Mathematics for Australia 10A Year 10 Advanced
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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
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
Practice test 2A Multiple choice CD
Practice test 2B Short response 45
Practice test 2C Extended response 45
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
Practice test 3A Multiple choice CD
Practice test 3B Short response 61
Practice test 3C Extended response 62
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 , =
Practice test 4A Multiple choice CD
Practice test 4B Short response 77
Practice test 4C Extended response 78
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
Practice test 5A Multiple choice CD
Practice test 5B Short response 95
Practice test 5C Extended response 96
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
Practice test 6A Multiple choice CD
Practice test 6B Short response 118
Practice test 6C Extended response 119
A Length and perimeter 122
B Area 125
C Surface area 130
D Volume 137
E Capacity 146
Review set 7 148
Practice test 7A Multiple choice CD
Practice test 7B Short response 149
Practice test 7C Extended response 151
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
Practice test 8A Multiple choice CD
Practice test 8B Short response 170
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
Practice test 8C Extended response 171
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
Practice test 9A Multiple choice CD
Practice test 9B Short response 194
Practice test 9C Extended response 195
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
Practice test 10B Short response 215
Practice test 10C Extended response 216
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
Practice test 11A Multiple choice CD
Practice test 11B Short response 243
Practice test 11C Extended response 244
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
Practice test 12A Multiple choice CD
Practice test 12B Short response 264
Practice test 12C Extended response 266
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
Practice test 13A Multiple choice CD
Practice test 13B Short response 300
Practice test 13C Extended response 301
A Business calculations 304
B Appreciation and depreciation 309
C Simple interest 311
D Compound interest 315
Review set 14 319
Practice test 14A Multiple choice CD
Practice test 14B Short response 319
Practice test 14C Extended response 320
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
Practice test 15A Multiple choice CD
Practice test 15B Short response 340
Practice test 15C Extended response 341
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
Practice test 16A Multiple choice CD
Practice test 16B Short response 356
Practice test 16C Extended response 357
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
Practice test 17A Multiple choice CD
Practice test 17B Short response 382
Practice test 17C Extended response 383
A Relations 386
B Functions 389
C Function notation 390
D Transforming 393
Review set 18 402
Practice test 18A Multiple choice CD
Practice test 18B Short response 404
Practice test 18C Extended response 405
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
Practice test 19A Multiple choice CD
Practice test 19B Short response 429
Practice test 19C Extended response 430
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
Practice test 20A Multiple choice CD
Practice test 20B Short response 446
Practice test 20C Extended response 447
A Circle theorems 451
B Further circle theorems 455
C Geometric proof 459
Review set 21 462
Practice test 21A Multiple choice CD
Practice test 21B Short response 463
Practice test 21C Extended response 464
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
Practice test 22A Multiple choice CD
Practice test 22B Short response 481
Practice test 22C Extended response 481
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
Practice test 23A Multiple choice CD
Practice test 23B Short response 501
Practice test 23C Extended response 502
A Circles 504
B Ellipses 511
C Hyperbolae 514
Review set 24 521
Practice test 24A Multiple choice CD
Practice test 24B Short response 522
Practice test 24C Extended response 523
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
Practice test 25A Multiple choice CD
Practice test 25B Short response 545
Practice test 25C Extended response 547
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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Y:\HAESE\AUS_10A\AUS10A_01\013AUS10A_01.cdr Monday, 19 November 2012 1:51:53 PM EMMA
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14 INDICES (Chapter 1)
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
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INDICES (Chapter 1) 15
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
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Y:\HAESE\AUS_10A\AUS10A_01\015AUS10A_01.cdr Wednesday, 31 October 2012 4:55:51 PM EMMA
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16 INDICES (Chapter 1)
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
3 Write the following as powers of 3:
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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Y:\HAESE\AUS_10A\AUS10A_01\016AUS10A_01.cdr Wednesday, 31 October 2012 5:02:44 PM EMMA
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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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Y:\HAESE\AUS_10A\AUS10A_01\017AUS10A_01.cdr Thursday, 8 November 2012 10:48:43 AM EMMA
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18 INDICES (Chapter 1)
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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Y:\HAESE\AUS_10A\AUS10A_01\018AUS10A_01.cdr Friday, 16 November 2012 1:58:41 PM BRIAN
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INDICES (Chapter 1) 19
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
6 Simplify the following, writing your answers in scientific notation:
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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Y:\HAESE\AUS_10A\AUS10A_01\019AUS10A_01.cdr Wednesday, 31 October 2012 4:58:24 PM EMMA
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20 INDICES (Chapter 1)
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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Y:\HAESE\AUS_10A\AUS10A_01\020AUS10A_01.cdr Friday, 16 November 2012 2:00:08 PM BRIAN
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INDICES (Chapter 1) 21
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
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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
310 Simplify the following, writing your answers in scientific notation:
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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Y:\HAESE\AUS_10A\AUS10A_01\022AUS10A_01.cdr Wednesday, 31 October 2012 5:12:29 PM EMMA
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INDICES (Chapter 1) 23
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
8 Write the following as powers of 2:
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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Y:\HAESE\AUS_10A\AUS10A_01\023AUS10A_01.cdr Wednesday, 31 October 2012 5:14:47 PM EMMA
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24 INDICES (Chapter 1)
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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Y:\HAESE\AUS_10A\AUS10A_01\024AUS10A_01.cdr Wednesday, 31 October 2012 5:16:11 PM EMMA
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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
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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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Y:\HAESE\AUS_10A\AUS10A_02\026AUS10A_02.cdr Monday, 19 November 2012 1:59:10 PM EMMA
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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)
2 Expand and simplify:
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
Expand and simplify:
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
3 Expand and simplify:
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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Y:\HAESE\AUS_10A\AUS10A_02\027AUS10A_02.cdr Monday, 19 November 2012 2:01:26 PM EMMA
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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
Expand and simplify:
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
1 Expand and simplify:
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)
2 Expand and simplify:
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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Y:\HAESE\AUS_10A\AUS10A_02\028AUS10A_02.cdr Monday, 19 November 2012 2:02:16 PM EMMA
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ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2) 29
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
Expand and simplify:
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
1 Expand and simplify:
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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Y:\HAESE\AUS_10A\AUS10A_02\029AUS10A_02.cdr Thursday, 1 November 2012 10:27:28 AM EMMA
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30 ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2)
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
Expand and simplify:
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
1 Expand and simplify:
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
2 Expand and simplify:
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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Y:\HAESE\AUS_10A\AUS10A_02\030AUS10A_02.cdr Monday, 19 November 2012 2:07:05 PM EMMA
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ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2) 31
3 Use the diagram alongside to show that
(a b)2 = a2 2ab+ b2.
4 Expand and simplify:
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
1 Expand and simplify:
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
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32 ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2)
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
2 Expand and simplify:
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)
4 Expand and simplify:
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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Y:\HAESE\AUS_10A\AUS10A_02\032AUS10A_02.cdr Thursday, 1 November 2012 9:41:43 AM EMMA
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ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2) 33
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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Y:\HAESE\AUS_10A\AUS10A_02\033AUS10A_02.cdr Monday, 19 November 2012 2:10:42 PM EMMA
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34 ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2)
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
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ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2) 35
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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Y:\HAESE\AUS_10A\AUS10A_02\035AUS10A_02.cdr Monday, 19 November 2012 2:12:47 PM EMMA
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36 ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2)
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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Y:\HAESE\AUS_10A\AUS10A_02\036AUS10A_02.cdr Monday, 19 November 2012 2:13:04 PM EMMA
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ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2) 37
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
ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2)
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
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ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2) 39
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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Y:\HAESE\AUS_10A\AUS10A_02\039AUS10A_02.cdr Thursday, 1 November 2012 10:15:31 AM EMMA
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40 ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2)
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
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ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2) 41
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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Y:\HAESE\AUS_10A\AUS10A_02\041AUS10A_02.cdr Monday, 19 November 2012 2:19:25 PM EMMA
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42 ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2)
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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Y:\HAESE\AUS_10A\AUS10A_02\042AUS10A_02.cdr Monday, 19 November 2012 2:21:17 PM EMMA
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ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2) 43
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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Y:\HAESE\AUS_10A\AUS10A_02\043AUS10A_02.cdr Monday, 19 November 2012 4:49:41 PM EMMA
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44
Puzzle
ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2)
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.
1 Expand and simplify:
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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Y:\HAESE\AUS_10A\AUS10A_02\044AUS10A_02.cdr Thursday, 1 November 2012 9:02:50 AM EMMA
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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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Y:\HAESE\AUS_10A\AUS10A_02\045AUS10A_02.cdr Monday, 19 November 2012 2:22:09 PM EMMA
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46 ALGEBRAIC EXPANSION AND FACTORISATION (Chapter 2)
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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Y:\HAESE\AUS_10A\AUS10A_02\046AUS10A_02.cdr Wednesday, 7 November 2012 3:37:06 PM EMMA
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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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Y:\HAESE\AUS_10A\AUS10A_03\047AUS10A_03.cdr Monday, 5 November 2012 12:04:54 PM EMMA
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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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Y:\HAESE\AUS_10A\AUS10A_03\048AUS10A_03.cdr Wednesday, 7 November 2012 3:45:55 PM EMMA
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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