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Edexcel AS and A level Mathematics
Pure Mathematics
Year 1 /AS
Series Editor: Harry Smith
Authors: Greg Attwood, Jack Barraclough, Ian Bettison, Alistair Macpherson, Bronwen/uni00A0Moran, Su Nicholson, Diane Oliver, Joe Petran, Keith Pledger, Harry Smith, Geoο¬ /uni00A0Staley, Robert Ward-Penny, Dave Wilkins11 β 19 PROGRESSION | [
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Edexcel AS and A level Mathematics
Pure Mathematics
Year 1 /AS
Series Editor: Harry Smith
Authors: Greg Attwood, Jack Barraclough, Ian Bettison, Alistair Macpherson, Bronwen/uni00A0Moran, Su Nicholson, Diane Oliver, Joe Petran, Keith Pledger, Harry Smith, Geoο¬ /uni00A0Staley, Robert Ward-Penny, Dave Wilkins11 β 19 PROGRESSION | [
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iiContents
Overarching themes iv
Extra online c
ontent vi
1 Algebraic e
xpressions 1
1.1 Index law
s 2
1.2 Expanding brack
ets 4
1.3 Factorising 6
1.4 Negative and fractional indic
es 9
1.5 Surds 12
1.6 Rationalising denominators 13
Mixed ex
ercise 1 15
2 Quadratics 18
2.1 Solving quadratic equations 19
2.2 Completing the squar
e 22
2.3 Functions 25
2.4 Quadratic graphs 27
2.5 The discriminant 30
2.6 Modelling with quadratics 32
Mixed ex
ercise 2 35
3 Equations and inequalities 38
3.1 Linear simultaneous equations 39
3.2 Quadratic simultaneous equations 41
3.3 Simultaneous equations on graphs 42
3.4 Linear inequalities 46
3.5 Quadratic inequalities 48
3.6 Inequalities on graphs 51
3.7 Regions 53
Mixed ex
ercise 3 56
4 Graphs and trans
formations 59
4.1 Cubic graphs 60
4.2 Quartic graphs 64
4.3 Reciprocal gr
aphs 66
4.4 Points of int
ersection 68
4.5 Translating gr
aphs 71 Contents
4.6 Stretching graphs 75
4.7 Trans
forming functions 79
Mixed ex
ercise 4 82
Review ex
ercise 1 85
5 Straight line gr
aphs 89
5.1 y = mx
+ c 90
5.2 Equations of st
raight lines 93
5.3 Parall
el and perpendicular lines 97
5.4 Length and area 100
5.5 Modelling with straight lines 103
Mixed ex
ercise 5 108
6 Circles 113
6.1 Midpoints and perpendicular
bisectors 114
6.2 Equation of a cir
cle 117
6.3 Intersections of st
raight lines
and circles 121
6.4 Use tangent and chord pr
operties 123
6.5 Circles and t
riangles 128
Mixed ex
ercise 6 132
7 Algebraic methods 137
7.1 Algebraic fr
actions 138
7.2 Dividing polynomials 139
7.3 The factor theorem 143
7.4 Mathematical proof 146
7.5 Methods of proo
f 150
Mixed ex
ercise 7 154
8 The binomial expansion 158
8.1 Pascal
βs triangle 159
8.2 Factorial notation 161
8.3 The binomial expansion 163
8.4 Solving binomial problems 165 | [
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iiiContents
8.5 Binomial estimation 167
Mixed ex
ercise 8 169
9 Trigonometric r
atios 173
9.1 The cosine rul
e 174
9.2 The sine rule 179
9.3 Areas o
f triangles 185
9.4 Solving triangle pr
oblems 187
9.5 Graphs of sine, c
osine and tangent 192
9.6 Trans
forming trigonometric graphs 194
Mixed ex
ercise 9 198
10 Trigonometric identities and
equations 202
10.1 Angles in all four quadr
ants 203
10.2 Exact values o
f trigonometrical ratios 208
10.3 Trigonomet
ric identities 209
10.4 Simple trig
onometric equations 213
10.5 Harder trig
onometric equations 217
10.6 Equations and identities 219
Mixed ex
ercise 10 222
Review ex
ercise 2 226
11 Vectors 230
11.1 Vectors 231
11.2 Representing v
ectors 235
11.3 Magnitude and direction 239
11.4 Position v
ectors 242
11.5 Solving geometric pr
oblems 244
11.6 Modelling with vectors 248
Mixed ex
ercise 11 251
12 Differentiation 255
12.1 Gradients of cur
ves 256
12.2 Finding the derivative 259
12.3 Differentiating
xn 262
12.4 Differentiating quadr
atics 26412.5 Differentiating functions with t
wo
or more terms 266
12.6 Gradients, tang
ents and normal 268
12.7 Increasing and decr
easing functions 270
12.8 Second order deriv
atives 271
12.9 Stationary points 273
12.10 Sketching gr
adient functions 277
12.11 Modelling with differentiation 279
Mixed ex
ercise 12 282
13 Integration 287
13.1 Integr
ating xn 288
13.2 Indefinite integr
als 290
13.3 Finding functions 293
13.4 Definite integr
als 295
13.5 Areas under cur
ves 297
13.6 Areas under the
x-axis 300
13.7 Areas bet
ween curves and lines 302
Mixed ex
ercise 13 306
14 Exponentials and logarithms 311
14.1 Exponential functions 312
14.2 y = ex 314
14.3 Exponential modelling 317
14.4 Logarithms 319
14.5 Law
s of logarithms 321
14.6 Solving equations using logarithms 324
14.7 Working with natur
al logarithms 326
14.8 Logarithms and non-linear data 328
Mixed ex
ercise 14 334
Review ex
ercise 3 338
Practic
e exam paper 342
Answ
ers 345
Index 399 | [
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ivOverarching themes
The following three overarching themes have been fully integrated throughout the Pearson Edexcel
AS and A level Mathematics series, so they can be applied alongside your learning and practice.
1. Mathematical argument, language and proof
β’ Rigorous and consistent approach throughoutβ’ Notation boxes explain key mathematical language and symbolsβ’ Dedicated sections on mathematical proof explain key principles and strategiesβ’ Opportunities to critique arguments and justify methods
2. Mathematical problem solving
β’ Hundreds of problem-solving questions, fully integrated
into the main exercises
β’ Problem-solving boxes provide tips and strategiesβ’ Structured and unstructured questions to build confi denceβ’ Challenge boxes provide extra stretch
3. Mathematical modelling
β’ Dedicated modelling sections in relevant topics provide plenty of practice where you need it β’ Examples and exercises include qualitative questions that allow you to interpret answers in the
context of the model
β’ Dedicated chapter in Statistics & Mechanics Year 1/AS explains the principles of modelling in
mechanics Overarching themes
Each chapter starts with
a list of objectives
The Prior knowledge check
helps make sure you are ready to start the chapterThe real world applications of the maths you are about to learn are highlighted at the start of the chapter with links to relevant questions in the chapterFinding your way around the book
Access an online digital edition using the code at the front of the book.The Mathematical Problem-solving cycle
specify the problem
interpret resultscollect information
process and
represent information | [
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vOverarching themes
Every few chapters a Review exercise
helps you consolidate your learning with lots of exam-style questionsEach section begins
with explanation and key learning points
Step-by-step worked
examples focus on the key types of questions youβll need to tackleExercise questions are
carefully graded so they increase in diffi culty and gradually bring you up to exam standard
Problem-solving boxes provide hints, tips and strategies, and Watch out boxes highlight areas where students oft en lose marks in their examsExercises are packed with exam-style questions to ensure you are ready for the exams
A full AS level practice paper at the back of the book helps you prepare for the real thing
Exam-style questions are ο¬ agged with
Problem-solving
questions are ο¬ agged withE
PEach chapter ends with a Mixed exercise and a Summary of key pointsChallenge boxes give you a chance to tackle some more diffi cult questions | [
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viExtra online content
Whenever you see an Online box, it means that there is extra online content available to support you.
SolutionBank
SolutionBank provides a full worked solution for
every question in the book.
Download all the solutions
as a PDF or quickly fi nd the solution you need online Extra online content
Full worked solutions are
available in SolutionBank.Online | [
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viiExtra online content
Access all the extra online content for FREE at:
www.pearsonschools.co.uk/p1maths
You can also access the extra online content by scanning this QR Code:
GeoGebra interactives
Explore topics in more detail,
visualise problems and consolidate your understanding with GeoGebra-powered interactives.
Interact with the maths
you are learning using GeoGebra's easy-to-use tools
Explore the gradient of the
chord AP using GeoGebra.Online
Casio calculator support
Our helpful tutorials will guide
you through how to use your calculator in the exams. They cover both Casio's scientific and colour graphic calculators.
See exactly which
buttons to press and what should appear on your calculator's screen
Work out each coefficient
qui
ckly using the nCr and power
functions on your calculator.Online | [
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viiiPublished by Pearson Education Limited, 80 Strand, London WC2R 0RL.
www.pearsonschoolsandfecolleges.co.uk Copies of official specifications for all Pearson qualifications may be found on the website:
qualifications.pearson.com
Text Β© Pearson Education Limited 2017
Edited by Tech-Set Ltd, GatesheadTypeset by Tech-Set Ltd, GatesheadOriginal illustrations Β© Pearson Education Limited 2017 Cover illustration Marcus@kja-artists
The rights of Greg Attwood, Jack Barraclough, Ian Bettison, Alistair Macpherson, Bronwen
Moran, Su Nicholson, Diane Oliver, Joe Petran, Keith Pledger, Harry Smith, Geoff Staley, RobertΒ Ward
-Penny, Dave Wilkins to be identified as authors of this work have been asserted
by them in accordance with the Copyright, Designs and Patents Act 1988.
First published 201720 19 18 17
10 9 8 7 6 5 4 3 2 1
British Library Cataloguing in Publication Data
A catalogue record for this book is available from the British Library
ISBN 978 1 292 20826 8 (Print)
Copyright notice
All rights reserved. No part of this publication may be reproduced in any form or by any means (including photocopying or storing it in any medium by electronic means and whether or not transiently or incidentally to some other use of this publication) without the written permission of the copyright owner, except in accordance with the provisions of the Copyright, Designs and Patents Act 1988 or under the terms of a licence issued by the Copyright Licensing Agency, Barnards Inn 86 Fetter Lane, London EC4A 1EN (www.cla.co.uk). Applications for the copyright ownerβs written permission should be addressed to the publisher.
Printed in Slovakia by NeografiaPicture Credits
The publisher would like to thank the following for their kind permission to reproduce their photographs:
(Key: b-bottom; c-centre; l-left; r-right; t-top)123RF.com: David Acosta Allely 287, 338cr; Alamy Images: Utah Images 113, 226l, Xinhua 38,
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All other images Β© Pearson EducationISBN 978 1 292 20759 9 (PDF) | [
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1
Algebraic expressions
After completing this chapter you should be able to:
β Multiply and divide integer po
wers β pages 2β3
β Expand a single term over brackets and collect like
terms
β pages 3β4
β Expand the product of two or three expressions β pages 4β6
β Factorise linear, quadratic and simple cubic expressions β pages 6β9
β Know and use the laws of indices β pages 9β11
β Simplify and use the rules of surds β pages 12β13
β Rationalise denominators β pages 13β16Objectives
1 Simplify:
a 4m2n + 5mn2 β 2m2n + mn2 β 3mn2
b 3x2 β 5x + 2 + 3x2 β 7x β 12
β GCSE Mathematics
2 Write as a single power of 2:a
25 Γ 23 b 26 Γ· 22
c (23)2 β GCSE Mathematics
3 Expand:a
3(x
+ 4) b 5(2 β 3
x)
c 6(2x
β 5y) β GCSE Mathematics
4 Write down the highest common factor of:a
24 and 16 b 6x
and 8x2
c 4xy2 and 3xy β GCSE Mathematics
5 Simplify:
a 10x ____ 5 b 20x ____ 2 c 40x ____ 24
β GCSE MathematicsPrior knowledge check
Computer scientists use indices to describe
very large numbers. A quantum computer with 1000 qubits (quantum bits) can consider 2
1000
values simultaneously. This is greater than the number of particles in the observable universe.1 | [
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2
Chapter 1
1.1 Index laws
β You can use the laws of indices to simplify powers of the same base.
β’ am Γ an = am + n
β’ am Γ· an = am β n
β’ (am)n = amn
β’ (ab)n = anbn
Example 1
Example 2
Expand these expressions and simplify if possible:
a β3x
(7x β 4) b y2(3 β 2y3)
c 4x
(3x β 2x2 + 5x3) d 2x (5x + 3) β 5(2x + 3)Simplify these expressions:a
x2 Γ x5 b 2r2 Γ 3r3 c b7
__ b4 d 6x5 Γ· 3x3 e (a3)2 Γ 2a2 f (3x2)3 Γ· x4
x5Notation
This is the base .
This is the index, power or
exponent.
a x2Β ΓΒ x5 =Β x2 + 5 =Β x7
b 2r2Β ΓΒ 3 r3 =Β 2Β Γ Β 3Β ΓΒ r2Β ΓΒ r3
=Β 6Β ΓΒ r2 + 3 =Β 6 r5
c b7 __ b4 = b7 β 4 =Β b3
d 6x5Β Γ·Β 3 x3 =Β 6 __ 3 Β ΓΒ x5 ____ x3
=Β 2Β Γ
Β x2 = 2x2
e (a3)2Β ΓΒ 2 a2 =Β a6Β ΓΒ 2 a2
=Β 2Β ΓΒ a6Β ΓΒ a2 =Β 2 a8
f (3x2)3 _____ x4 =Β 33 Γ (x2)3 ____ x4
= 27
Γ x6 __ x4 =Β 27 x2Use the rule amΒ ΓΒ anΒ =Β am + n to simplify the index.
x5Β Γ·Β x3Β =Β x5 β 3Β =Β x2Rewrite the expression with the numbers
together and the r terms together.
2Β ΓΒ 3Β =Β 6
r2Β ΓΒ r3Β =Β r2 + 3
A min us sign outside
brackets changes the sign of every term inside the brackets.Watch outUse the rule am Γ· an = am β n to simplify the index.
Use the rule (am)nΒ =Β amn to simplify the index.
a6Β Γ a2 = a6 + 2 = a8
Use the rule (ab)nΒ =Β anbn to simplify the numerator.
(x2)3Β =Β x2 Γ 3Β =Β x6
x6 __ x4 Β =Β x6 β 4Β =Β x2 | [
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3Algebraic expressions
a β3x(7xΒ β 4 ) =Β β21x2Β +Β 12 x
b y2(3Β βΒ 2y3) =Β 3 y2Β βΒ 2y5
c 4x(3xΒ βΒ 2 x2Β +Β 5 x3)
=Β 12 x2Β βΒ 8 x3Β +Β 20 x4
d 2x(5xΒ +Β 3 )Β βΒ 5(2 xΒ +Β 3)
=Β 10 x2Β +Β 6 xΒ βΒ 10 xΒ βΒ 15
=Β 10 x2Β βΒ 4 xΒ βΒ 15
a x7 + x4 _______ x3 = x7 ___ x3 + x4 ___ x3
=
x7 β 3Β + x4 β 3 = x4Β + x
b 3x2 β 6x5 __________ 2x = 3 x 2 ____ 2x β 6 x 5 ____ 2x
= 3 __ 2 x 2 β 1 β 3x5 β 1 = 3x ___ 2 β 3 x 4
c 20x7 + 15x3 ____________ 5x2 = 20 x 7 _____ 5 x 2 + 15 x 3 _____ 5 x 2
= 4x7 β 2 + 3 x3 β 2 = 4 x 5 + 3 xExample 3β3xΒ ΓΒ 7xΒ ξ΅Β β21x1 +Β 1Β ξ΅Β β21x2
β3xΒ ΓΒ (β4)Β ξ΅Β ξ±12x
Remember a minus sign outside the brackets
changes the signs within the brackets.
Simplify 6xΒ βΒ 10x to give β4x.
Simplify these expressions:
a x 7 + x 4 ______ x 3 b 3 x 2 β 6 x 5 ________ 2x c 20 x 7 + 15 x 3 __________ 5 x 2 y2Β ΓΒ (β2y3)Β ξ΅Β β2y2 +Β 3Β ξ΅Β β2y5
Divide each term of the numerator by x 3 .
x1 is the same as x.
Divide each term of the numerator by 2x.
Simplify each fraction:
3 x 2 ____ 2x = 3 __ 2 Γ x 2 ___ x = 3 __ 2 Γ x2 β 1
β 6 x 5 ____ 2x = β 6 __ 2 Γ x 5 ___ x = β3 Γ x5 β 1
Divide each term of the numerator by 5x2.
1 Simplify these expressions:
a x3 Γ x4 b 2x3 Γ 3x2 c k3
__ k2
d 4p3
___ 2p e 3x3 ___ 3x2 f (y2)5
g 10x5 Γ· 2x3 h ( p3)2 Γ· p4 i (2a3)2 Γ· 2a3
j 8p4 Γ· 4p3 k 2a4 Γ 3a5 l 21a3b7 ______ 7ab4
m 9x2 Γ 3(x2)3 n 3x3 Γ 2x2 Γ 4x6 o 7a 4 Γ (3a 4)2
p (4y 3)3 Γ· 2y3 q 2a3 Γ· 3a2 Γ 6a5 r 3a4 Γ 2a5 Γ a3Exercise 1A | [
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4
Chapter 1
1.2 Expanding brackets
To find the product of two expressions you multiply each term in one expression by each term in the
other expression.
(x + 5)(4x β 2y + 3)x Γ
5 Γ= x(4x β 2y + 3) + 5(4x β 2y + 3)= 4x
2 β 2xy + 3x + 20x β 10y + 15
= 4x2 β 2xy + 23x β 10y + 15Multiplying each of the 2 terms in the first expression by each of the
3 terms in the second expression gives 2 Γ 3 = 6 terms.
Simplify your answer by collecting like terms.2 Expand and simplify if possible:
a 9(x β
2) b x(x
+ 9) c β3y
(4 β 3y)
d x(y
+ 5) e βx(3
x + 5) f β5x
(4x + 1)
g (4x
+ 5)x h β3y
(5 β 2y2) i β2x (5x β 4)
j (3x β
5)x2 k 3(x + 2) + (x β 7) l 5x β 6 β (3x β 2)
m 4(c +
3d 2) β 3(2c + d 2) n (r2 + 3t2 + 9) β (2r2 + 3t2 β 4)
o x(3x2 β 2x + 5) p 7y2(2 β 5y + 3y2) q β2y2(5 β 7y + 3y2)
r 7(x β
2) + 3(x + 4) β 6(x β 2) s 5x β
3(4 β 2x) + 6
t 3x2 β x(3 β 4x) + 7 u 4x( x + 3) β 2x(3x β 7) v 3x2(2x + 1) β 5x2(3x β 4)
3 Simplify these fractions:
a 6 x 4 + 10 x 6 _________ 2x b 3 x 5 β x 7 _______ x c 2 x 4 β 4 x 2 ________ 4x
d 8 x 3 + 5x ________ 2x e 7 x 7 + 5 x 2 ________ 5x f 9 x 5 β 5 x 3 ________ 3x
a (x + 5)(x + 2)
= x2 + 2x + 5 x + 10
=
x2 + 7x + 10
b (x
β 2y)(x2 + 1)
= x3 + x β 2x2y β 2 yExample 4
Expand these expressions and simplify if possible:
a (x
+ 5)(x + 2) b (x
β 2y)(x2 + 1) c (x β y)2 d (x + y)(3x β 2y β 4)
Multiply x by (x + 2) and then multiply 5 by (x + 2).
Simplify your answer by collecting like terms.
β2y Γ x2 = β2x2y
There are no like terms to collect. | [
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-0.0230973232537508,
-0.06870998442173004,
0.01709219068288803,
0.00... |
5Algebraic expressions
c (x β y)2
= (x β y)(x β y)
=
x2 β xy β xy + y2
= x2 β 2xy + y2
d (x + y)(3x β 2 y β 4)
= x(3x
β 2y β 4) + y (3x β 2 y β 4)
= 3x2 β 2xy β 4 x + 3 xy β 2 y2 β 4y
= 3x2 + xy β 4 x β 2 y2 β 4y
a x(2x + 3)(x β 7)
= (2x2 + 3 x)(x β 7)
= 2
x3 β 14 x2 + 3 x2 β 21x
= 2
x3 β 11 x2 β 21x
b x(5x β
3y)(2x β y + 4)
= (5x2 β 3 xy)(2x β y + 4)
= 5x2(2x β y + 4) β 3 xy(2x β y + 4)
= 10x3 β 5 x2y + 20 x2 β 6 x2y + 3 xy2
β 12 xy
= 10
x3 β 11 x2y + 20 x2 + 3 xy2 β 12 xy
c (x
β 4)( x + 3)( x + 1)
= (x2 β x β 12)( x + 1)
=
x2(x + 1) β x (x + 1) β 12( x + 1)
=
x3 + x2 β x2 β x β 12 x β 12
=
x3 β 13 x β 12Example 5
Expand these expressions and simplify if possible:
a x(2x
+ 3)(x β 7) b x(5x
β 3y)(2x β y + 4) c (x
β 4)(x + 3)(x + 1)
Be careful with minus signs. You need to change
every sign in the second pair of brackets when you multiply it out.
Choose one pair of brackets to expand first, for example:
(x β 4)(x + 3) = x
2 + 3x β 4x β 12
= x2 β x β 12
You multiplied together three linear terms, so the final answer contains an x
3 term.βxy β xy = β2xy
Multiply x by (3x β 2y β 4) and then multiply y by (3x β 2y β 4).
Start by expanding one pair of brackets:x(2x + 3) = 2x
2 + 3x
You could also have expanded the second pair of brackets first: (2x + 3)(x β 7) = 2x
2 β 11x β 21
Then multiply by x.(x β y)2 means (x β y) multiplied by itself.
1 Expand and simplify if possible:
a (x
+ 4)(x + 7) b (x
β 3)(x + 2) c (x
β 2)2
d (x β y)(2x + 3) e (x + 3y)(4x β y) f (2x β 4y)(3x + y)
g (2x
β 3)(x β 4) h (3x
+ 2y)2 i (2x + 8y)(2x + 3)
j (x
+ 5)(2x + 3y β 5) k (x
β 1)(3x β 4y β 5) l (x
β 4y)(2x + y + 5)
m (x
+ 2y β 1)(x + 3) n (2x
+ 2y + 3)(x + 6) o (4 β
y)(4y β x + 3)
p (4y
+ 5)(3x β y + 2) q (5y
β 2x + 3)(x β 4) r (4y
β x β 2)(5 β y)Exercise 1B | [
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0.0203... |
6
Chapter 1
1.3 Factorising
You can write expressions as a product of their factors.
β Factorising is the opposite of expanding
brack
ets.4x(2x + y)
(x + 5)3
(x + 2y)(x β 5y)= 8x2 + 4xy
= x3 + 15x2 + 75x + 125
= x2 β 3xy β 10y2Expanding brackets
FactorisingExpand and simplify ( x + y )4. You can use the binomial expansion to expand
ex
pressions like ( x + y )4 quickly. β Section 8.3LinksChallenge2 Expand and simplify if possible:
a 5(x
+ 1)(x β 4) b 7(x
β 2)(2x + 5) c 3(x
β 3)(x β 3)
d x(x
β y)(x + y) e x(2x
+ y)(3x + 4) f y(x
β 5)(x + 1)
g y(3x
β 2y)(4x + 2) h y(7 β
x)(2x β 5) i x(2x
+ y)(5x β 2)
j x(x
+ 2)(x + 3y β 4) k y(2x
+ y β 1)(x + 5) l y(3x
+ 2y β 3)(2x + 1)
m x(2x
+ 3)(x + y β 5) n 2x
(3x β 1)(4x β y β 3) o 3x
(x β 2y)(2x + 3y + 5)
p (x
+ 3)(x + 2)(x + 1) q (x
+ 2)(x β 4)(x + 3) r (x
+ 3)(x β 1)(x β 5)
s (x
β 5)(x β 4)(x β 3) t (2x
+ 1)(x β 2)(x + 1) u (2x
+ 3)(3x β 1)(x + 2)
v (3x
β 2)(2x + 1)(3x β 2) w (x
+ y)(x β y)(x β 1) x (2x
β 3y)3
3 The diagram shows a rectangle with a square cut out.
The rectangle has length 3
x β y + 4 and width x + 7.
The square has length x β 2.Find an expanded and simplified expression for the shaded area.
x β 2x + 7
3x β y + 4
4 A cuboid has dimensions x + 2 cm, 2x β 1 cm and 2x + 3 cm.
Show tha
t the volume of the cuboid is 4x3 + 12x2 + 5x β 6 cm3.
5 Given tha
t (2x + 5y)(3x β y)(2x + y) = ax3 + bx2y + cxy2 + dy3, where a, b, c and d are
constants, find the values of a, b, c and d. (2 marks)P
Use the same strategy as you would use
if the lengths were given as numbers:
3cm6cm
10cmProblem-solving
P
E/P | [
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0.0... |
7Algebraic expressions
An ex pression in the form x2 β y2 is
called the difference of two squares.Notation= (x + 3)(2x β 1)β A quadratic expression has the form
ax2 + bx + c where a, b and c are real
numbers and a β 0.
To factorise a quadratic expression:
β’Find two fact
ors of ac that add up to b
β’Rewrite the
b term as a sum of these two
factors
β’Factorise each p
air of terms
β’Take out the c
ommon factor
β x2 β y2 = (x + y)(x β y)a 3x + 9 = 3( x + 3)
b x2 β 5 x = x(x β 5)
c 8x2 + 20 x = 4 x(2x + 5)
d 9x2y + 15 xy2 = 3 xy(3x + 5 y)
e 3x2 β 9 xy = 3 x(x β 3 y)Example 6
Factorise these expressions completely:
a 3x
+ 9 b x2 β 5 x c 8x2 + 20x d 9x2y + 15xy2 e 3x2 β 9xy
3 is a common factor of 3x and 9.
For the expression 2x2 + 5x β 3, ac = β6 = β1 Γ 6
and β1 + 6 = 5 = b.
2x2 β x + 6x β 3
= x(2x β 1) + 3(2x β 1)x is a common factor of x2 and β5x.
4 and x are common factors of 8x2 and 20x.
So take 4x outside the brackets.
3, x and y are common factors of 9x2y and 15xy2.
So take 3xy outside the brackets.
x and β3y have no common factors so this
expression is completely factorised.
Real n umbers are all the positive and
negative numbers, or zero, including fractions and surds.Notation
Example 7
Factorise:
a x2Β βΒ 5xΒ βΒ 6 b x2Β +Β 6xΒ +Β 8 c 6x2Β βΒ 11xΒ βΒ 10 d x2Β βΒ 25 e 4x2Β βΒ 9y2
a x2 β 5 x β 6
ac =
β6 and b = β 5
So x2 β 5 x β 6 ξ΅ x2 + x β 6 x β 6
= x(x
+ 1) β 6( x + 1)
= (x
+ 1)( x β 6)Here aΒ =Β 1, bΒ =Β β 5 and cΒ =Β β 6.
1 Work out the two factors of ac =Β β 6 which add
t
o give you bΒ =Β β5. β6 + 1Β =Β β5
2 Rewrite the b term using these two factors.
3 Factorise first two terms and last two terms.
4 xΒ + 1 is a factor of both terms, so take that
outside the brackets. This is now completely
factorised. | [
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8
Chapter 1
Example 8
Factorise completely:
a x3 β 2x2 b x3 β 25x c x3 + 3x2 β 10xb x2 + 6 x + 8
= x2 + 2 x + 4 x + 8
= x(x
+ 2) + 4( x + 2)
= (x
+ 2)( x + 4)
c 6x2 β 11 x β 10
= 6x2 β 15 x + 4 x β 10
= 3x(2x
β 5) + 2(2 x β 5)
= (2 x
β 5)(3 x + 2)
d x2 β 25
= x2 β 52
= (x + 5)( x β 5)
e 4x2 β 9 y2
= 22x2 β 32y2
= (2 x + 3y)(2x β 3 y)
a x3 β 2x2 = x2(x β 2)
b x3 β 25 x = x(x2 β 25)
= x(x2 β 52)
= x(x + 5)( x β 5)
c x3 + 3 x2 β 10x = x (x2 + 3 x β 10)
= x(x + 5)( x β 2)
1 Factorise these expressions completely:a
4x + 8 b 6x β 24 c 20x + 15
d 2x2 + 4 e 4x2 + 20 f 6x2 β 18x
g x2 β 7x h 2x2 + 4x i 3x2 β x
j 6x2 β 2x k 10y2 β 5y l 35x2 β 28x
m x2 + 2x n 3y2 + 2y o 4x2 + 12x
p 5y2 β 20y q 9xy2 + 12x2y r 6abΒ β 2ab2
s 5x2 β 25xy t 12x2y ξ± 8xy2 u 15y β 20yz2
v 12x2 β 30 w xy2 β x2y x 12y2 β 4yxExercise 1Cx2 β 25 is the difference of two squares.This is the difference of two squares as the two
terms are x2 and 52.
The two x terms, 5x and β 5x, cancel each other out.acΒ =Β β60 and 4Β βΒ 15Β =Β β11Β =Β b.
Factorise.
This is the same as (2x)2Β βΒ (3y)2.
You canβt factorise this any further.
x is a common factor of x3 and β25x.
So take x outside the brackets.
Write the expression as a product of x and a
quadratic factor.
Factorise the quadratic to get three linear factors.acΒ =Β 8 and 2Β +Β 4Β =Β 6Β =Β b.
Factorise. | [
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9Algebraic expressions
Write 4x4 β 13x2 + 9 as the product of four linear factors.Challenge2 Factorise:
a x2 + 4x b 2x2 + 6x c x2 + 11x + 24
d x2 + 8x + 12 e x2 + 3xΒ β 40 f x2 β 8x + 12
g x2 + 5x + 6 h x2 β 2xΒ β 24 i x2 β 3xΒ β 10
j x2 +Β xΒ β 20 k 2x2 + 5xΒ + 2 l 3x2 + 10x β 8
m 5x2 β 16xΒ + 3 n 6x2 β 8x β 8
o 2x2 + 7xΒ β 15 p 2x4 + 14x2 + 24
q x2 β 4 r x2 β 49
s 4x2 β 25 t 9x2 β 25y2 u 36x2 β 4
v 2x2 β 50 w 6x2 β 10xΒ + 4 x 15x2 + 42xΒ β 9
3 Factorise completely:a
x3 + 2x b x3 β x2 + x c x3 β 5x
d x3 β 9x e x3 β x2 β 12x f x3 + 11x2 + 30x
g x3 β 7x2 + 6x h x3 β 64x i 2x3 β 5x2 β 3x
j 2x3 + 13x2 + 15x k x3 β 4x l 3x3 + 27x2 + 60x
4 Factorise completel
y x4 β y4. (2 marks)
5 Factorise completel
y 6x3 + 7x2 β 5x. (2 marks) For part n , ta ke 2 out as a common
factor first. For part p , let yΒ =Β x2.Hint
Watch out for terms that can be written as a
function of a function: x4 = (x2)2Problem-solving P
E
1.4 Negative and fractional indic es
Indices can be negative numbers or fractions.
x 1 _ 2 Γ x 1 _ 2 = x 1 _ 2 + 1 _ 2 = x 1 = x,
similarly x 1 __ n Γ x 1 __ n Γ . . . Γ x 1 __ n = x 1 __ n + 1 __ n +...+ 1 __ n = x 1 = x
n terms
β You can use the laws of indices with any rational power.
β’ a 1 __ m = m ββ―__
a
β’ a n __ m = m ββ―___ a n
β’ a βm = 1 ___ a m
β’ a 0 = 1 β«
βͺβͺβ¬βͺβͺβ Ratio nal
numbers are those that
can be written as a __ b where
a and b are integers.Notation
a 1 _ 2 = ββ―__
a is the
positive square root of a .
For example 9 1 _ 2 = ββ―__
9 = 3
but
9 1 _ 2 β β3 .Notation | [
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10
Chapter 1
Example 9
Simplify:
a x 3 ___ x β3 b x1
2 Γ x32
c (x3)23
d 2x1.5Β Γ·Β 4xβ0.25 e 3 ββ―______ 125 x 6 f 2 x 2 β x _______ x 5
a x 3 ____ x β3 = x3 β (β3) = x6
b x1
2 Γ x3
2 = x1
2 ξ±Β 32 = x2
c (x3)23 =Β x3 ξ³Β 23 =Β x2
d 2x1.5Β ξ΄Β 4 xβ0.25 = 1 __ 2 x1.5Β β (β0 .25) = 1 __ 2 x1.75
e 3 β _____ 125 x 6 = ( 125 x6 ) 1 __ 3
= (12
5 ) 1 __ 3 (x6 ) 1 __ 3 = 3 ββ―_____ 125 ( x 6 Γ 1 __ 3 ) = 5 x2
f 2 x 2 β x ______ x 5 = 2 x 2 ____ x 5 β x ___ x 5
= 2 Γ
x2 β 5 β x1 β 5 = 2xβ3 β xβ4
= 2 ___ x 3 β 1 ___ x 4 Use the rule amΒ Γ·Β anΒ =Β am β n.
Evaluate:
a 9 1 _ 2 b 6 4 1 _ 3 c 4 9 3 _ 2 d 2 5 β 3 _ 2 Example 10
a 9 1 __ 2 = ββ―__
9 = 3
b 6 4 1 __ 3 = 3 ββ―___ 64 = 4
c 4 9 3 __ 2 = ( ββ―___ 49 ) 3
73 = 343
d 2 5 β 3 __ 2 = 1 ____
2 5 3 __ 2 = 1 ______
( ββ―___ 25 ) 3
= 1 ___ 53 = 1 _____ 125 Using a 1 __ m = m ββ―__
a . 9 1 _ 2 = ββ―__
9
Using a n __ m = m ββ―__ an .
This means the square root o
f 49, cubed.
Using aβm = 1 ___ am This could also be written as ββ―__
x .
Use the rule amΒ ΓΒ anΒ =Β am + n.
Use the rule (am)nΒ =Β amn.
Use the rule amΒ Γ·Β anΒ =Β am β n.
1.5Β βΒ (β0.25)Β =Β 1.75
Using a 1 __ m = m ββ―__
a .
Divide each term of the numerator by x5.
Using a βm = 1 ___ a m
This means the cube root of 64.
Use your calculator to enter
ne
gative and fractional powers.Online | [
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11Algebraic expressions
1 Simplify:
a x3 Γ· xβ2 b x5 Γ· x7 c x 3 _ 2 Γ x 5 _ 2
d (x2 ) 3 _ 2 e (x3 ) 5 _ 3 f 3x0.5 Γ 4xβ0.5
g 9 x 2 _ 3 Γ· 3 x 1 _ 6 h 5 x 7 _ 5 Γ· x 2 _ 5 i 3x4 Γ 2xβ5
j ββ―__
x Γ 3 ββ―__
x k ( ββ―__
x )3 Γ ( 3 ββ―__
x )4 l ( 3 ββ―__
x )2 _____ ββ―__
x
2 Eva
luate:
a 2 5 1 _ 2 b 8 1 3 _ 2 c 2 7 1 _ 3
d 4β2 e 9 β 1 _ 2 f (β 5)β3
g ( 3 _ 4 ) 0 h 129 6 3 _ 4 i ( 25 __ 16 ) 3
2
j ( 27 __ 8 ) 2
3 k ( 6 _ 5 ) β1 l ( 343 ___ 512 ) β2
3
3 Simplify:
a (64x10)1
2 b 5 x 3 β 2 x 2 ________ x 5 c (125x12)1
3 d x + 4 x 3 _______ x 3
e 2x + x 2 _______ x 4 f ( 4 __ 9 x4) 3
2 g 9 x 2 β 15 x 5 _________ 3 x 3 h 5x + 3 x 2 ________ 15 x 3
4 a Find the value of
8 1 1 _ 4 . (1 mark)
b Simplify x(2 x β 1 _ 3 )4. (2 marks)
5 Given tha
t y = 1 __ 8 x 3 express each of the following in the form k x n , where k and n are constants.
a y 1
3 (2 marks)
b 1 __ 2 y β2 (2 marks)E
EExercise 1Da y 1 _ 2 = ( 1 __ 16 x 2 ) 1 _ 2
= 1 ___ ββ―___ 16 x 2 Γ 1 _ 2 = x __ 4
b 4yβ1 = 4 ( 1 __ 16 x 2 ) β1
= 4 ( 1 __ 16 ) β1
x 2 Γ (β1) = 4 Γ 16 xβ2
= 64 xβ2Substitute y = 1 ___ 16 x 2 into y 1 _ 2 .
( 1 ___ 16 ) 1 _ 2
= 1 ____
β ___ 16 and ( x 2 ) 1 _ 2 = x 2 Γ 1 _ 2 Given that y = 1 __ 16 x2 express each of the following in the form k x n , where k and n are constants.
a y 1 __ 2 b 4 y β1 Example 11
Check that your answers are in the correct form.
If k and n are constants they could be positive or negative, and they could be integers, fractions or surds.Problem-solving ( 1 ___ 16 ) β1
= 16 and x2 Γ β1 = xβ2 | [
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