Complete

Pure Mathematics
2/3 for Cambridge International
AS & A Level
Second Edition

Jean Linsky
James Nicholson
Brian Western

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Contents
1 Algebra 2 7.3 Binomial expansions of the form (a + x)n
1.1 The modulus function 3 where n is not a positive integer 147
1.2 Sketching linear graphs of the form y = a|x| + b 6 7.4 Binomial expansions and partial fractions 149
1.3 Division of polynomials 8 8 Further integration 154
1.4 The remainder theorem 10 8.1 Integration using partial fractions 155
1.5 The factor theorem 12 1
8.2 Integration of 2 160
x + a2
2 Logarithms and exponential functions 18
kf′(x)
2.1 Continuous exponential growth and decay 19 8.3 Integration of 164
f(x)
2.2 The logarithmic function 21 8.4 Integration by parts 167
2.3 ex and logarithms to base e 25 8.5 Integration using substitution 173
2.4 Equations and inequalities using logarithms 29 9 Vectors 182
2.5 Using logarithms to reduce equations to linear form 33 9.1 Vector notation 183
3 Trigonometry 40 9.2 The magnitude of a vector 187
3.1Secant, cosecant, and cotangent 41 9.3 Addition and subtraction of vectors:
3.2Further trigonometric identities 47 a geometric approach 190
3.3Addition formulae 50 9.4 The vector equation of a straight line 193
3.4Double angle formulae 55 9.5 Intersecting lines 197
3.5Expressing a sin θ + b cos θ in the form 9.6 Scalar products 201
R sin(θ ± α) or R cos(θ ± α) 58 9.7 The angle between two straight lines 205
Maths in real-life: Predicting tidal behaviour 72 9.8 The distance from a point to a line 206
4 Differentiation 68 10 Differential equations 215
4.1 Differentiating the exponential function 69 10.1 Forming simple differential equations (DEs) 216
4.2 Differentiating the natural logarithmic function 71 10.2 Solving first-order differential equations
4.3 Differentiating products 73 with separable variables 221
4.4 Differentiating quotients 76 10.3 Finding particular solutions to
4.5 Differentiating sin x, cos x, and tan x 79 differential equations 224
4.6 Implicit differentiation 81 10.4 Modelling with differential equations 229
4.7 Parametric differentiation 84 11 Complex numbers 241
5 Integration 91 11.1 Introducing complex numbers 243
5.1 Integration of eax+b 92 11.2 Calculating with complex numbers 248
1
5.2 Integration of 95 11.3 Solving equations involving complex numbers 251
ax + b
5.3 Integration of sin(ax + b), cos(ax + b), sec (ax + b) 99
2 11.4 Representing complex numbers geometrically 255
5.4 Extending integration of trigonometric functions 102 11.5 Polar form and exponential form 259
5.5 Numerical integration using the 11.6 Loci in the Argand diagram 264
trapezium rule – Pure 2 107 Maths in real-life: Electrifying, magnetic
6 Numerical solution of equations 117 and damp: how complex mathematics
6.1 Finding approximate roots by change makes life simpler 276
of sign or graphical methods 118 Exam-style paper 2A – Pure 2 278
6.2 Finding roots using iterative relationships 123 Exam-style paper 2B – Pure 2 280
6.3 Convergence behaviour of iterative functions 129 Exam-style paper 3A – Pure 3 282
Maths in real-life: Nature of mathematics 134 Exam-style paper 3B – Pure 3 284
7 Further algebra 136 Answers 286
7.1 Partial fractions 137 Glossary 323
7.2 Binomial expansions of the form (1 + x)n
when n is not a positive integer 144

iii
 Introduction
About this book
This book has been written to cover the Throughout the book, you will encounter
Cambridge AS & A level International worked examples and a host of rigorous
Mathematics (9709) course, and is fully aligned exercises. The examples show you the important
to the syllabus. The first six chapters of the book techniques required to tackle questions. The
cover material applicable to both Pure 2 and exercises are carefully graded, starting from
Pure 3, and the final five chapters cover Pure 3 a basic level and going up to exam standard,
material only. allowing you plenty of opportunities to practise
In addition to the main curriculum content, you your skills. Together, the examples and exercises
will find: put maths in a real-world context, with a truly
international focus.
●● ‘Maths in real-life’, showing how principles
learned in this course are used in the real At the start of each chapter, you will see a list
world. of objectives that are covered in the chapter.
●● Chapter openers, which outline how each These objectives are drawn from the Cambridge
topic in the Cambridge 9709 syllabus is used AS and A level syllabus. Each chapter begins
in real-life. with a Before you start section and finishes with
a Summary exercise and Chapter summary,
●● ‘Did you know?’ boxes (as shown below),
ensuring that you fully understand each topic.
which give interesting side-notes beyond the
scope of the syllabus. Each chapter contains key mathematical terms
The book contains the following features: to improve understanding, highlighted in colour,
with full definitions provided in the Glossary of
Note Did you know? terms at the end of the book.
The answers given at the back of the book are
concise. However, when answering exam-style
questions, you should show as many steps
Advice on Pure 2
calculator in your working as possible. All exam-style
use questions, as well as Exam-style papers 2A, 2B,
3A and 3B, have been written by the authors.
Pure 3

iv
About the authors
Brian Western has over 40 years of experience in teaching mathematics
up to A Level and beyond, and is also a highly experienced examiner.
He taught mathematics and further mathematics, and was an Assistant
Headteacher in a large state school. Brian has written and consulted on a
number of mathematics textbooks.
James Nicholson is an experienced teacher of mathematics at secondary
level, having taught for 12 years at Harrow School and spent 13 years as
Head of Mathematics in a large Belfast grammar school. He is the author
of several A Level texts, and editor of the Concise Oxford Dictionary
of Mathematics. He has also contributed to a number of other sets of
curriculum and assessment materials, is an experienced examiner and has
acted as a consultant for UK government agencies on accreditation of new
specifications.
Jean Linsky has been a mathematics teacher for over 30 years, as well
as Head of Mathematics in Watford, Herts, and is also an experienced
examiner. Jean has authored and consulted on numerous mathematics
textbooks.

A note from the authors

The aim of this book is to help students prepare for the Pure 2 and Pure 3
units of the Cambridge International AS and A Level mathematics syllabus,
although it may also be useful in providing support material for other AS
and A Level courses. The book contains a large number of practice questions,
many of which are exam-style.
In writing the book we have drawn on our experiences of teaching
mathematics and Further mathematics to A Level over many years as well
as on our experiences as examiners, and our discussions with mathematics
educators from many countries at international conferences.

v
 Student book & Cambridge syllabus
matching grid

Student Book: Complete Pure Mathematics

2 & 3 for Cambridge International AS & A Level
Syllabus: Cambridge International AS & A Level
Mathematics: Pure Mathematics 2 & 3 (9709)

PURE MATHEMATICS 2 & 3 Student Book

Syllabus overview

Unit P2: Pure Mathematics 2 (Paper 2)

Knowledge of the content of unit P1 is assumed and candidates may be required to demonstrate such knowledge
in answering questions.

1. Algebra
• understand the meaning of |x|, sketch the graph of y = |ax + b| and use relations such Pages 2–17
as |a| = |b| ⇔ a2 = b2 and |x − a| < b ⇔ a − b < x < a + b when solving equations and
inequalities;
• divide a polynomial, of degree not exceeding 4, by a linear or quadratic polynomial, and Pages 2–17
identify the quotient and remainder (which may be zero);
• use the factor theorem and the remainder theorem, e.g. to find factors, solve polynomial Pages 2–17
equations or evaluate unknown coefficients.

2. Logarithmic and exponential functions

• understand the relationship between logarithms and indices, and use the laws of Pages 18–39
logarithms (excluding change of base);
• understand the definition and properties of ex and ln x, including their relationship as Pages 18–39
inverse functions and their graphs;
• use logarithms to solve equations and inequalities in which the unknown appears in Pages 18–39
indices;
• use logarithms to transform a given relationship to linear form, and hence determine Pages 18–39
unknown constants by considering the gradient and/or intercept.

vi
 Student book & Cambridge syllabus
matching grid

3. Trigonometry
• understand the relationship of the secant, cosecant and cotangent functions to cosine, Pages 40–65
sine and tangent, and use properties and graphs of all six trigonometric functions for
angles of any magnitude;
• use trigonometrical identities for the simplification and exact evaluation of expressions Pages 40–65
and in the course of solving equations, and select an identity or identities appropriate to
the context, showing familiarity in particular with the use of:
– sec2 θ ≡ 1 + tan2 θ and cosec2 θ ≡ 1 + cot2 θ,
– the expansions of sin(A ± B), cos(A ± B) and tan(A ± B),
– the formulae for sin 2A, cos 2A and tan 2A,
– the expressions of a sin θ + b cos θ in the forms R sin(θ ± a) and R cos(θ ± a).

4. Differentiation
• use the derivatives of ex, in x, sin x, cos x, tan x, together with constant multiples, sums, Pages 68–90
differences and composites;
• differentiate products and quotients; Pages 68–90
• find and use the first derivative of a function which is defined parametrically or implicitly. Pages 68–90

5. Integration
1
• extend the idea of ‘reverse differentiation’ to include the integration of eax+b, , Pages 91–116
ax + b
sin(ax + b), cos(ax + b) and sec (ax + b) (knowledge of the general method
2

of integration by substitution is not required);

• use trigonometrical relationships (such as double-angle formulae) to facilitate the Pages 91–116
integration of functions such as cos2 x;
• use the trapezium rule to estimate the value of a definite integral, and use sketch graphs Pages 91–116
in simple cases to determine whether the trapezium rule gives an over-estimate or an
under-estimate.

6. Numerical solution of equations

• locate approximately a root of an equation, by means of graphical considerations and/or Pages 117–133
searching for a sign change;
• understand the idea of, and use the notation for, a sequence of approximations which Pages 123–140
converges to a root of an equation;
• understand how a given simple iterative formula of the form xn + 1 = F( xn) relates to Pages 123–140
the equation being solved, and use a given iteration, or an iteration based on a given
rearrangement of an equation, to determine a root to a prescribed degree of accuracy
(knowledge of the condition for convergence is not included, but candidates should
understand that an iteration may fail to converge).

vii
 Student book & Cambridge syllabus
matching grid

Unit P3: Pure Mathematics (Paper 3)

Knowledge of the content of unit P1 is assumed and candidates may be required to
demonstrate such knowledge in answering questions.

1. Algebra
• understand the meaning of |x|, sketch the graph of y = |ax + b| and use relations such Pages 2–17
as
|a| = |b| ⇔ a2 = b2 and
|x – a| < b ⇔ a – b < x < a + b
when solving equations and inequalities;
• divide a polynomial, of degree not exceeding 4, by a linear or quadratic polynomial, Pages 2–17
and identify the quotient and remainder (which may be zero);
• use the factor theorem and the remainder theorem, e.g. to find factors, solve Pages 2–17
polynomial equations or evaluate unknown coefficients;
• recall an appropriate form for expressing rational functions in partial fractions, and Pages 136–153
carry out the decomposition, in cases where the denominator is no more complicated
than:
- (ax + b)(cx + d)(ex + f),
- (ax + b)(cx + d)2,
- (ax + b)(x2 + c2),
and where the degree of the numerator does not exceed that of the denominator;
• use the expansion of (1 + x)n, where n is a rational number and IxI < 1 (finding a general Pages 152–169
1
term is not included, but adapting the standard series to expand e.g. (2 – x)–1 is
2
included).

2. Logarithmic and exponential functions

• understand the relationship between logarithms and indices, and use the laws of Pages 18–39
logarithms (excluding change of base);
• understand the definition and properties of ex and In x, including their relationship as Pages 18–39
inverse functions and their graphs;
• use logarithms to solve equations of the form ax = b, and similar inequalities; Pages 18–39
• use logarithms to transform a given relationship to linear form, and hence determine Pages 18–39
unknown constants by considering the gradient and/or intercept.

viii
 Student book & Cambridge syllabus
matching grid

3. Trigonometry
• understand the relationship of the secant, cosecant and cotangent functions to cosine, Pages 40–65
sine and tangent, and use properties and graphs of all six trigonometric functions for
angles of any magnitude;
• use trigonometrical identities for the simplification and exact evaluation of expressions Pages 40–65
and in the course of solving equations, and select an identity or identities appropriate to
the context, showing familiarity in particular with the use of:
- sec2 θ ≡ 1 + tan2 θ and cosec2 θ ≡ 1 + cot2 θ,
- the expansions of sin(A ± B), cos(A ± B) and tan(A ± B),
- the formulae for sin 2A, cos 2A and tan 2A,
- the expressions of a sin θ + b cos θ in the forms R sin(θ ± a) and R cos(θ ± a).

4. Differentiation
• use the derivatives of ex, ln x, sin x, cos x, tan x, tan–1 x, together with constant Pages 68–90
multiples, sums, differences and composites;
• differentiate products and quotients; Pages 68–90
• find and use the first derivative of a function which is defined parametrically or implicitly. Pages 68–90

5. Integration
1
• extend the idea of ‘reverse differentiation’ to include the integration of eax + b, , Pages 91–116 and
1 ax +b
sin(ax + b), cos(ax + b), sec2(ax + b) and 2 ; Pages 170–197
x + a2
• use trigonometrical relationships (such as double-angle formulae) to facilitate the Pages 97–122
integration of functions such as cos2 x;
• integrate rational functions by means of decomposition into partial fractions (restricted Pages 154–181
to the types of partial fractions specified in paragraph 1 above);
kf’ (x)
• recognise an integrand of the form , and integrate, for example, 2 x or tan x; Pages 170–197
f’ (x) x +1
• recognise when an integrand can usefully be regarded as a product, and use integration Pages 170–197
by parts to integrate, for example, x sin 2x, x2 ex or ln x;
• use a given substitution to simplify and evaluate either a definite or an indefinite integral; Pages 170–197
• use the trapezium rule to estimate the value of a definite integral, and use sketch graphs Pages 97–122
in simple cases to determine whether the trapezium rule gives an over-estimate or an
under-estimate.

ix
 Student book & Cambridge syllabus
matching grid

6. Numerical solution of equations

• locate approximately a root of an equation, by means of graphical considerations and/or Pages 117–133
searching for a sign change;
• understand the idea of, and use the notation for, a sequence of approximations which Pages 117–133
converges to a root of an equation;
• understand how a given simple iterative formula of the form xn + 1 = F(xn) relates to the Pages 117–133
equation being solved, and use a given iteration, or an iteration based on a given
rearrangement of an equation, to determine a root to a prescribed degree of accuracy
(knowledge of the condition for convergence is not included, but candidates should
understand that an iteration may fail to converge).

7. Vectors
 x
 x 
• use standard notations for vectors, i.e.   , xi+ yj,  y  , xi+ yj+ zk, AB, a ; Pages 182-214
 y  
z
• carry out addition and subtraction of vectors, and multiplication of a vector by a scalar, Pages 182-214
and interpret these operations in geometrical terms;
• calculate the magnitude of a vector, and use unit vectors, displacement vectors and Pages 182-214
position vectors;
• understand the significance of all the symbols used when the equation of a straight Pages 182-214
line is expressed in the form r = a + tb, and find the equation of a line, given sufficient
information;
• determine whether two lines are parallel, intersect or are skew, and find the point of Pages 182-214
intersection of two lines when it exists;
• use formulae to calculate the scalar product of two vectors, and use scalar products in Pages 182-214
problems involving lines and points.

8. Differential equations
• formulate a simple statement involving a rate of change as a differential equation, Pages 215–240
including the introduction if necessary of a constant of proportionality;
• find by integration a general form of solution for a first order differential equation in which Pages 215–240
the variables are separable;
• use an initial condition to find a particular solution; Pages 215–240
• interpret the solution of a differential equation in the context of a problem being Pages 215–240
modelled by the equation.

x
 Student book & Cambridge syllabus
matching grid

9. Complex numbers
• understand the idea of a complex number, recall the meaning of the terms real part, Pages 241–275
imaginary part, modulus, argument, conjugate, and use the fact that two complex
numbers are equal if and only if both real and imaginary parts are equal;
• carry out operations of addition, subtraction, multiplication and division of two complex Pages 241–275
numbers expressed in cartesian form x + iy;
• use the result that, for a polynomial equation with real coefficients, any non-real roots Pages 241–275
occur in conjugate pairs;
• represent complex numbers geometrically by means of an Argand diagram; Pages 241–275
• carry out operations of multiplication and division of two complex numbers expressed in Pages 241–275
polar form r(cos θ + i sin θ) ≡ r eiθ;
• find the two square roots of a complex number; Pages 241–275
• understand in simple terms the geometrical effects of conjugating a complex number Pages 241–275
and of adding, subtracting, multiplying and dividing two complex numbers;
• illustrate simple equations and inequalities involving complex numbers by means of loci Pages 241–275
in an Argand diagram, e.g. I z – a I < k, I z – a I = I z – b I, arg(z – a) = α.

xi
 1 Algebra
Algebra is used extensively in mathematics, chemistry, physics, economics
and social sciences. For example, the study of polynomials in astrophysics
has led to our understanding of gravitational lensing.
Gravitational lensing occurs when light from a distant source bends around
a massive object (such as a galaxy) between a source and an observer.
Multiple images of the same object may be seen. Here, the ‘Einstein Cross’,
four images of a very distant supernova, is seen in a photograph taken by
the Hubble telescope. The supernova is at a distance of approximately 8
billion light years, and is 20 times further away than the galaxy, which is at
a distance of 400 million light years. The light from the supernova is bent
in its path by the gravitational field of the galaxy. This bending produces
the four bright outer images. The bright central region of the galaxy is seen
as the central object. This phenomena was predicted by Einstein’s general
theory of relativity published in 1915, but was not observed until 1979.

Objectives
● Understand the meaning of |x|, sketch the graph of y = |ax + b| and use relations such as
|a| = |b| ⇔ a2 = b2 and |x − a| < b ⇔ a – b < x < a + b when solving equations and inequalities.
● Divide a polynomial, of degree not exceeding 4, by a linear or quadratic polynomial,
and identify the quotient and remainder (which may be zero).
● Use the factor theorem and the remainder theorem, e.g. to find factors, solve
polynomial equations or evaluate unknown coefficients.

Before you start

You should know how to: Skills check:
1. Do long division, 1. Find the following using long division.
e.g. 357 ÷ 21 a) 608 ÷ 19
17
21)357 b) 2774 ÷ 38
21 c) 1081 ÷ 23
147 d) 1392 ÷ 24
147 Therefore 357 = 17
0 21
2. Find the remainder, 2. Find the remainder of the following after
e.g. 461 ÷ 37 doing long division.
12 a) 923 ÷ 21
37)461
b) 742 ÷ 32
37
91 c) 1527 ÷ 43
74 Remainder = 17 d) 4258 ÷ 26
17

2
 Pure 2 Pure 3

1.1 The modulus function

The modulus of a real number is the magnitude of that number.
If we have a real number x, then the modulus of x is written as x . We say this as ‘mod x’.
Thus 2 = 2 and −2 = 2, and if we write x < 2 this means that −2 < x < 2.

The modulus function f(x) = x is defined as

x =x for x≥0
x = −x for x<0

Consider the impact that the modulus function has by looking at the graphs
of y = x − 1 and y = |x − 1|.
y y
y = |x – 1|
y=x−1
1

0 1 x
0 1 x
–1 Note: For f(x) < 0, f ( x ) = −f(x).

When graphing y = |f(x)|, we reflect the graph of

y = f(x) in the x-axis whenever f(x) < 0.

Example 1
Solve the equation x + 2 = 3x .

Method 1: Using a graph

y Sketch the graphs and find where they
y = –(3x) y = 3x
6 intersect.
y = –(x + 2) y=x+2
4 The lines cannot be drawn below the x-axis.
B
A 2 For x < 0, 3x = −(3x)
–4 –3 –2 –1 0 1 2 3 4 x For x < −2, x + 2 = −(x + 2)
–2
–4
Graphs intersect at A and B.

At A, x + 2 = −3x At A, the line y = x + 2 intersects the line y = −(3x).

4x = −2 ⇒ x = − 1
2
At B, x + 2 = 3x At B, the line y = x + 2 intersects the line y = 3x.
2x = 2 ⇒ x = 1
x = − 1 or x = 1
2
Continued on the next page
▲

Algebra 3
 Method 2: Squaring both sides of the equation
(x + 2)2 = (3x)2
We do this to ensure both sides of the equation are
x2 + 4x + 4 = 9x2 positive.
8x2 – 4x – 4 = 0
2x2 – x – 1 = 0
(2x + 1)(x – 1) = 0
Note: You can only use this method if the variable
x = − 1 or x = 1 (e.g. x) is inside the modulus expression.
2

Method 3: Removing modulus signs by equating the left-hand side

with both ‘plus and minus’ the right-hand side
x + 2 = 3x or x + 2 = −3x
We get the same result if we say 3x = ±(x + 2).
2x = 2 or 4x = −2
x=1 or x = − 1
2

Example 2
Solve the inequality 4 x + 3 > 2 x − 1 .

Method 1: Using a graph

y = –(4x + 3)
y Sketch the graphs and find where they intersect.
8 y = 4x + 3
y = –(2x – 1) The lines cannot be drawn below the x-axis.
6
y = 2x – 1 3
A 4 For x < − , 4 x + 3 = −(4x + 3)
4
2
For x < 1 , 2 x − 1 = −(2x – 1)
B

–4 –3 –2 –1 0 1 2 3 4 x
2
–2
Graphs meet at A and B.
–4

At A, the line y = −(4x + 3) intersects the line

At A, −(4x + 3) = −(2x – 1)
y = −(2x – 1).
2x = −4, x = −2

At B, the line y = 4x + 3 intersects the line

At B, 4x + 3 = −(2x – 1)
y = −(2x – 1).
1
6x = −2, x=−
3
We want the region where 4 x + 3 > 2 x − 1 .
This is where the ‘blue’ lines are above the ‘green’ lines.
1
x < −2 and x > −
3
Continued on the next page
▲

4 The modulus function

 Pure 2 Pure 3

Method 2: Squaring both sides of the inequality

We can do this because we know both
(4x + 3)2 > (2x – 1)2
|4x + 3| and |2x − 1| are positive so
16x2 + 24x + 9 > 4x2 – 4x + 1 squaring them will not change the inequality.
12x2 + 28x + 8 > 0
3x2 + 7x + 2 > 0
(3x + 1)(x + 2) > 0 Draw a rough sketch to see where y > 0.
There are two separate blue regions that
satisfy y > 0.
–2 1
–
3
1 There are two separate regions that satisfy
x < −2 or x > −
3 y > 0.

Method 3: Removing modulus signs by equating the left-hand side with

both ‘plus and minus’ the right-hand side
First find where the graphs of y = |4x + 3| and y = |2x – 1| intersect
(the values of x at these points are called the critical values).
4x + 3 = 2x – 1 or 4x + 3 = −(2x – 1)
2x = −4 or 6x = −2
x = −2 or x = −1 This gives us the two critical values.
3
To find the correct inequalities you need to take values
of x on either side of the critical values.

Take values on either side of x = −2 and

–3 –2 –1 10 x
–
3 x = − 1.
3
e.g.
x = −3, 4 x + 3 = 9 and 2 x − 1 = 7

Thus 4 x + 3 > 2 x − 1 , so x < −2 One of the solution regions is less than −2.

is one region where 4 x + 3 > 2 x − 1

When x = −1, 4 x + 3 = 1 and 2 x − 1 = 3

4 x + 3 > 2 x − 1 is false so there is no solution
between −2 and − 1 , so we do not require this region.
3
When x = 0, 4 x + 3 = 3 and 2 x − 1 = 1
One of the solution regions is greater
Thus 4 x + 3 > 2 x − 1 so x > − 1 than − .
1
3 3
1
x < −2 or x>−
3

Algebra 5
 Exercise 1.1
1. Solve each of these equations algebraically.
a) 1 − 2 x = 3 b) x − 3 = x + 1

c) 5x − 2 = 2x d) 5 − 4 x = 4

e) 2 x − 1 = x + 2 f) x = 4 − 2x

g) 3x + 1 = 4 − 2 x h) 2 x − 6 = 3x + 1

i) x + 4 = 3x + 1 j) 1 − 3x = 5x − 3

k) 3 x − 4 = x + 2 l) 5 2 x − 3 = 4 x − 5

2. Solve each of these inequalities algebraically.

a) 2 x − 3 < x b) x − 1 ≥ 4

c) x + 3 ≥ 2x + 2 d) 2 x + 3 > x + 6

3. Solve each of these inequalities graphically.

a) x + 6 ≤ 3 x − 2 b) 3x − 2 < x + 4

c) 2x < 1 − x d) 5 ≤ 2 x − 1

e) 2 x − 1 < x + 3 f) 2 x + 1 ≥ 1 − 4 x

g) x + 2 < 2 x + 1 h) 3x − 1 ≤ x + 3

1.2 Sketching linear graphs of the form y = a|x| + b

We have sketched graphs of the form y = |ax + b| in section 1.1. Using the information given in P1
Chapter 3 on transformations, we can sketch graphs of the form y = a|x| + b.

Example 3
Sketch the following graphs.
a) y = |2x − 4| b) y = 2|x| − 4
a) y y = 2x − 4
4
2
First sketch the graph without the modulus.
2 2 4 x
–2
–4

Continued on the next page

▲

6 Sketching linear graphs of the form y = a|x| + b

 Pure 2 Pure 3

y
4 y = |2x – 4| Then reflect the negative y values in the x-axis.
2

–2 2 4 x

b)
y
4 y = |x|
First sketch the graph of |x|.
2

–4 –2 0 2 4 x

y
y = 2|x|
4
Then stretch the graph by a stretch factor of 2 in the direction
2 of the y-axis.

–2 0 2 x

y  0
Then translate the graph by the vector   .
y = 2|x| – 4  −4 
2

–2 0 2 x When y = 0, x = 2 or −2.
–2

–4

Exercise 1.2
1. Sketch the following.
a) i) y = |x + 1| ii) y = |x| + 1
b) i) y = |3x + 2| ii) y = 3|x| + 2
c) i) y = |2x − 2| ii) y = 2|x − 2|
d) i) y = | 12 x + 3| 1
ii) y = |x| + 3
2
e) i) y = |−x| ii) y = −|x|
f) i) y = |3 − x| ii) y = 3 − |x|

Algebra 7
 1.3 Division of polynomials
We can use long division to divide a polynomial by another polynomial.

Example 4
Divide x3 – 5x2 + x + 10 by (x – 2).

x2 – 3x − 5
x – 2) x – 5x2 + x + 10
3
x3 ÷ x = x2, so multiply (x – 2) by x2.
x3 – 2x2
–3x2 + x + 10 Subtract (x3 – 2x2 ) from (x3 – 5x2 ) and bring down (+ x + 10).
–3x2 + 6x
−5x + 10 –3x2 ÷ x = 3x, so multiply (x – 2) by –3x.
−5x + 10
Subtract (3x2 + 6x) from (3x2 + x) and bring down + 10.
We cannot continue the process because (−5x + 10) − (−5x + 10) = 0.
Thus (x3 – 5x2 + x + 10) ÷ (x – 2) = x2 – 3x − 5

The expression (x3 – 5x2 + x + 10) is called the dividend, (x – 2) is called

the divisor, and (x2 – 3x − 5) is called the quotient. When we subtract
(−5x + 10) from (−5x + 10) we are left with nothing, so we say there is
no remainder. Because there is no remainder, we can say that (x – 2) is a
factor of x3 – 5x2 + x + 10.

Example 5
Find the remainder when 4x3 – 7x – 1 is divided by (2x + 1).

2x2 – x − 3
As there is no term in x2 in the dividend, it is useful to write
2x + 1) 4x + 0x2 − 7x − 1
3
0x2 as part of the dividend.
4x3 + 2x2
–2x2 − 7x − 1 (4x3 ÷ 2x) = 2x2, so multiply (2x + 1) by 2x2.
–2x2 − x
−6x − 1 Subtract (4x3 + 2x2 ) from (4x3 + 0x2) and bring down −7x − 1.
−6x − 3
(−2x2 ÷ 2x) = −x, so multiply (2x + 1) by −x.
+2
Subtract (−2x2 – x) from (–2x2 − 7x) and bring down −1.

Subtract (−6x – 3) from (−6x − 1) to get a remainder of +2.

We cannot continue the process as 2 cannot be divided by (2x + 1).

Thus, when (4x3 – 7x – 1) is divided by (2x + 1), the remainder is 2.

8 Division of polynomials
 Pure 2 Pure 3

Looking at Example 5 we can write

(4x3 – 7x – 1) = (2x2 – x − 3)(2x + 1) + 2

In general:
f(x) = quotient × divisor + remainder

Exercise 1.3
1. Divide
a) x3 + 3x2 + 3x + 2 by (x + 2)
b) x3 – 2x2 + 6x + 9 by (x + 1)
c) x3 – 3x2 + 6x − 8 by (x – 2)
d) x3 + x2 − 3x − 2 by (x + 2)
e) 2x3 – 6x2 + 7x − 21 by (x – 3)
Hint: In part (g), use the same
f) 3x3 – 20x2 + 10x + 12 by (x – 6)
method as when dividing by a linear
g) 6x4 + 5x3 + 5x2 + 10x + 7 by (3x2 – 2x + 4). expression. State any remainder.

2. Find the remainder when

a) 6x3 + 28x2 − 7x + 10 is divided by (x + 5)
b) 2x3 + x2 + 5x − 4 is divided by (x − 1)
c) x3 + 2x2 − 17x − 2 is divided by (x – 3)
d) 2x3 + 3x2 − 4x + 5 is divided by (x2 + 2)
e) 4x3 – 5x + 4 is divided by (2x – 1)
f) 3x3 – x2 + 1 is divided by (x + 2).

3. Show that (2x + 1) is a factor of 2x3 – 3x2 + 2x + 2.

4. a) Show that (x − 1) is a factor of x3 – 6x2 + 11x − 6.

b) Hence factorise x3 – 6x2 + 11x − 6.

5. Show that when 4x3 – 6x2 + 5 is divided by (2x – 1) the remainder is 4.

6. Divide x3 + 1 by (x + 1).

7. Find the quotient and the remainder when x4 + 2x3 + 3x2 + 7

is divided by (x2 + x + 1).

8. Find the quotient and the remainder when 2x3 + 3x2 − 4x + 5 is divided
by (x + 2).

Algebra 9
 9. a) Show that (2x − 1) is a factor of 12x3 + 16x2 − 5x − 3.
b) Hence factorise 12x3 + 16x2 − 5x − 3.

10. The expression 2x3 – 5x2 − 16x + k has a remainder of −6 when

divided by (x – 4).
Find the value of k.

11. Find the quotient and the remainder when 2x4 − 8x3 − 3x2 + 7x − 7
is divided by (x2 − 3x − 5).

12. The polynomial x4 + x3 − 5x2 + ax − 4 is denoted by p(x). It is

given that p(x) is divisible by (x2 + 2x − 4).
Find the value of a.

1.4 The remainder theorem

You can find the remainder when a polynomial is divided by (ax – b) by using
the remainder theorem.
We know that if f(x) is divided by (x − a) then f(x) = quotient × (x − a) + remainder.
When x = a, f(a) = quotient × (a − a) + remainder = remainder.
Thus f(a) = remainder.

When a polynomial f(x) is divided by (x – a), the remainder is f(a).

When a polynomial f(x) is divided by (ax – b), the remainder is f b .()
a

Example 6
Find the remainder when 4x3 + x2 − 3x + 7 is divided by (x + 2).

f(x) = 4x3 + x2 − 3x + 7 Write the polynomial as a function.

f(−2) = 4(−2)3 + (−2)2 −3(−2) + 7 (x + 2) = 0 ⇒ x = −2, so calculate f(−2).

= −32 + 4 + 6 + 7 = −15
The remainder is −15.

10 The remainder theorem

 Pure 2 Pure 3

Example 7
When 16x4 − ax3 + 8x2 − 4x − 1 is divided by (2x −1), the remainder is 3.
Find the value of a.

f(x) = 16x4 − ax3 + 8x2 − 4x − 1 Write the polynomial as a function.

(2) (2) (2) (2) (2)

4 3 2
f 1 = 16 1 − a 1 + 8 1 − 4 1 − 1 1
(2x − 1) = 0 ⇒ x = , so calculate f
2
1
2
.()
=1− 1 a+2−2−1=3 Equate this to 3, since the remainder is 3.
8
− 1 a = 3 ⇒ a = −24
8

Exercise 1.4
1. Find the remainder when
a) 2x3 + 8x2 − x + 4 is divided by (x − 3)
b) 5x4 − 3x3 − 2x2 + x − 1 is divided by (x + 1)
c) x3 + 4x2 + 8x − 3 is divided by (2x + 1)
d) 3x3 − 2x2 − 5x − 7 is divided by (2 − x)
e) 9x3 – 8x + 3 is divided by (1 − x)
f) 243x4 − 27x3 + 6x + 4 is divided by (3x − 2).

2. When ax3 + 16x2 − 5x − 5 is divided by (2x – 1) the remainder is −2.

Find the value of a.

3. The polynomial 4x3 – 4x2 + ax + 1, where a is a constant, is denoted

by p(x). When p(x) is divided by (2x − 3) the remainder is 13.
Find the value of a.

4. The polynomial x3 + ax2 + bx + 1, where a and b are constants, is denoted

by p(x). When p(x) is divided by (x − 2) the remainder is 9 and when
p(x) is divided by (x + 3) the remainder is 19. Find the value of a and the
value of b.

5. When 5x3 + ax + b is divided by (x – 2), the remainder is equal to the

remainder obtained when the same expression is divided by (x + 2).

Algebra 11
 a) Explain why b can take any value.
b) Find the value of a.

6. The polynomial 2x4 + 3x2 − x + 2 is denoted by p(x). Show that

the remainder when p(x) is divided by (x + 2) is 8 times the remainder
when p(x) is divided by (x – 1).

7. The polynomial x3 + ax + b, where a and b are constants, is denoted

by p(x). When p(x) is divided by (x − 1) the remainder is 14 and
when p(x) is divided by (x − 4) the remainder is 56. Find the
values of a and b.

8. The polynomial x3 + ax2 + 2, where a is a constant, is denoted by p(x).

When p(x) is divided by (x + 1) the remainder is one more than when p(x)
is divided by (x + 2). Find the value of a.

9. When 6x2 + x + 7 is divided by (x – a), the remainder is equal to the

remainder obtained when the same expression is divided by (x + 2a),
where a ≠ 0. Find the value of a.

1.5 The factor theorem

We can deduce the factor theorem directly from the remainder theorem (section 1.4).

For any polynomial f(x), if f(a) = 0 then the remainder when f(x) is divided
by (x – a) is zero. Thus (x – a) is a factor of f(x).
For any polynomial f(x), if f ( ) = 0, then (ax – b) is a factor of f(x).
b
a

Example 8
The polynomial x3 − ax2 + 2x + 8, where a is a constant, is denoted by p(x).
It is given that (x – 2) is a factor of p(x).
a) Evaluate a.
b) When a has this value, factorise p(x) completely.
a) p(2) = 8 – 4a + 4 + 8 = 0
4a = 20 ⇒ a = 5
b) We can factorise x3 – 5x2 + 2x + 8 using either (i) long division
or (ii) testing other factors using the factor theorem.
Continued on the next page
▲

12 The factor theorem

 Pure 2 Pure 3

i) x2 – 3x − 4
x – 2 )x – 5x2 + 2x + 8
3
Put a = 5.
x3 – 2x2
– 3x2 + 2x + 8
– 3x2 + 6x
−4x + 8
You would expect there to be no
−4x + 8 remainder since x − 2 is a factor.

and x2 – 3x – 4 = (x – 4)(x + 1) Factorise the quotient.

So p(x) = (x – 2)(x – 4)(x + 1)

ii) f(+1) = 1 − 5 + 2 + 8 ≠ 0 (x − 1) is not a factor Try a value of x.

f(−1) = −1 − 5 − 2 + 8 = 0 (x + 1) is a factor
f(4) = 64 − 80 + 8 + 8 = 0 (x − 4) is a factor
So p(x) = (x – 2)(x – 4)(x + 1)
Note: Instead of performing this last trial we could
work out that the final factor is (x – 4) as we know
x × x × x = x3 and (−2) × (+1) × (−4) = +8.

Example 9
Solve x3 – 3x2 – 4x + 12 = 0.

Let f(x) = x3 – 3x2 – 4x + 12.

To solve, we must first factorise x3 – 3x2 – 4x + 12.
f(1) = 1 – 3 – 4 + 12 ≠ 0
so (x – 1) is not a factor. Trial any value of x that is a factor of 12.
f(2) = 8 – 12 – 8 + 12 = 0
so (x – 2) is a factor. Alternatively, at this stage you could also do a long
division to find the other two factors since you
f(−2) = −8 – 12 + 8 + 12 = 0 already know one factor.
so (x + 2) is a factor.
We can deduce that the third factor is (x – 3). 12 ÷ 2 ÷ −2 = −3 (and the coefficient of x3 is 1).
f(3) = 27 – 27 – 12 + 12
so (x – 3) is a factor.
Thus (x – 2)(x + 2)(x – 3) = 0
x=2 or x = −2 or x=3

Algebra 13
 Example 10
The polynomial ax3 + x2 + bx + 6, where a and b are constants, is denoted by p(x). It is given
that (2x – 1) is a factor of p(x) and that when p(x) is divided by (x − 1) the remainder is −4.
Find the values of a and b.
p(x) = ax3 + x2 + bx + 6
⎛ 1⎞
() a 1 b
p 1 = + + +6=0
2 8 4 2
p ⎜ ⎟ = 0 as (2x – 1) is a factor.
⎝2⎠

a + 2 + 4b + 48 = 0
Multiply each term by 8.
a + 4b = −50 (1)

p(1) = a + 1 + b + 6 = −4 p(1) = −4 as the remainder is −4.

a + b = −11 (2)
(1) – (2) ⇒ 3b = −39 Solve the two equations simultaneously.
b = −13
(2) ⇒ a − 13 = −11
a=2
a = 2 and b = −13

Exercise 1.5
1. Factorise the following as a product of three linear factors.
In each case, one of the factors has been given.
a) 2x3 – 5x2 – 4x + 3 One factor is (x – 3).
b) x3 – 6x2 + 11x − 6 One factor is (x – 2).
c) 5x + 14x + 7x − 2
3 2
One factor is (5x – 1).
d) 2x3 + 3x2 − 18x + 8 One factor is (x + 4).
e) x + x − 4x – 4
3 2
One factor is (x + 2).
f) 6x3 + 13x2 − 4 One factor is (3x + 2).

2. Solve the following equations.

a) 2x3 + 7x2 – 7x − 12 = 0 b) 2x3 − 5x2 – 14x + 8 = 0
c) x3 − 6x2 + 3x + 10 = 0 d) x3 + 3x2 – 6x − 8 = 0
e) 2x3 − 15x2 + 13x + 60 = 0 f) 3x3 − 2x2 – 7x − 2 = 0

3. Show that (x – 3) is a factor of x5 – 3x4 + x3 – 4x − 15.

4. Factorise x4 + x3 – 7x2 – x + 6 as a product of four linear factors.

14 The factor theorem

 Pure 2 Pure 3

5. (x – 2) is a factor of x3 − 3x2 + ax − 10. Evaluate the coefficient a.

6. a) Show that (2x – 5) is a factor of 4x3 − 20x2 + 19x + 15.

b) Hence factorise 4x3 − 20x2 + 19x + 15 as a product of
three linear factors.

7. The polynomial ax3 – 3x2 − 5ax − 9 is denoted by p(x) where a is a real number.
It is given that (x – a) is a factor of p(x). Find the possible values of a.

8. The polynomial 3x3 + 2x2 − bx + a, where a and b are constants, is denoted

by p(x). It is given that (x – 1) is a factor of p(x) and that when p(x) is
divided by (x + 1) the remainder is 10. Find the values of a and b.

9. The polynomial ax3 + bx2 − 5x + 3, where a and b are constants, is

denoted by p(x). It is given that (2x – 1) is a factor of p(x) and that
when p(x) is divided by (x − 1) the remainder is −3. Find the remainder
when p(x) is divided by (x + 3).

10. Factorise 2x4 + 5x3 – 5x − 2 as a product of four linear factors.

Summary exercise 1
1. Solve algebraically the equation | 5 − 2x | = 7. 9. Divide 2x4 – 9x3 + 13x2 − 15x + 9 by (x − 3).
2. Solve algebraically the equation 10. Find the quotient and the remainder when
| 3x − 4 | = | 5 − 2x |. x3 – 3x2 + 6x + 1 is divided by (x – 2).
3. Sketch the following graphs: Exam-Style Questions
a) y = 2|x| + 5 11. a) Show that (x − 4) is a factor of
b) y = 2 − |x|. x3 – 3x2 − 10x + 24.
b) Hence factorise x3 – 3x2 − 10x + 24.
4. Solve graphically the inequality
2 | x − 2 | < | x |. 12. The expression x3 + 3x2 + 6x + k has a
remainder of −3 when divided by (x + 1).
5. Solve graphically the inequality Find the value of k.
| 2 x − 1 | < | 3x − 4 |.
13. The polynomial ax4 + bx3 – 8x2 + 6 is denoted
Exam-Style Question by p(x). When p(x) is divided by (x2 – 1)
6. Solve the inequality | x + 3 | ≥ 2 | x − 3 |. the remainder is 2x + 1. Find the value of a
and the value of b.
7. Solve the inequality | x − 2 | ≤ 3 | x + 1 |.
14. The polynomial x4 + ax3 + bx2 – 16x − 12 is
Exam-Style Question denoted by p(x).
8. Solve the inequality 2 | x − a | > | 2x + a | (x + 1) and (x – 2) are factors of p(x).
where a is a constant and a > 0. a) Evaluate the coefficients a and b.
b) Hence factorise p(x) fully. Algebra 15
 15. The polynomial x4 + x3 – 22x2 − 16x + 96 is 18. The polynomial x3 − 15x2 + Ax + B, where A
denoted by p(x). and B are constants, is denoted by p(x). (x – 16)
a) Find the quotient when p(x) is divided is a factor of p(x). When p(x) is divided by (x − 2)
by x2 + x − 6. the remainder is −56.
b) Hence solve the equation p(x) = 0. a) Find the value of A and the value of B.
16. The polynomial 6x3 − 23x2 + ax + b is b) i) Find all 3 roots of the equation p(x) = 0.
denoted by p(x). When p(x) is divided by ii) Find the 4 real roots of the equation
(x + 1) the remainder is −21. When p(x) is p(x4) = 0.
divided by (x − 3) the remainder is 11. 19. i) Find the quotient and remainder when
a) Find the value of a and the value of b. x4 + 2x3 + x2 + 20x − 25
b) Hence factorise p(x) fully. is divided by (x2 + 2x − 5).
17. A polynomial is defined by ii) It is given that, when
p(x) = x3 + Ax2 + 49x – 36, where A is a x4 + 2x3 + x2 + px + q
constant. (x – 9) is a factor of p(x). is divided by (x2 + 2x − 5), there is no remainder.
a) Find the value of A. Find the values of the constants p and q.
b) i) Find all the roots of the equation iii) When p and q have these values, show
p(x) = 0. that there are exactly two real values of
ii) Find all the roots of the equation x satisfying the equation
p(x2) = 0. x4 + 2x3 + x2 + px + q = 0
and state what these values are. Give your
answer in the form a ± b .

16 Summary
The factor exercise
theorem 1
 Pure 2 Pure 3

Chapter summary
The modulus function
●● The modulus of a real number is the magnitude of that number.
●● The modulus function f(x) = x is defined as
x =x for x≥0
x = −x for x < 0

Sketching graphs of the modulus function

●● When sketching the graph of y = |f(x)| we reflect the section of the graph where y < 0 in the
x-axis.
●● When sketching the graph of y = f(| x |) we sketch the section of the graph where x > 0 and
then reflect this in the y-axis.

Division of polynomials
●● When dividing algebraic expressions, for example (4x3 – 7x − 3) ÷ (2x + 1) = (2x2 – x – 3),
you need to know the following terms:
●● (4x3 – 7x − 3) is called the dividend.
●● (2x + 1) is called the divisor.
●● (2x2 – x – 3) is called the quotient,
and there is no remainder.
●● (2x + 1) is a factor of (4x3 – 7x – 3).
●● f(x) = quotient × divisor + remainder

The remainder theorem

●● When a polynomial f(x) is divided by (x – a), the remainder is f(a).
●● When a polynomial f(x) is divided by (ax – b), the remainder is f b . ()
a

The factor theorem

●● For any polynomial f(x), if f(a) = 0 then the remainder when f(x) is divided
by (x – a) is zero. Thus (x – a) is a factor of f(x).
●● For any polynomial f(x), if f ( ) = 0, then (ax – b) is a factor of f(x).
b
a

Algebra 17
Complete
Pure Mathematics
2/3 for Cambridge International
AS & A Level
Jean Linsky
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Brian Western

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