notice that 51

2 51

2 = (51

2 )2 = 51

22

= 51 = 5 and (p5)2 = 5.

a Copy and complete the following:

i 31

2 31

2 = :::::: = :::::: = :::::: ii (p3)2 = ::::::

iii 131

2 131

2 = :::::: = :::::: = :::::: iv (p13)2 = ::::::

b Copy and complete: a1

2 = ::::::

2 Notice that (71

3 )3 = 71

33

= 71 = 7 and ( 3p7)3 = 7.

a Copy and complete the following:

i (81

3 )3 = :::::: = :::::: = :::::: ii ( 3p8)3 = ::::::

iii (271

3 )3 = :::::: = :::::: = :::::: iv ( 3p27)3 = ::::::

b Copy and complete: a1

3 = ::::::

3 Suggest a rule for the general case: a1

n = ::::::

RATIONAL (FRACTIONAL) INDICESB

Remember that

(am)n = amn.

3p7 is read as

the cube root of 7.

This Investigation will help you discover the meaning of numbers raised to rational indices.


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From the Investigation, we can conclude that a1

2 =pa and a

1

3 = 3pa .

a1

n = npa where n

pa is the nth root of a.

Self Tutor

Evaluate:

a 161

2 b 81

3 c 16 12 d 8

13

a 161

2

=p16

= 4

b 81

3

=3p8

= 2

c 16 12

=1

1612

=1p16

=1

4

d 8 13

=1

813

=13p8

=1

2

Self Tutor

Write the following in index form:

ap3 b 3

p7 c

14p7

ap3

= 31

2

b3p7

= 71

3

c14p7

=1

714

= 7 14

EXERCISE 1B.1

1 Evaluate the following without using a calculator:

a 41

2 b 4 12 c 9

1

2 d 912

e 361

2 f 36 12 g 27

1

3 h 27 13

i 10001

3 j 1000 13 k 125

1

3 l 125 13

2 Write the following in index form:

ap11 b

1p11

cp12 d

1p12

e 3p26 f

13p26

g 4p7 h

15p7

Example 5

Example 4


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Y:\HAESE\AUS_10A\AUS10A_01\014AUS10A_01.cdr Wednesday, 31 October 2012 5:02:35 PM EMMA


Self Tutor

Use your calculator to find 4p50, rounded to 2 decimal places.

4p50 = 50

1

4

4p50 2:66

3 Use your calculator to evaluate the following, rounded to 2 decimal places:

a 4p20 b 5

p300 c

14p80

d1

6p12

RATIONAL INDICES OF THE FORM mn

So far we have seen that: 71

2 is the square root of 7,

71

3 is the cube root of 7, and so on.

But what about values such as 72

3 ?

Using the index laws, 72

3 = (71

3 )2

= (3p7)2

and also 72

3 = (72)1

3

=3p72

In general, am

n = ( npa)m = n

pam

When dealing with indices of this form, it is often easiest to write the base number as a prime

raised to a power. We simplify the result using the index laws.

Self Tutor

Evaluate without using a calculator: a 84

3 b 32 25

a 84

3

= (23)4

3

= 23 4

3

= 24

= 16

b 32 25

= (25) 25

= 25 2

5

= 22

= 14

Example 7

Example 6

TI-84 Plus

TI- spiren

Casio fx-9860G Plus


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Self Tutor

Write the following as powers of 2: a 3p4 b

1p8

c16p2

a3p4

= (22)1

3

= 22 1

3

= 22

3

b1p8

= 8 12 fas 1p

a= a

12 g

= (23) 12

= 2 32

c16p2

=24

212

= 24 1

2

= 27

2

EXERCISE 1B.2

1 Evaluate without using a calculator:

a 82

3 b 43

2 c 45

2 d 85

3 e 163

4

f 93

2 g 9 32 h 4

12 i 32

1

5 j 322

5

k 323

5 l 16 34 m 8

23 n 27

43 o 25

32

2 Write the following as powers of 2:

ap8 b 3

p32 c 4

p4 d 3

p16 e

14p8

f1

3p16

g17p8

h1

5p64

i 8p2 j 4

p32 k

23p4

l4p32

8


a 3p9 b

p27 c

14p27

d1

5p81

e 9p3 f 3

p27 g

95p3

h3p81

9

4 Write with a prime number base:

a3p25 b 4

p32 c 5

p125 d 7

p121

e1

3p49

f 5p64 g

17p625

h1

6p243

i 16p8 j 25

p125 k

133p169

l81p27

5 Use your calculator to evaluate, rounded to 3 significant figures where necessary:

a 253

2 b 272

3 c 87

3 d 92

5 e 103

7

f 155

3 g 102

7 h 187

3 i 163

11 j 1464

9

k 452 l 27

53 m 15

25 n 53

37 o 3

75

6 Without using your calculator, evaluate3p9 4p2712p243

.

Example 8


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INDICES

Puzzle

(Chapter 1) 17

What to do:

Taking alternate turns, each player selects 3 squares on the board to create a statement ofthe form ab = c.

For example, the shaded squares can be used to create the statement 34 = 81. Thesesquares are then crossed out and cannot be used again.

The last player who is able to make a valid selection is the winner.Single player variant:

Try to use all 81 squares in 27 selections.

Observe the pattern: 10 000 = 104

1000 = 103

100 = 102

10 = 101

1 = 100

1

10= 101

1

100= 102

1

1000= 103

FORM)

C

(STANDARD

SCIENTIFIC NOTATION

10

10

10

1

1

1

10

10

10

10

1

1

1

1

This game can be played by 2 players.

14 5 3

13 2 4

18 2

45

2 9 4 2523 81 3 12 81

32 3 127 125 116 64 12 243 54 3 18 2 3 32 7 34 25 0 2 12 23 1 6 2 3649 25 12 3 0 2 64 7 1525 181 4 27 2 3 5 4 271 6 13 16 2 116 3 125 12125 16 343

12 5 3 1 9

15

PRINTABLEBOARD

As we divide by 10,

the exponent or power

of 10 decreases by one.


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We can use this pattern to simplify the writing of very large and very small numbers.

For example, 5 000 000

= 5 1 000 000= 5 106

and 0:000 003

=3

1 000 000

= 3 11 000 000

= 3 106

Scientific notation involves writing any given number as a number between

1 inclusive and 10, multiplied by a power of 10. The result has the form

a 10k where 1 6 a < 10 and k is an integer.

Self Tutor

Write in scientific notation:

a 23 600 000 b 0:000 023 6

a 23 600 000

= 2:36 107b 0:000 023 6

= 2:36 105

Self Tutor

Write as an ordinary decimal number:

a 2:57 104 b 7:853 103

a 2:57 104= 2:5700 10 000= 25 700

b 7:853 103= 0007:853 103= 0:007 853

EXERCISE 1C

1 Write using scientific notation:

a 230 b 53 900 c 0:0361 d 0:006 80

e 3:26 f 0:5821 g 361 000 000 h 0:000 001 674

2 Write as an ordinary decimal number:

a 2:3 103 b 2:3 102 c 5:64 105 d 7:931 104e 9:97 100 f 6:04 107 g 4:215 101 h 3:621 108

3 Express the following quantities using scientific notation:

a There are approximately 4 million red blood cells in adrop of blood.

b The thickness of a coin is about 0:0008 m.

c Earths radius is about 6:38 million metres.

d A Rubiks Cube has approximately

43 252 000 000 000 000000 possible arrangements.

Example 10

Example 9

Remember that

103 = 1103

.


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4 Express the following quantities as ordinary decimal numbers:

a The Murray River is approximately 2:38106 m long.b A piece of paper is about 1:8 102 cm thick.c A test tube holds 3:2 107 bacteria.d A mushroom weighs 8:2 106 tonnes.

Self Tutor

Simplify, writing your answer in scientific notation:

a (3 104) (8 103) b 2 103

5 108

a (3 104) (8 103)= 24 104+3= (2:4 101) 107= 2:4 108

b2 1035 108

=2

5 103(8)

= 0:4 105= (4 101) 105= 4 104

5 Simplify the following, writing your answers in scientific notation:

a (3 103) (2 107) b (4 103) (7 105)c (8 104) (7 105) d (9 105) (6 102)e (3 105)2 f (4 107)2g (2 103)4 h (5 103)3i (6 101) (4 103) (5 104) j (6 103)2 (8 1011)k (4 103)1 l (5 104)2


a8 1064 103 b

9 1033 101 c

4 1062 102

d2:5 104(5 107)2 e

(8 102)22 106 f

(5 103)2(2 104)1

7 a How many times larger is 3 1011 than 3 108 ?b i Which is smaller, 5 1016 or 5 1021 ?

ii By how many times is it smaller than the other number?

c How many times larger is 4 106 than 8 105 ?

Example 11


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Self Tutor

Use your calculator to find:

a (2:58 107) (1:5 106) b 6:5 102

1:04 105

a (2:58 107) (1:5 106) = 3:87 1013

b6:5 1021:04 105 = 6:25 10

7

8 Calculate the following, giving each answer in scientific notation. The decimal part should be

rounded to 3 significant figures.

a (4:7 105) (8:53 107) b (2:7 103) (9:6 1014)

c3:4 1074:8 1015 d

7:3 1071:5 104

e (2:83 103)2 f (5:96 105)2

g(3:56 104)28:05 105 h

2:9 102(7:62 107)3

9 Use your calculator to answer the following:

a A rocket travels in space at 4 105 km h1. Assuming 1 year 365:25 days, howfar will it travel in:

i 30 days ii 20 years?

b A bullet travelling at an average speed of 2 103 km h1 hits a target 500 m away.Find the time of the bullets flight, in seconds.

c Mars has volume 1:31 1021 m3 whereasPluto has volume 4:93 1019 m3.How many times bigger is Mars than Pluto?

d Microbe C has mass 2:63 105 grams whereasmicrobe D has mass 8 107 grams.

i Which microbe is heavier?

ii How many times heavier is it, than the other

microbe?

Example 12

GRAPHICSCALCULATOR

INSTRUCTIONS

TI- spiren

Casio fx-9860G Plus TI-84 Plus


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10 The table alongside shows the land area of each Australian state

and territory.

a Find the total land area of Australia.

b Place the states and territories in order, from largest to

smallest.

c How many times larger is:

i New South Wales than the ACT

ii Queensland than Tasmania?

d What percentage of the land area of Australia, is included in

Western Australia?

Land area

ACT 2:0 105 haNSW 8:0 107 haNT 1:3 108 ha

QLD 1:7 108 haSA 9:8 107 ha

TAS 6:8 106 haVIC 2:3 107 haWA 2:5 108 ha

Historical note

The ancient Indians explored the concept of expressing very

large and very small numbers. In the Lalitavistara Sutra, a

Sanskrit text dating from around the 4th century, it is writtenthat the Buddha gave a description of the size of an atom.

In terms of the length of a finger bone, the Buddha stated that:

.... each was the length of

seven grains of barley, each of which was the length of

seven mustard seeds, each of which was the length of

seven poppy seeds, each of which was the length of

seven particles of dust stirred up by a cow, each of which was the length of

seven specks of dust disturbed by a ram, each of which was the length of

seven specks of dust stirred up by a hare, each of which was the length of

seven specks of dust carried away by the wind, each of which was the length of

seven tiny specks of dust, each of which was the length of

seven minute specks of dust, each of which was the length of

seven particles of the first atoms.

1 Assuming a finger bone is 4 cm long, use the Buddhas description to estimate the lengthof an atom, in metres. Write your answer in scientific notation.

2 Research the size of a carbon atom. How accurate is the estimate in 1?

1 Simplify using the index laws:

a k5 k3 b p6

pc (m6)8

2 Remove the brackets of:

a (3w)2 b (2x2y)3 c

a

b

6d

1

5n

3

Review set 1


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Y:\HAESE\AUS_10A\AUS10A_01\021AUS10A_01.cdr Tuesday, 20 November 2012 9:21:01 AM BRIAN

22

Practice test 1A Multiple Choice

INDICES (Chapter 1)

3 Simplify, giving answers in simplest rational form:

a 71 b4

3

1c 110 111 d

13

4

24 Write using scientific notation:

a 59 000 b 0:009 c 6 085 000 d 0:000 007 71

5 Evaluate without using a calculator:

a 491

2 b 64 13 c 125

4

3 d 27 23

6 Write as an ordinary decimal number:

a 6:23 105 b 3:008 104 c 4:597 1007 Write with a prime number base:

a 5p16 b

13p9

c625p5

d 8p32

8 Use your calculator to evaluate the following correct to 3 significant figures:

a 3p20 b

16p100

c 105

4 d 15 37

9 Write without brackets or negative indices:

a (mn)2 b mn2 c

x

5y2


a (6 105) (3 106) b (8 109) (5 104)

c8 1072 103 d

9 1056 103 e (7 10

3)2 f (8 107)1

Click on the icon to obtain this printable test.

1 Express in simplest form with a prime number base:

a 82 b25x

125c

49k+3

7k1

2 Simplify using one or more of the index laws:

a 5c3 3c4 b 14x5y2

2x2yc

3p

q3

2

Practice test 1B Short response

PRINTABLETEST


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3 Evaluate 274

3 without using a calculator.

4 Write without brackets or negative indices: a

a

b

2b

3a1

2b2

35 a Approximately 58 400 000 vehicles cross the Sydney Harbour Bridge each year.

Write this number in scientific notation.

b Approximately how many vehicles cross the Sydney Harbour Bridge each day?

Write your answer using scientific notation.

6 a Write using index notation: i 3p13 ii

15p40

b Use your calculator to evaluate1

5p40

correct to 3 significant figures.

7 Use your calculator to find:

a (2:7 105) (3:3 109) b 2:97 104

4:5 107


a 4p8 b

15p16

c4p32

9 How many times larger is 3:5 1011 than 5 109 ?10 Find the smaller of 260 and 720 without using a calculator.

1 Write the answers to the following in scientific notation:

a The speed of light in a vacuum is about 2:998 108 m/s.Assuming 1 year 365:25 days, determine how far light travels in:

i 1 hour ii 1 day iii 1 year.

b How long does it take for light to travel:

i 1 m ii 1 cm iii 1 mm?

c In air, light travels at 2:989 108 m/s and sound travels at 343:2 m/s.How many times faster is light than sound?

2 a i Evaluate 112 and 122.

ii Hence, explain why 53

2 lies between 11 and 12.

iii Calculate 53

2 correct to 4 significant figures.

b Without using a calculator, evaluate 125 23 .

3 a Write ( 5p7 4p7)20 as a power of 7.

b Hence, show that 5p7 4p7 = 7

9

20 .

c Use the index laws to show that mpa npa = a

m+n

mn .

d Hence, write 3p11 5p11 as a power of 11.

Practice test 1C Extended response


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4 The table alongside shows the diameters of the planets in

the solar system.

a Find the diameter of Saturn in:

i kilometres ii centimetres.

b Find the radius of Venus.

c Write the planets in order of size, from smallest to

largest.

d How many times greater is the diameter of:

i Uranus than Mercury

ii Jupiter than Mars?

5 Mary calculated the following powers of 2:

20 = 1, 21 = 2, 22 = 4, 23 = 8

She noticed that as the index gets larger, the resulting values also get larger.

She wonders whether this is true for all bases, not just 2.

a Copy and complete the table below. As the index increases for each base, indicate

whether the values:

increase decrease remain constant, or alternate between increasing and decreasing.

b Describe the conditions under which, as the index increases, the values:

i increase ii decrease iii remain constant

iv alternate between increasing and decreasing.

Planet Diameter

Mercury 4:88 106 mVenus 1:21 107 mEarth 1:27 107 mMars 6:79 106 m

Jupiter 1:40 108 mSaturn 1:21 108 mUranus 5:11 107 mNeptune 4:95 107 m

Index

0 1 2 3

2 1 2 4 8 increase

5

12

Base 21

1223

3


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2Chapter2

Algebraic expansion

and factorisation

Contents:

A Expansion laws

B Further expansion

C The binomial expansion

D Revision of factorisationE Factorising expressions with four terms

F Factorising quadratic trinomials

H Miscellaneous factorisation

G Factorising ax2 + bx+ c, a 6= 1


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Y:\HAESE\AUS_10A\AUS10A_02\025AUS10A_02.cdr Wednesday, 7 November 2012 3:28:42 PM EMMA

26

Opening problem

ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2)

Jody showed her friend Leanne a trick for performing multiplications of 2 digit numbers, suchas 42 83:

Step 1: Multiply the digits in the units column.

2 3 = 64 2

8 36

Step 2: Multiply the digits along the diagonals, then add the results.

(4 3) + (8 2) = 28, so we write 8 and carry the 2.4 2

82

3

8 6

Step 3: Multiply the digits in the tens column.

4 8 = 32, adding the 2 gives 34.4 2

8 33 4 8 6

So, 42 83 = 3486.Things to think about:

Can you use algebra to explain why this trick works?

The study of algebra is vital for many areas of mathematics. We need it to manipulate equations,

solve problems for unknown variables, and also to develop higher level mathematical theories.

In this chapter we revise the expansion of expressions which involve brackets, and the reverse

process which is called factorisation.

DISTRIBUTIVE LAW

We use the distributive law to expand expressions of the form a(b + c). Each term inside thebrackets is multiplied by the factor outside the brackets.

a(b+ c) = ab+ ac

Self Tutor

Expand and simplify:

a 2(3x 1) b 3x(x+ 2)

a 2(3x 1)= 2 3x + 2 (1)= 6x 2

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

EXPANSION LAWSA

Example 1


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ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2) 27

EXERCISE 2A.1

1 Expand and simplify:

a 3(2x+ 5) b 4x(x 3) c 2(3 + x) d 3x(x+ y)e 2x(x2 1) f x(1 x2) g ab(b a) h x2(x 3)


a 3(a2 + 3a+ 1) b 5(b2 3b+ 2) c 4(2c2 3c 7)d d(d2 2d+ 1) e 2e(e2 + 3e 5) f 3a(2a2 3a+ 1)

Self Tutor


a a(a+ 2) + 2a(3a 2) b y(3y 1) 3y(2y 5)

a a(a+ 2) + 2a(3a 2)= a a + a 2 + 2a 3a + 2a (2)= a2 + 2a+ 6a2 4a= 7a2 2a

b y(3y 1) 3y(2y 5)= y 3y + y (1) + 3y 2y + 3y (5)= 3y2 y 6y2 + 15y= 14y 3y2


a 3(x+ 2) + 4x(x 1) b a(a 2) a(4 + a)c 2(p+ q) 3(q p) d x(x2 + 1) 3x2(1 2x)e x2(x 8) 3x(2 + x2) f 6(a b+ 3) 2(2 + a 3b)

4 Simplify:

a x(x+ 5) + 3(x+ 5) b x(x 2) 7(x 2)

THE PRODUCT (a+ b)(c+ d)

Consider the following rectangle which is 8 units long and 6 units wide.

Comparing the total number of squares on each side of the equals sign, we notice that:

(4 + 2)(5 + 3) = 4 5 + 4 3 + 2 5 + 2 3.

Example 2

+ 2

3

= 4 44

++

+

+ 2

2

55

33

5

8

6


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28 ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2)

We generalise this result by considering a rectangle with sides (a+ b) and (c+ d).

The original rectangle has area = (a+ b)(c+ d).

The sum of the areas of the smaller rectangles = ac+ ad+ bc+ bd.

So, (a+ b)(c+ d) = ac+ ad+ bc+ bd.

This expansion rule is called the FOIL rule as:

inners

(a+ b)(c+ d)

outers

= ac

Firsts

+ ad

Outers

+ bc

Inners

+ bd

Lasts

Self Tutor


a (x+ 4)(x 3) b (2x 5)(x+ 3)

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

b (2x 5)(x+ 3)= 2x (x) + 2x 3 5 (x) 5 3= 2x2 + 6x+ 5x 15= 2x2 + 11x 15

EXERCISE 2A.2


a (x+ 2)(x+ 6) b (x 3)(x+ 7) c (x+ 5)(x 3)d (x 2)(x 10) e (2x+ 1)(x 3) f (3x 4)(2x 5)g (2x+ y)(x y) h (x+ 3)(2x 1) i (x+ 2y)(x 1)j (9 2x)(1 4x) k (3k 7)(10 k) l (x2 + 8)(9 5x)


a (x+ 3)(x 1) + 3(x 5) b (x+ 7)(x 5) + (x+ 1)(x+ 4)c (2x+ 3)(x 2) (x+ 1)(x+ 6) d (4t 3)(t+ 1) (2t 1)(2t+ 5)e (4x 1)(3 x) + (2x 3)(3x 2) f 5(3x 4)(x+ 2) (7 x)(8 5x)

Example 3

cc

caa a

dd

d

b

b b= + + +

(c + d)

(a + b)(a + b)


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3 Copy and complete: (a+ b)(c+ d)

= (a+ b) ::::::+ (a+ b) :::::: f...... lawg= ::::::(a+ b) + ::::::(a+ b)

= :::::: a+ :::::: b+ :::::: a+ :::::: b f...... lawg= ac+ ad+ bc+ bd fon rearrangingg

4 Answer the Opening Problem on page 26.

Hint: The 2 digit number ab represents the value 10a+ b.

DIFFERENCE OF TWO SQUARES

If we expand expressions of the form (a+ b)(a b) using the FOIL rule, we get(a+ b)(a b) = a2 ab+ ab b2

= a2 b2

Since a2 and b2 are perfect squares, a2 b2 is called the difference of two squares.

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

Self Tutor


a (x+ 4)(x 4) b (3x 2)(3x+ 2)

a (x+ 4)(x 4)= x2 42= x2 16

b (3x 2)(3x+ 2)= (3x)2 22= 9x2 4

EXERCISE 2A.3


a (x+ 1)(x 1) b (b+ 2)(b 2) c (a 7)(a+ 7)d (t 4)(t+ 4) e (6 b)(6 + b) f (5 x)(5 + x)g (8 + a)(8 a) h (2 + 3y)(2 3y) i (7 2a)(7 + 2a)j (3x+ 1)(3x 1) k (5 3y)(5 + 3y) l (x+ 2)(x 2)

m (5x+ y)(5x y) n (7m 3n)(7m+ 3n) o (x2 + 5y)(5y x2)2 Expand and simplify:

a (x+ 3)(x 3) (x+ 6)(x 6) b (5p 2)(5p+ 2) p(3p 1)c (3y z)(3y + z) (2y + 3z)(2y 3z)d (10 x2)(10 + x2) (10 3x2)(10 + 3x2)

Example 4


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PERFECT SQUARES

Perfect squares have the form (a + b)2 or (a b)2. We can use the following rule forexpanding them:

(a+ b)2 = a2 + ab+ ab+ b2 fusing the FOIL ruleg= a2 + 2ab+ b2

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

This rule can be demonstrated using areas.

The overall square alongside has area = (a+ b)2.

The sum of the areas of the 4 smaller rectangles

= a2 + ab+ ab+ b2

= a2 + 2ab+ b2

So, (a+ b)2 = a2 + 2ab+ b2.

The following is a useful way of remembering the perfect square rule:

(a+ b)2 = a2

square of the

first term

+ 2ab

twice the product

of the terms

+ b2

last term

Self Tutor


a (2x+ 1)2 b (3 4y)2

a (2x+ 1)2

= (2x)2 + 2 2x 1 + 12= 4x2 + 4x+ 1

b (3 4y)2= 32 + 2 3 (4y) + (4y)2= 9 24y + 16y2

EXERCISE 2A.4


a (x+ 5)2 b (2x+ 3)2 c (7 + x)2

d (3x+ 4)2 e (5 + x2)2 f (3x2 + 2)2

g (5x+ 3y)2 h (2x2 + 7y)2 i (x3 + 8x)2


a (x 3)2 b (2 x)2 c (3x 1)2d (6 5p)2 e (2x 5y)2 f (ab 2)2g (x2 5)2 h (4x2 3y)2 i (x2 y2)2

Example 5

a

a b

b ab

aba2

b2

a + b

a + b

square of the


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3 Use the diagram alongside to show that

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


a (x+ 9)2 + (x 2)2 b (3x+ 1)2 (2x 3)2c (x+ 8)2 (x+ 2)(x 5) d (5 p)2 + (p2 4)2e (3x2 1)2 4(1 x)2 f (5x+ y2)2 x(x2 y)2

When expressions containing more than two terms are multiplied together, we can still use the

distributive law to expand the brackets. Each term in the first set of brackets is multiplied by each

term in the second set of brackets.

If there are 2 terms in the first brackets and 3 terms in the second brackets, there will be2 3 = 6 terms in the expansion. However, when we simplify by collecting like terms, thefinal answer may contain fewer terms.

Self Tutor

Expand and simplify: (x+ 3)(x2 + 2x+ 4)

(x+ 3)(x2 + 2x+ 4)

= x3 + 2x2 + 4x fx each term in 2nd bracketg+ 3x2 + 6x+ 12 f3 each term in 2nd bracketg

= x3 + 5x2 + 10x+ 12 fcollecting like termsg

EXERCISE 2B


a (x+ 2)(x2 + x+ 4) b (x+ 3)(x2 + 2x 3)c (x+ 3)(x2 + 2x+ 1) d (x+ 1)(2x2 x 5)e (2x+ 3)(x2 + 2x+ 1) f (2x 5)(x2 2x 3)g (x+ 5)(3x2 x+ 4) h (4x 1)(2x2 3x+ 1)

FURTHER EXPANSIONB

Example 6

ba - b

a

Each term in the first

bracket is multiplied

by each term in

the second bracket.


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Y:\HAESE\AUS_10A\AUS10A_02\031AUS10A_02.cdr Tuesday, 20 November 2012 12:36:55 PM BRIAN


Self Tutor

Expand and simplify: (x+ 1)(x 3)(x+ 2)

(x+ 1)(x 3)(x+ 2)= (x2 3x+ x 3)(x+ 2) fexpand first two factorsg= (x2 2x 3)(x+ 2) fcollect like termsg= x3 + 2x2 2x2 4x 3x 6 fexpand remaining factorsg= x3 7x 6 fcollect like termsg


a (x+ 4)(x+ 3)(x+ 2) b (x 3)(x 2)(x+ 4) c (x 3)(x 2)(x 5)d (2x 3)(x+ 3)(x 1) e (4x+ 1)(3x 1)(x+ 1) f (2 x)(3x+ 1)(x 7)g (x 2)(4 x)(3x+ 2) h (x+ 3)3 i (x 2)3

3 State how many terms you would obtain by expanding:

a (a+ b)(c+ d) b (a+ b+ c)(d+ e) c (a+ b)(c+ d+ e)

d (a+ b+ c)(d+ e+ f) e (a+ b)(c+ d)(e+ f) f (a+ b+ c)(d+ e)(f + g)


a (x2 + 3x+ 1)(x2 x+ 3) b (2x2 + x 1)(x2 + 3x 2)c (3x2 + x 4)(2x2 3x+ 1) d (x2 3x+ 2)(x+ 5)(x 3)

Consider (a+ b)n where n is a positive integer.

a+ b is called a binomial as it contains two terms.

The binomial expansion of (a+ b)n is obtained by writing the expression without brackets.

In this Investigation we discover the binomial

expansion of (a+ b)3.

What to do:

1 Find a large potato and cut it to obtain a 4 cm by4 cm by 4 cm cube.

2 By making 3 cuts parallel to the cubes surfaces,divide the cube into 8 rectangular prisms as shown.

3 How many prisms are:

a 3 by 3 by 3 b 3 by 3 by 1

c 3 by 1 by 1 d 1 by 1 by 1?

THE BINOMIAL EXPANSIONC

Investigation The binomial expansion of (a+ b)3

Example 7

1 cm

3 cm

3 cm 1 cm

3 cm1 cm


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4 Now instead of the 4 cm 4 cm 4 cm potato cube, suppose you had a cube withedge length (a+ b) cm.

a Explain why the volume of the cube is given by (a+ b)3.

b Suppose you made cuts so each edge was divided into a cm and b cm.

How many prisms would be:

i a by a by a ii a by a by b

iii a by b by b iv b by b by b?

c By adding the volumes of the 8 rectangular prisms, find an expression for the total

volume. Hence write down the binomial expansion of (a+ b)3.

Another method of finding the binomial expansion of (a+ b)3 is to expand the brackets:

(a+ b)3 = (a+ b)2(a+ b)

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

= a3 + 2a2b+ ab2 + a2b+ 2ab2 + b3

= a3 + 3a2b+ 3ab2 + b3

So, (a+ b)3 = a3 + 3a2b+ 3ab2 + b3.

The binomial expansion of (a+ b)3 can be used to expand other perfect cubes.

Self Tutor

Expand and simplify using the rule (a+ b)3 = a3 + 3a2b+ 3ab2 + b3 :

a (x+ 4)3 b (3x 2)3

a We substitute a = x and b = 4.

) (x+ 4)3 = x3 + 3 x2 4 + 3 x 42 + 43= x3 + 12x2 + 48x+ 64

b We substitute a = (3x) and b = (2).) (3x 2)3 = (3x)3 + 3 (3x)2 (2) + 3 (3x) (2)2 + (2)3

= 27x3 54x2 + 36x 8

EXERCISE 2C

1 Use the binomial expansion for (a+ b)3 to expand and simplify:

a (x+ 1)3 b (x+ 3)3 c (x+ 5)3 d (x+ y)3

e (x 1)3 f (x 5)3 g (x 4)3 h (x y)3i (2 + y)3 j (2x+ 1)3 k (3x+ 1)3 l (2y + 3x)3

m (2 y)3 n (2x 1)3 o (3x 1)3 p (2y 3x)3

2 By expanding and simplifying (a+ b)3(a+ b), show that

(a+ b)4 = a4 + 4a3b+ 6a2b2 + 4ab3 + b4.

Example 8

Notice the use

of brackets.

DEMO


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3 Use the binomial expansion (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 to expand andsimplify:

a (x+ y)4 b (x+ 1)4 c (x+ 2)4 d (x+ 3)4

e (x y)4 f (x 1)4 g (x 2)4 h (2x 1)4

4 Consider: (a+ b)1 =

(a+ b)2 =

(a+ b)3 =

(a+ b)4 =

a + b

a2 + 2ab + b2

a3 + 3a2b + 3ab2 + b3

a4 + 4a3b + 6a2b2 + 4ab3 + b4

1 1

1 2 11 3 3 1

1 4 6 4 1

This triangle of numbers is called Pascals triangle.

a Predict the next two rows of Pascals triangle, and explain how you found them.

b Hence, write down the binomial expansion for:

i (a+ b)5 ii (a b)5 iii (a+ b)6 iv (a b)6c i Expand and simplify (x 2)5.

ii Check your answer by substituting x = 1 into your expansion.

Factorisation is the process of writing an expression as a product of its factors.

Factorisation is the reverse process of expansion, so we use the expansion laws in reverse.

FACTORISING WITH COMMON FACTORS

If every term in an expression has the same common factor, then we can place this factor in front

of a set of brackets. This is the reverse of the distributive law for expansion.

Self Tutor

Fully factorise:

a 6x2 + 4x b 4(a+ 1) + (a+ 2)(a+ 1)

a 6x2 + 4x

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

b 4(a+ 1) + (a+ 2)(a+ 1)= (a+ 1)[4 + (a+ 2)]= (a+ 1)(a 2)

REVISION OF FACTORISATIOND

Example 9

The expressions on the right hand side of

each identity contain the coefficients:


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Y:\HAESE\AUS_10A\AUS10A_02\034AUS10A_02.cdr Wednesday, 28 November 2012 11:14:12 AM BRIAN


Self Tutor

Fully factorise by removing a common factor:

a (x 5)2 2(x 5) b (x+ 2)2 + 2x+ 4

a (x 5)2 2(x 5)= (x 5)(x 5) 2(x 5) fHCF = (x 5)g= (x 5)[(x 5) 2]= (x 5)(x 7)

b (x+ 2)2 + 2x+ 4

= (x+ 2)(x+ 2) + 2(x+ 2) fHCF = (x+ 2)g= (x+ 2)[(x+ 2) + 2]

= (x+ 2)(x+ 4)

EXERCISE 2D.1

1 Fully factorise by first removing a common factor:

a 3x2 + 5x b 2x2 7x c 3x2 + 6xd 4x2 8x e 2x2 + 9x f 3x2 15xg 4x+ 8x2 h 5x 10x2 i 12x 4x2j x3 + x2 + x k 2x3 + 11x2 + 4x l ab+ ac+ ad

m ax2 + 2ax n ab2 + a2b o ax3 + ax2

2 Fully factorise:

a 3(x+ 5) + x(x+ 5) b a(b+ 3) 5(b+ 3)c x(x+ 4) + x+ 4 d x(x+ 2) + (x+ 2)(x+ 5)

e a(c d) + b(c d) f y(2 + y) y 2g ab(x 1) + c(x 1) h a(x+ 2) x 2

3 Fully factorise by removing a common factor:

a (x+ 2)2 5(x+ 2) b (x 1)2 3(x 1) c (x+ 1)2 + 2(x+ 1)d (x 2)2 + 3x 6 e x+ 3 + (x+ 3)2 f (x+ 4)2 + 8 + 2xg (x 3)2 x+ 3 h (x+ 4)2 2x 8 i (x 4)2 5x+ 20j 3x+ 6 + (x+ 2)2 k (x+ 1)3 + (x+ 1)2 l (a+ b)3 + a+ b

m 2(x+ 1)2 + x+ 1 n 3(x 2)2 (x 2) o 4(a+ b)2 2a 2b

DIFFERENCE OF TWO SQUARES FACTORISATION

We know the expansion of (a+ b)(a b) is a2 b2.

Thus, the factorisation of a2 b2 is: a2 b2 = (a+ b)(a b)

Example 10

Look for the highest

common factor

(HCF) of the terms.


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Self Tutor

Fully factorise: a 9 x2 b 4x2 25

a 9 x2= 32 x2 fdifference of squaresg= (3 + x)(3 x)

b 4x2 25= (2x)2 52 fdifference of squaresg= (2x+ 5)(2x 5)

EXERCISE 2D.2

1 Fully factorise:

a x2 4 b 4 x2 c x2 81 d 25 x2e 4x2 1 f 9x2 16 g 4x2 9 h 36 49x2

Self Tutor

Fully factorise: a 2x2 8 b x3 + 36x

a 2x2 8= 2(x2 4) fHCF is 2g= 2(x2 22) fdifference of squaresg= 2(x+ 2)(x 2)

b x3 + 36x= x(x2 36) fHCF is xg= x(x2 62) fdifference of squaresg= x(x+ 6)(x 6)

2 Fully factorise:

a 3x2 27 b 2x2 + 8 c 3k2 75d 5x2 + 5 e 8t2 18 f 27x2 + 75g x3 49x h 64n2 n4 i 28x3 63x

Self Tutor

a (3x+ 2)2 9 b (x+ 2)2 (x 1)2

a (3x+ 2)2 9= (3x+ 2)2 32= [(3x+ 2) + 3][(3x+ 2) 3]= (3x+ 5)(3x 1)

b (x+ 2)2 (x 1)2= [(x+ 2) + (x 1)][(x+ 2) (x 1)]= [x+ 2 + x 1][x+ 2 x+ 1]= [2x+ 1][3]

= 3(2x+ 1)

Example 13

Example 12

Example 11

Always look to

remove a common

factor first.

Factorise using the difference of two squares:


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3 Factorise using the difference of two squares:

a (x+ 1)2 4 b (2x+ 1)2 9 c (1 x)2 16d (x+ 3)2 4x2 e 4x2 (x+ 2)2 f 9x2 (3 x)2g (2x+ 1)2 (x 2)2 h (3x 1)2 (x+ 1)2 i 16x2 (2x+ 3)2

PERFECT SQUARE FACTORISATION

We know the expansion of (a+ b)2 is a2 + 2ab+ b2,

so the factorisation of a2 + 2ab+ b2 is: a2 + 2ab+ b2 = (a+ b)2

Similarly, the expansion of (a b)2 is a2 2ab+ b2,so the factorisation of a2 2ab+ b2 is: a2 2ab+ b2 = (a b)2

Self Tutor

Fully factorise:

a x2 + 10x+ 25 b x2 14x+ 49

a x2 + 10x+ 25

= x2 + 2 x 5 + 52= (x+ 5)2

b x2 14x+ 49= x2 2 x 7 + 72= (x 7)2

EXERCISE 2D.3

1 Fully factorise:

a x2 + 6x+ 9 b x2 + 8x+ 16 c x2 6x+ 9d x2 8x+ 16 e x2 + 2x+ 1 f x2 10x+ 25g y2 + 18y + 81 h m2 20m+ 100 i t2 + 12t+ 36

Self Tutor

Fully factorise:

a 9x2 6x+ 1 b 8x2 24x 18

a 9x2 6x+ 1= (3x)2 2 3x 1 + 12= (3x 1)2

b 8x2 24x 18= 2(4x2 + 12x+ 9) fHCF = 2g= 2([2x]2 + 2 2x 3 + 32)= 2(2x+ 3)2

2 Fully factorise:

a 9x2 + 6x+ 1 b 4x2 4x+ 1 c 9x2 + 12x+ 4d 25x2 10x+ 1 e 16x2 + 24x+ 9 f 25x2 20x+ 4

Example 15

Example 14

(x+ 5)2 and (x 7)2are perfect squares!


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38

Discussion


3 Fully factorise:

a x2 + 2x 1 b 2x2 + 8x+ 8 c 3x2 30x 75d 32x2 16x+ 2 e 36x2 + 120x+ 100 f 2x3 + 28x2 98x

What constant term must be added to these expressions to create a perfect square?I x2 6x+ :::::: I x2 + 12x+ :::::: I x2 + 16x+ ::::::

If we know that x2 + bx + c is a perfect square, what is the relationship between band c?

Sometimes we can factorise an expression containing four terms by grouping them in two pairs.

For example, ax2 + 2x+ 2 + ax can be rewritten as

ax2 + ax + 2x+ 2| {z }= ax(x+ 1) + 2(x+ 1) ffactorising each pairg= (x+ 1)(ax+ 2) f(x+ 1) is a common factorg

Self Tutor

Factorise:

a 3ab+ d+ 3ad+ b b x2 + 2x+ 5x+ 10

a 3ab+ d+ 3ad+ b

= 3ab+ b| {z } + 3ad+ d| {z } fputting terms containing b togetherg= b(3a+ 1) + d(3a+ 1) ffactorising each pairg= (3a+ 1)(b+ d) f(3a+ 1) is a common factorg

b x2 + 2x| {z } + 5x+ 10| {z }= x(x+ 2) + 5(x+ 2) ffactorising each pairg= (x+ 2)(x+ 5) f(x+ 2) is a common factorg

EXERCISE 2E

1 Factorise:

a 2a+ 2 + ab+ b b 4d+ ac+ ad+ 4c c ab+ 6 + 2b+ 3a

d mn+ 3p+ np+ 3m e x2 + 3x+ 7x+ 21 f x2 + 5x+ 4x+ 20

g 2x2 + x+ 6x+ 3 h 3x2 + 2x+ 12x+ 8 i 20x2 + 12x+ 5x+ 3

WITH FOUR TERMSE

Example 16

Sometimes we

need to reorder

the terms first.

FACTORISING EXPRESSIONS


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Y:\HAESE\AUS_10A\AUS10A_02\038AUS10A_02.cdr Friday, 30 November 2012 10:24:40 AM BRIAN


2 Factorise:

a x2 4x+ 5x 20 b x2 7x+ 2x 14 c x2 3x 2x+ 6d x2 5x 3x+ 15 e x2 + 7x 8x 56 f 2x2 + x 6x 3g 3x2 + 2x 12x 8 h 4x2 3x 8x+ 6 i 9x2 + 2x 9x 2

A quadratic trinomial is an algebraic expression of the form ax2 + bx+ c where x is avariable and a, b, c are constants, a 6= 0.

For example, x2 + 7x+ 6 and 3x2 13x 10 are both quadratic trinomials.Consider the expansion of the product (x+ 1)(x+ 6):

(x+ 1)(x+ 6) = x2 + 6x+ x+ 1 6 fusing FOILg= x2 + (6 + 1)x+ (1 6)= x2 + (sum of 1 and 6)x+ (product of 1 and 6)

= x2 + 7x+ 6

x2 + px+ q = (x+ a)(x+ b)

where a and b are two numbers whose sum is p, and whose product is q.

So, to factorise x2 + 7x + 6, we need two numbers with a sum of 7 and a product of 6.

These numbers are 1 and 6, and so x2 + 7x+ 6 = (x+ 1)(x+ 6).

We call this the sum and product method.

Self Tutor

Use the sum and product method to fully factorise:

a x2 + 5x+ 4 b x2 x 12

a We need two numbers with sum 5, and product 4.

The numbers are 1 and 4.

) x2 + 5x+ 4 = (x+ 1)(x+ 4)

b We need two numbers with sum 1 and product 12.The numbers are 4 and 3.) x2 x 12 = (x 4)(x+ 3)

EXERCISE 2F

1 Find two numbers which have:

a product 12 and sum 7 b product 15 and sum 8

c product 16 and sum 10 d product 18 and sum 11

e product 36 and sum 9 f product 36 and sum 9g product 12 and sum 4 h product 30 and sum 13

FACTORISING QUADRATIC TRINOMIALSF

Example 17The of the numbers is

the coefficient of .

The of the numbers

is the constant term.

sum

product

x


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2 Fully factorise:

a x2 + 3x+ 2 b x2 + 5x+ 6 c x2 x 6 d x2 + 3x 10e x2 + 4x 21 f x2 + 8x+ 16 g x2 14x+ 49 h x2 + 3x 28i x2 + 7x+ 10 j x2 11x+ 24 k x2 + 15x+ 44 l x2 + x 42

m x2 x 56 n x2 18x+ 81 o x2 4x 32 p x2 + 4x 45

Self Tutor

Fully factorise by first removing a common factor:

a 3x2 9x+ 6 b 2x2 + 2x+ 12

a 3x2 9x+ 6= 3(x2 3x+ 2)= 3(x 2)(x 1)

b 2x2 + 2x+ 12= 2(x2 x 6)= 2(x 3)(x+ 2)

3 Fully factorise by first removing a common factor:

a 2x2 + 10x+ 8 b 3x2 21x+ 18 c 2x2 + 14x+ 24d 5x2 30x 80 e 4x2 8x 12 f 3x2 42x+ 99g 2x2 2x 180 h 3x2 6x 24 i 2x2 + 18x+ 40j x3 7x2 8x k 4x2 24x+ 36 l 3x2 + 18x 81

m 2x2 44x+ 240 n x3 3x2 28x o x4 + 2x3 + x24 Fully factorise:

a x2 3x+ 54 b x2 7x 10 cd 4x x2 3 e 4 + 4x x2 fg x2 + 2x+ 48 h 6x x2 9 ij 2x2 + 4x+ 126 k 20x 2x2 50 l

5 Given that x2 + bx+ c = (x+m)(x+ n), factorise x2 bx+ c.

So far we have considered the factorisation of quadratic expressions of the form ax2 + bx + cwhere:

a = 1, for example x2 + 5x+ 6 = (x+ 3)(x+ 2) a is a common factor, for example 2x2 + 10x+ 12 = 2(x2 + 5x+ 6)

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

Example 18

x2 10x 213 x2 2x30x 3x2 63x3 + x2 + 2x

G FACTORISING ax2 + bx+ c, a 6= 1

fHCF = 3gfsum = 3 and product = 2) the numbers are 2 and 1g

fHCF = 2gfsum = 1 and product = 6) the numbers are 3 and 2g


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Y:\HAESE\AUS_10A\AUS10A_02\040AUS10A_02.cdr Friday, 16 November 2012 10:46:52 AM BRIAN


we have a perfect square or difference of two squares,for example 4x2 9 = (2x)2 32

= (2x+ 3)(2x 3)Factorising a quadratic expression such as 8x2 + 22x + 15 appears to be more complicatedbecause it does not fall into any of these categories.

We need to develop a method for factorising this type of quadratic expression.

FACTORISATION BY SPLITTING THE MIDDLE TERM

Using the FOIL rule, we see that (2x+ 3)(4x+ 5)

= 8x2 + 10x+ 12x+ 15

= 8x2 + 22x+ 15

We will now reverse the process to factorise the quadratic expression 8x2 + 22x+ 15.

8x2 + 22x+ 15

= 8x2 + 10x+ 12x+ 15 fsplitting the middle termg= (8x2 + 10x) + (12x+ 15) fgrouping in pairsg= 2x(4x+ 5) + 3(4x+ 5) ffactorising each pair separatelyg= (4x+ 5)(2x+ 3) fcompleting the factorisationg

But how do we correctly split the middle term? How do we determine that 22x must be writtenas +10x+ 12x rather than 15x+ 7x or 20x+ 2x?

When looking at 8x2 +10x+12x+15, we notice that 8 15 = 120 and 10 12 = 120.

So, for 8x2 + 22x + 15, we need two numbers whose sum is 22 and whose product is8 15 = 120. These numbers are 10 and 12.Likewise, for 6x2 + 19x+ 15 we need two numbers with sum 19 and product 6 15 = 90.These numbers are 10 and 9, so 6x2 + 19x+ 15

= 6x2 + 10x+ 9x+ 15

= (6x2 + 10x) + (9x+ 15)

= 2x(3x+ 5) + 3(3x+ 5)

= (3x+ 5)(2x+ 3)

The following procedure is recommended for factorising ax2 + bx+ c by splittingthe middle term:

Step 1: Find two numbers whose sum is b and whose product is ac.

Let the numbers be p and q.

Step 2: Replace bx by px+ qx.

Step 3: Complete the factorisation.

The order in

which the split

terms are written

does not matter.


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Self Tutor

Show how to split the middle term of the following so that factorisation can occur:

a 3x2 + 7x+ 2 b 10x2 23x 5

a In 3x2 + 7x+ 2, ac = 3 2 = 6 and b = 7.We need two numbers with a product of 6 and a sum of 7. These are 1 and 6.

So, the split is 7x = x+ 6x.

b In 10x2 23x 5, ac = 105 = 50 and b = 23.We need two numbers with a product of 50 and a sum of 23. These are 25 and 2.So, the split is 23x = 25x+ 2x.

EXERCISE 2G

1 Show how to split the middle term so that factorisation can occur:

a 3x2 + 11x+ 6 b 2x2 + 9x+ 4 c 4x2 + 4x 15d 5x2 11x 12 e 3x2 8x 3 f 12x2 25x+ 12

Self Tutor

Factorise by splitting the middle term:

a 6x2 + 19x+ 10 b 3x2 x 10

a In 6x2 + 19x+ 10, ac = 60 and b = 19.

We need two numbers with a product of 60 and a sum of 19. These are 4 and 15.

) 6x2 + 19x+ 10

= 6x2 + 4x+ 15x+ 10 fsplitting the middle termg= 2x(3x+ 2) + 5(3x+ 2) ffactorising in pairsg= (3x+ 2)(2x+ 5) ftaking out the common factorg

b In 3x2 x 10, ac = 30 and b = 1.We need two numbers with a product of 30 and a sum of 1. These are 5 and 6.) 3x2 x 10= 3x2 + 5x 6x 10 fsplitting the middle termg= x(3x+ 5) 2(3x+ 5) ffactorising in pairsg= (3x+ 5)(x 2) ftaking out the common factorg

2 Fully factorise:

a 2x2 + 5x+ 3 b 2x2 + 13x+ 18 c 7x2 + 9x+ 2

d 3x2 + 13x+ 4 e 3x2 + 8x+ 4 f 3x2 + 16x+ 21

g 8x2 + 14x+ 3 h 21x2 + 17x+ 2 i 6x2 + 5x+ 1

j 6x2 + 19x+ 3 k 10x2 + 17x+ 3 l 14x2 + 37x+ 5

Example 20

Example 19

Check your

factorisations

by expansion!


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3 Fully factorise:

a 2x2 9x 5 b 3x2 + 5x 2 c 3x2 5x 2d 2x2 + 3x 2 e 2x2 + 3x 5 f 5x2 8x+ 3g 11x2 9x 2 h 2x2 3x 9 i 3x2 17x+ 10j 5x2 13x 6 k 3x2 + 10x 8 l 2x2 + 17x 9

m 2x2 + 9x 18 n 15x2 + x 2 o 21x2 62x 3p 9x2 12x+ 4 q 12x2 + 17x 40 r 16x2 + 34x 15

Self Tutor

Fully factorise: 5x2 7x+ 6

We remove 1 as a common factor first.5x2 7x+ 6

= 1[5x2 + 7x 6]= [5x2 + 10x 3x 6]= [5x(x+ 2) 3(x+ 2)]= [(x+ 2)(5x 3)]= (x+ 2)(5x 3)

Here, ac = 30 and b = 7. The twonumbers with a product of 30 and a sumof 7 are 10 and 3.

4 Fully factorise by first removing 1 as a common factor:a 3x2 x+ 14 b 5x2 + 11x 2 c 4x2 9x+ 9d 9x2 + 12x 4 e 8x2 14x 3 f 12x2 + 16x+ 3

5 a Show that (3x+ 5)2 (2x 3)2 = 5x2 + 42x+ 16.b Factorise 5x2 + 42x+ 16 by splitting the middle term.

c Factorise (3x+ 5)2 (2x 3)2 using the difference of two squares.

In the following Exercise you will need to determine which factorisation method to use.

The following flowchart may prove useful:

Expression to be factorised.

Remove any common factors.

Look for the

difference of

two squares.

Look for perfect

squares.

For four terms,

look for

grouping in

pairs.

Look for the

sum and

product type.

Look for

splitting the

middle term.

MISCELLANEOUS FACTORISATIONH

Example 21


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44

Puzzle


EXERCISE 2H

1 Fully factorise:

a 3x2 + 9x b 81x2 4 c x2 7x 60d 3x 5x2 e x2 + 3x 40 f 2x2 32g 4x 3y + xy 12 h x2 + 10x+ 25 i 2x2 + 2x 12j x2 16x+ 39 k 49x x3 l x2 2x 8

2 Fully factorise:

a 4x2 8x 60 b x2 + 6x 16 c 4x2 + 8x 5d 3x2 + 6x 72 e 16x2 8x+ 1 f 9x2 (x+ 3)2g 6x2 + x 12 h x2 13x 36 i 2x2 14x+ 36j 14x2 + 19x+ 3 k 3x2 36x+ 108 l 16x2 + 44x 10

Click on the icon to obtain a printable puzzle for factorisation.


a 5(4x 5) b 4x(x 3) c 2(x+ 6) + x(3x 7)2 Expand and simplify:

a (x+ 5)(x 6) b (2x+ 5)(3x 1)c (x+ 3)(x+ 2) (2x 1)(x 6)

3 Fully factorise:

a 7x2 4x b x3 + 5x2 6x c x(x 8) + 5(x 8)4 Expand and simplify:

a (x+ 5)(x 2)(x+ 1) b (2x 3)(x2 + 4x+ 2)5 Fully factorise:

a 16 9m2 b x3 81x c (x+ 7)2 256 Expand and simplify:

a (t+ 7)(t 7) b (2y + 5)(2y 5) c (2m 5n)27 Fully factorise:

a 2x2 + 20x+ 50 b 2b dc+ 2d bc8 Use the binomial expansion of (a+ b)3 to expand and simplify:

a (2k + 3)3 b (r 4t)3

Review set 2

PUZZLE


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ALGEBRAIC

Practice test 2B Short response

Practice test 2A Multiple Choice

EXPANSION AND FACTORISATION (Chapter 2) 45

9 Fully factorise:

a x2 + 7x 18 b 3x2 9x 30 c 64 2x2 + 8x10 Fully factorise:

a 8x2 + 10x+ 3 b 5x2 13x+ 6 c 9x2 + 3x+ 2

Click on the icon to obtain this printable test.

1 Use the diagram alongside to show that

a(b+ c) = ab+ ac.

2 Expand and simplify: x(x2 3) + 5(x 4)3 Fully factorise: a 2x2 98 b (3x+ 1)2 (x 4)2

4 Expand and simplify: (a+ b)(a b) (a+ 2b)(a 2b)5 Fully factorise: 3x2 + 24x+ 48

6 How many terms would you obtain by expanding (a+ b+ c+ d)(e+ f)(g + h)?

7 Expand and simplify: (3x2 5)2

8 Fully factorise:

a x2 5x 66 b 2x2 + 20x 78 c 4x2 8x 219 Expand and simplify: (x2 x+ 4)(x2 + 2x+ 3)

10 Fully factorise: 6x2 5x+ 50

1 a Show that (2x+ 9)2 (x 3)2 = 3x2 + 42x+ 72.b Factorise 3x2 + 42x+ 72 by first taking out a common factor.

c Factorise (2x+ 9)2 (x 3)2 using the difference of two squares.

Practice test 2C Extended response

PRINTABLETEST

a

b c


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2 a Write down the binomial expansion of:

i (a+ b)2 ii (a+ b)3 iii (a+ b)4 iv (a+ b)5

b In (a + b)2 = a2 + 2ab + b2, the sum of the coefficients of the expansion is1 + 2 + 1 = 4. Find the sum of the coefficients in the expansion of:

i (a+ b)3 ii (a+ b)4 iii (a+ b)5

c What do you suspect is the sum of the coefficients in the expansion of (a+ b)n ?

d Prove your result by letting a = b = 1.

3 a Use your calculator to write as a mixed number:

i1122

ii2122

iii3122

iv4122

b What do you suspect is the value ofn+ 12

2, where n is a positive integer?

c Prove your result by expandingn+ 12

2.

d Hence, find: i1012

2ii1912

24 Consider factorising the expression 6x2 + 17x+ 12.

a Explain why the middle term 17x should be split into 9x and 8x.

b By writing 17x as 9x+ 8x, factorise 6x2 + 17x+ 12.

c Now factorise 6x2 + 17x+ 12 by writing 17x as 8x+ 9x. Check that you getthe same answer as in b.

5 a Use your calculator to find:

i 232 and 272 ii 182 and 322 iii 112 and 392 iv 142 and 362

b If a and b are two integers whose sum is 50, what can we say about the last 2 digits

of the squares a2 and b2?

c Prove that your answer to b is correct.

Hint: Write b in terms of a, then find the difference between the two squares.


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3Chapter3

Radicals

Contents:

A Radicals and surdsB Simplifying radicals

C Adding and subtracting radicals

D Multiplications involving radicals

E Division by radicals


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Opening problem

Weisshape

Jeremys shape

48 RADICALS (Chapter 3)

Jeremy and Wei were each asked to draw a shape with an area of 15 cm2. Jeremy drew a5 cm 3 cm rectangle, while Wei drew a square. The two shapes were placed on top of oneanother as shown.

Things to think about:

a Can you explain why the side length of the square

isp15 cm?

b Can you find the area covered by:

i Jeremys shape, but not Weis shape

ii at least one of the shapes?

c Can you find the perimeter of the region formed

by the shapes?

In previous years we have seen that the set of real numbers R can be divided into the set of

rational numbers Q , and the set of irrational numbers Q 0.

A rational number is a real number which can be written in the forma

bwhere a and b are

integers, b 6= 0.An irrational number cannot be written in the form

a

bwhere a and b are integers, b 6= 0.

The decimal expansion of an irrational number will neither terminate nor recur.

In Chapter 1 we encountered values such asp3,p5, and 3

p8. These numbers are known as

radicals.

Radical numbers may be rational or irrational. An irrational radical is called a surd.

Examples of rational radicals include:

p9 = 3 =

3

1r1

25=

s1

5

2=

1

5r4

9=

s2

3

2=

2

3

RADICALS AND SURDSA

If the number under the radical

sign can be written as a perfect

square, then the radical is rational.

A radical is a number that is written using the radical signp

.#bluebox#101.97935#7.40742


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RADICALS

Research

(Chapter 3) 49

Examples of surds include:p2 1:414 213 56 ::::

p19 4:358 898 94 ::::r1

3 0:577 350 269 ::::

Self Tutor

Determine whether each of the following numbers is a rational radical or a surd. If it is a

surd, find its value correct to 4 decimal places.

ap49 b

p53

ap49 = 7 =

7

1

)p49 is a rational radical.

bp53 is not rational, so it is a surd.

p53 7:2801

EXERCISE 3A

1 Determine whether each of the following numbers is a rational radical or a surd. If it is a

surd, find its value correct to 4 decimal places.

ap5 b

p36 c

p72 d

p81

e

r25

49f

r1

10g

r1

64h

r3

14

2 Consider the 1000 radicalsp1,p2,p3, ....,

p1000. How many of these radicals are surds?

Where did the names radical and surd come from? Why do we use the word irrational to describe some numbers?

In previous years we have established some rules which can be used to simplify radicals:

papa = pa2 = a papb = p

mathematics for australia 10a year 10 advanced

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