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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 PRO... | [
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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 PRO... | [
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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 ... | [
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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 ... | [
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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 box... | [
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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 ... | [
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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... | [
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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 interactiv... | [
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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-... | [
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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 expre... | [
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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)Simplif... | [
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-0.022316208109259605,
-0.03964490815997124,
0.06053663045167923,
-0.... |
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 __________ 2... | [
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-0.038605... |
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 firs... | [
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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 ... | [
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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 expa... | [
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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... | [
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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 ... | [
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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 1... | [
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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 ... | [
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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... | [
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