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OLYMPIAD
MATHEMATICS

Pathfinder for

Vikash Tiwari | V. Seshan

EXTENSIVE PEDAGOGY



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Vikash Tiwari has been teaching students for Mathematical Olympiads (Pre RMO, RMO, INMO, and IMOTC)
and other examinations like KVPY and JEE for the last 20 years. He is a renowned resource person in the field
of Mathematics in India. He conducts Olympiad training camps for several organizations such as Kendriya
Vidyalaya Sangathan, Delhi Public School (DPS), etc. He has devoted himself to the service of mankind via the
medium of mathematics and has come up with this “first of its kind” book, which is an inventory of all
essential concepts required to ace the Mathematics Olympiad at various levels. The students have found his
methods of problem-solving and teaching to be both insightful and intriguing. He has been instrumental in
the success of several medal-winning students who have made our country proud in various International
Mathematics Olympiads.





Fulbright Teacher Awardee (USA-1970)
Presidential Awardee (1987)
Advisor to National Science Olympiad
Foundation (Since 1989)





Rotary (Int.) Awardee (1992)
PEE VEE National Awardee (2000)
Ramanujan Awardee (2008)

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OLYMPIAD
MATHEMATICS

Vikash Tiwari
V. Seshan

Tiwari
Seshan


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Cover Image: vlastas/Click Bestsellers. Shutterstock

V. Seshan is one of the key resource persons nominated by CBSE to provide Olympiad training across India.
He is a popular teacher and retd. Principal and Director of Bhartiya Vidya Bhavan, Baroda Centre. He is well
known for his unique ability in teaching Mathematics with utmost conceptual clarity. With teaching
experience spanning over 40 years, he has been instrumental in setting-up the Olympiad Centre in Tata
Institute of Fundamental Research, Mumbai. He has also been awarded with various medals, honors and
recognitions by prestigious universities and institutions from across the globe. These bear testament to his
immense contribution to the field of mathematics. Many of his students have taken part in both national
and international Mathematics Olympiad and have also won gold and silver medals to their credits. A few of
his key recognitions are listed below.

Pathfinder for O LYMPIAD MATHEMATICS

This book has been prepared in line with the requirements of national and international Olympiad examinations. The
questions are carefully chosen to suit the needs of Olympiad aspirants and to provide highest level of clarity for
mathematical concepts. This book also provides deep insights about the origin of important formulae and equations
by eminent Mathematicians. The exercises are designed and graded from simple to difficult level to enable the
students’ to build, check and challenge their understanding.

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Pathfinder for Olympiad

MATHEMATICS

Vikash Tiwari
V. Seshan

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Copyright © 2017 Pearson India Education Services Pvt. Ltd
Published by Pearson India Education Services Pvt. Ltd, CIN: U72200TN2005PTC057128,

formerly known as TutorVista Global Pvt. Ltd, licensee of Pearson Education in South Asia.
No part of this eBook may be used or reproduced in any manner whatsoever without
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Brief Contents
Preface ������������������������������������������������������������������������������������������������������������������������������������������������  xi
Acknowledgements ����������������������������������������������������������������������������������������������������������������������������� xii
About the Authors������������������������������������������������������������������������������������������������������������������������������� xii

1. Polynomials ���������������������������������������������������������������������������������������������������������� 1.1
2. Inequalities ���������������������������������������������������������������������������������������������������������� 2.1
3.  Mathematical Induction ������������������������������������������������������������������������������������� 3.1
4.  Recurrence Relation ������������������������������������������������������������������������������������������� 4.1
5.  Functional Equations������������������������������������������������������������������������������������������� 5.1
6.  Number Theory���������������������������������������������������������������������������������������������������� 6.1

7. Combinatorics ����������������������������������������������������������������������������������������������������� 7.1
8. Geometry ������������������������������������������������������������������������������������������������������������� 8.1

Answer Keys ���������������������������������������������������������������������������������������������������������������������������������� AK.1
Appendix ���������������������������������������������������������������������������������������������������������������������������������������� AP.1
Logarithms Table ���������������������������������������������������������������������������������������������������������������������������� LT.1
Photo Credits ��������������������������������������������������������������������������������������������������������������������������������� PC.1

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Contents
Prefacexi
Acknowledgementsxii
About the Authors

Chapter 1  Polynomials
Polynomial Functions
Division in Polynomials
Remainder Theorem and Factor Theorem
Fundamental Theorem of Algebra

Polynomial Equations
Vieta’s Relations
Symmetric Functions
Common Roots of Polynomial Equations
Irreducibility of Polynomials
Chapter 2  Inequalities
Basic Rules
Weirstras’s Inequality
Modulus Inequalities
Sum of Squares (SOS)
Arithmetic Mean ≥ Geometric Mean ≥ Harmonic Mean
Weighted Means
Power Mean Inequality
Rearrangement Inequality
Chebyshev’s Inequality
Cauchy–Schwarz Inequality 
Hölders Inequality
Some Geometrical Inequalities
Jensen’s Inequality

A01_Olympiad Mathematics_FM.indd 7

xii

1.1
1.1
1.2
1.3
1.3
1.7

1.9
1.16
1.22
1.24

2.1
2.1
2.3
2.4
2.6
2.11
2.22
2.24
2.26
2.27
2.29
2.33
2.35
2.36

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viii  Contents

Chapter 3   Mathematical Induction

3.1

Introduction3.1

First (or Weak) Principle of Mathematical Induction 
3.2
Second (or Strong) Principle of Mathematical Induction
3.13

Chapter 4   Recurrence Relation

4.1

Introduction4.1
Classification4.1
First Order Linear Recurrence Relation
4.3
First Order Non-linear
Linear Homogeneous Recurrence Relation
with Constant Coefficient of Order ‘2’
General Form of Linear Homogeneous Recurrence Relation
with Constant Coefficients
General Method for Non-Homogeneous
Linear Equation

Chapter 5   Functional Equations

4.7
4.12
4.14
4.15

5.1


Function5.1
Functional Equation
5.3

Chapter 6   Number Theory

6.1

Divisibility of Integers
6.1
Euclids Division Lemma
6.4
Greatest Common Divisor (GCD)
6.4
Primes6.8
Fundamental Theorem of Arithmetic
6.13
Number of Positive Divisors of a Composite Number
6.13
6.21
Modular Arithematic 
Complete Residue System (Modulo n)6.27
Some Important Function/Theorem
6.28

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Contents  ix

Scales of Notation
Greatest Integer Function
Diophantine Equations

6.35
6.39
6.45

Chapter 7  Combinatorics

7.1

Definition of Factorial
7.1
Basic Counting Principles
7.2
Combinations7.13
The Bijection Principle
7.33
7.34
Combinations with Repetitions Allowed
Definition of Permutation (Arrangements)
7.39
Introduction to Circular Permutation
7.57
Division and Distribution of Non-identical Items in Fixed Size
7.64
Number of Integral Solutions 

7.69
Binomial, Multinomial and Generating Function
7.72
Application of Recurrence Relations
7.78
Principle of Inclusion and Exclusion (PIE)
7.81
Derangement7.93
Classical Occupancy Problems
7.98
Dirichlet’s (Or Pigeon Hole) Principle (PHP)
7.104

Chapter 8  Geometry

8.1

Angle8.1
Congruent Triangles
8.7
Triangle Inequality
8.16
Ratio and Proportion Theorem (or Area Lemma)
8.22
Mid-point Theorem
8.26
Basic Proportionality Theorem (Thales’ Theorem)
8.29
Similar Triangles
8.38

Baudhayana (Pythagoras) Theorem
8.44
Quadrilaterals8.55

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x  Contents

Concurrency and Collinearity
8.66
Circles8.90
Quadrilaterals (Cyclic and Tangential)
8.110
Application of Trigonometry in Geometry
8.127
Construction of Triangles
8.169

Answer Keys 

AK.1

Appendix  

AP.1

Logarithms Table 


LT.1

Photo Credits

PC.1

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Preface
“For another hundred years, School will teach children ‘to do’ rather than ‘to think’”
observed Bertrand Russell. This statement is still seen to be true without being even
remotely contradicted.
NCF 2005 (National Curriculum Framework) provides a vision for perspective
planning ofschool education in scholastic and non-scholastic domains. It also emphasizes on ‘mathematisation’ of the child’s thought and processes by recognizing
­mathematics as an integral part of development of the human potential. The higher aim
of teaching mathematics is to enhance the ability to visualize, logically understand,
build arguments, prove statements and in a sense, handle abstraction. For motivated
and talented students, there is a need to widen the horizon as these students love challenges and always look beyond the curriculum at school.
Hence, we created this book to cater to the needs of these students. With numerous
problems designed to develop thinking and reasoning, the book contains statements,
definitions, postulates, formulae, theorems, axioms, and propositions, which normally
do not appear in school textbooks. These are spelt out and interpreted to improve the
student’s conceptual knowledge.
The book also presents ‘non-routine problems’ and detailed, step-by-step solutions
to these problems to enable the reader to acquire a better understanding of the concepts
as well as to develop analytical and reasoning (logical) abilities. Thus, ­readers get the

‘feel’ of problem-solving as an activity which, in turn, reveals the innate pleasure of
successfully solving a challenging problem. This ‘pleasure’ is permanent and helps
to build-in them a positive attitude towards the subject. Developing ability for critical analysis and problem solving is an essential requirement if one wants to become
­successful in life.
No one has yet discovered a way of learning mathematics better than, by ­solving
problems in the subject. This book helps students to face competitive examinations
such as the Olympiads (RMO, INMO, IMO), KVPY and IIT-JEE confidently w
­ ithout
being befuddled by the intricacies of the subject. It has been designed to enable s­ tudents
and all lovers of mathematics to master the subject at their own pace.
We have made efforts to provide solutions along with the problems in an error-free
and unambiguous manner as far as possible. However, if any error is detected by the
reader, it may please be brought to our notice, so that we may make necessary corrections in the future editions of the book. We look forward to your suggestions and shall
be grateful for them.
Lastly, we share the observation made by Pundit Jawaharlal Nehru: “Giving
­opportunity to potential creativity is a matter of life and death for an enlightened
­society because the contributions of a few creative individuals are the mankind’s
­ultimate capital asset.”
We wish best of luck at all times to all those using this book.
Vikash Tiwari
V. Seshan

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Acknowledgements
First and foremost, we thank the Pearson group for motivating us and rendering all
possible assistance in bringing out this book in its present form. We are grateful to the

Pearson group for having consented to publish the book on our behalf.
We would also like to thank Ajai Lakheena, who has been instrumental in g­ iving
this book its present shape. He has made invaluable contributions to “Geometry”
chapter of this book.This section would not have been as effective without his efforts.
We also express our gratitude to Bhupinder Singh Tomar and Abir Bhowmick,
who have helped us with their discerning inputs and suggestions for making this book
­error-free. We are indebted to R.K. Thakur for his inputs and constant encouragement
to write this book.
This book is dedicated to my wife Priyanka for her kindness, devotion and
endless-support in managing household chores and to my two adorable daughters
Tanya and Manya who sacrificed their vacation umpteen times for my (our) sake.
Vikash Tiwari

About the Authors
Vikash Tiwari has been teaching students for Mathematical Olympiads (Pre RMO,
RMO, INMO and IMOTC) and other examinations like KVPY and JEE for the last
20 years. He is a renowned figure in the field of Mathematics across the geography of
the country. His students have always founds his methods of teaching insightful and his
approach to problem solving very intriguing. He has guided several of the medal winning
students that have done India proud at the International Mathematical Olympiad over
the years. He use to conduct Olympiad training camps for several organizations such
as Kendriya Vidyalaya Sangathan, Delhi Public School (DPS) etc. He has immersed
himself into the service of mankind via the medium of Mathematics for the past couple
of decades and has come with this first book of his which is basically an inventory of all
the concepts required to ace the Mathematics Olympiad at various levels.
V. Seshan is the key resource person shortlisted by CBSE to provide Olympiad training
across India. He is a popular teacher and retd. Principal & Director of Bhartiya Vidya
Bhavan, Baroda Centre. He is well known for his unique ability in teaching Mathematics
with conceptual clarity. With teaching experience spanning over 40 years, he has been instrumental in setting-up the Olympiad Centre in Tata Institute of Fundamental Research,
Mumbai. He has also been awarded with various medals, honors and recognitions by

prestigious universities and institutions from across the globe. These bear testament to
his immense contribution to the field of Mathematics. Many of his students have taken
part in both National & International Mathematics Olympiad and have also won gold
and silver medals to their credits. A few of his recognition are listed below.
••
••
••
••
••
••

A01_Olympiad Mathematics_FM.indd 12

Fulbright Teacher Awardee (USA-1970)
Presidential Awardee (1987)
Advisor to National Science Olympiad Foundation (Since 1989)
Rotary (Int) Awardee (1992)
PEE VEE National Awardee (2000)
Ramanujan Awardee (2008)

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Chapter

1

Niccolò Fontana Tartaglia
(1499/1500–13 Dec 1557) Tartaglia was an Italian mathematician.The name “Tartaglia” is actually a nickname meaning “stammerer”,
a reference to his injury-induced speech impediment. He was largely self-taught, and was the first person to translate Euclid’s

Elements into a modern European language. He is best remembered for his contributions to algebra, namely his discovery of a
formula for the solutions to a cubic equation. Such a formula was also found by Gerolamo Cardano at roughly the same time, and
the modern formula is known as the Cardano-Tartaglia formula. Cardano also found a solution to the general quartic equation.
Évariste Galois
(25 Oct 1811–31 May 1832) Galois was a very gifted young French mathematician, and his story is one of the most tragic in the history of mathematics. He was killed at the age of 20 in a duel that is still veiled in mystery. Before that, he made huge contributions
to abstract algebra. He helped to found group theory as we know it today, and he was the first to use the term “group”. Perhaps
most importantly, he proved that it is impossible to solve a fifth-degree polynomial (or a polynomial of any higher degree) using
radicals by studying permutation groups associated to polynomials. This area of algebra is still important today, and it is known as
Galois theory in his honor.
Niels Henrik Abel
(5 Aug 1802–6 Apr 1829) Abel was a Norwegian mathematician who, like Galois, did seminal work in algebra before dying at a
very young age. Strangely enough, he proved similar results regarding the insolvability of the quintic independently from Galois. In
honor of his work in group theory, abelian groups are named after him. The Abel Prize in mathematics, sometimes thought of as
the “Nobel Prize in Mathematics,” is also named for him.

Joseph-Louis Lagrange
(25 Jan 1736–10 Apr 1813) Despite his French-sounding name, Lagrange was an Italian mathematician. Like many of the great
mathematicians of his time, he made contributions to many different areas of mathematics. In particular, he did some early work
in abstract algebra.

Polynomials
1.1 Polynomial FuncTions
Any function, f (x) = anxn + an−1xn−1 + … + a1x + a0, is a polynomial function in ‘x’ where
ai(i = 0, 1, 2, 3, …, n) is a constant which belongs to the set of real numbers and sometimes to the set of complex numbers, and the indices, n, n − 1, …, 1 are natural numbers.
If an ≠ 0, then we can say that f (x) is a polynomial of degree n. an is called leading coefficient of the polynomial. If an = 1, then polynomial is called monic polynomial. Here,
if n = 0, then f (x) = a0 is a constant polynomial. Its degree is 0, if a0 ≠ 0. If a0 = 0, the
polynomial is called zero polynomial. Its degree is defined as −∞ to preserve the first
two properties listed below. Some people prefer not to defined degree of zero polynomial. The domain and range of the function are the set of real numbers and complex
numbers, respectively. Sometimes, we take the domain also to be complex numbers. If z
is a complex number and f (z) = 0, then z is called ‘a zero of the polynomial’.

If f(x) is a polynomial of degree n and g(x) is a polynomial of degree m then
1. f(x) ± g(x) is polynomial of degree ≤ max {n, m}
2. f(x) ⋅ g(x) is polynomial of degree m + n
3. f(g(x)) is polynomial of degree m ⋅ n, where g(x) is a non-constant polynomial.
Illustrations
1. x4 − x3 + x2 − 2x + 1 is a polynomial of degree 4 and 1 is a zero of the polynomial as
14 − 13 + 12 − 2 × 1 + 1 = 0.
3
2
2. x − ix + ix + 1 = 0 is a polynomial of degree 3 and i is a zero of his polynomial
as i3 − i ⋅ i2 + i ⋅ i + 1 = −i + i − 1 + 1 = 0.
3. x 2 − ( 3 − 2 ) x − 6 is a polynomial of degree 2 and

3 is a zero of this poly-

− ( 3 − 2 ) 3 − 6 = 3 − 3 + 6 − 6 = 0.
nomial as (
Note: The above-mentioned definition and examples refer to polynomial functions in
one variable. Similarly, polynomials in 2, 3, …, n variables can be defined. The domain
3 )2

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1.2  Chapter 1
for polynomial in n variables being the set of (ordered) n tuples of complex numbers
and the range is the set of complex numbers.
Illustration  f(x, y, z) = x2 − xy + z + 5 is a polynomial in x, y, z of degree 2 as both x2

and xy have degree 2 each.
Note:  In a polynomial in n variables, say, x1, x2,…, xn, a general term is x1k1 ⋅ x2k2 xnkn .
Degree of the term is k1 + k2 + … + kn where ki ∈0, i = 1, 2, …, n. The degree of a
polynomial in n variables is the maximum of the degrees of its terms.

1.2  Division in Polynomials
If P(x) and f(x) (f(x) ≡/ 0) are any two polynomials, then we can find unique polynomials Q(x) and R(x), such that P(x) = f(x) × Q(x) + R(x) where the degree of R(x) < degree
of f(x) or R(x) ≡ 0. Q(x) is called the quotient and R(x), the remainder.
In particular, if P(x) is a polynomial with complex coefficients, and a is a complex
number, then there exists a polynomial Q(x) of degree 1 less than P(x) and a complex
number R, such that P(x) = (x − a)Q(x) + R.
Illustration  x5 = (x − a)(x4 + ax3 + a2x2 + a3x + a4) + a5.
Here,P(x) = x5, Q(x) = x4 + ax3 + a2x2 + a3x + a4,
and
R = a5.
9

25

Example 1  What is the remainder when x + x + x

+ x49 + x81 is divided by x3 − x.

Solution:  We have,

x + x9 + x25 + x49 + x81 = x(1 + x8 + x24 + x48 + x80)
= x[(x80 − 1) + (x48 − 1) + (x24 − 1) + (x8 − 1) + 5]
= x(x80 − 1) + x (x48 − 1) + x(x24 − 1) + x(x8 − 1) + 5x
Now, x3 − x = x(x2 − 1) and all, but the last term 5x is divisible by x(x2 − 1). Thus, the
remainder is 5x.

Example 2  Prove that the polynomial x 9999 + x 8888 + x 7777 +

by x9 + x8 + x 7 +

+ x1111 + 1 is divisible

+ x + 1.

Solution:  Let,

 P = x 9999 + x8888 + x 7777 +
Q = x 9 + x8 + x 7 +
P −Q =

x 9 ( x 9990

− 1) +

+ x1111 + 1

+ x +1

x8 ( x8880

− 1) + x 7 ( x 7770 − 1) +

= x 9 [( x10 )999 − 1] + x8 [( x10 )888 − 1] + x 7 [( x10 )777 − 1] +

+ x( x1110 − 1)
+ x[( x10 )111 − 1] (1)


But, (x10)n - 1 is divisible by x10 - 1 for all n ≥ 1.
∴  RHS of Eq. (1) divisible by x10 - 1.
∴ 
P - Q is divisible by x10 - 1
As x9 + x8 + … + x + 1 | (x10 - 1)
⇒ x9 + x8 + x7 + … + x + 1 | (P - Q)
⇒ x9 + x8 + x7 + … + x + 1 | P

M01_Polynomials_C01.indd 2

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Polynomials  1.3

1.3  Remainder Theorem and Factor Theorem
1.3.1  Remainder Theorem
If a polynomial f (x) is divided by (x − a), then the remainder is equal to f (a).
Proof:
f (x) = (x − a)Q(x) + R
and so,
f (a) = (a − a)Q(a) + R = R.
If R = 0, then f (x) = (x − a)Q(x) and hence, (x − a) is a factor of f (x).
Further, f (a) = 0, and thus, a is a zero of the polynomial f (x). This leads to the factor theorem.

1.3.2  Factor Theorem
(x − a) is a factor of polynomial f (x), if and only if, f (a) = 0.
Example 3  If f (x) is a polynomial with integral coefficients and, suppose that f (1) and


f (2) both are odd, then prove that there exists no integer n for which f (n) = 0.

Solution:  Let us assume the contrary. So, f (n) = 0 for some integer n.

Then, (x − n) divides f (x).
Therefore, f (x) = (x − n)g(x)
where g(x) is again a polynomial with integral coefficients.
Now, f (l) = (1 − n) g(1) and f (2) = (2 − n) g(2) are odd numbers but one of (1 − n)
and (2 − n) should be even as they are consecutive integers.
Thus one of f (l) and f (2) should be even, which is a contradiction. Hence, the result.

Aliter: See the Example (41) on page 6.24 in Number Theory chapter.
Example 4  If f is a polynomial with integer coefficients such that there exists four dis-

tinct integer a1, a2, a3 and a4 with f (a1) = f (a2) = f (a3) = f (a4) = 1991, show that there
exists no integer b, such that f (b) = 1993.

Solution:  Suppose, there exists an integer b, such that f (b) = 1993, let g(x) = f (x) −
1991.
Now, g is a polynomial with integer coefficients and g (a1) = 0 for i = 1, 2, 3, 4.
Thus (x − a1)(x − a2)(x − a3) and (x − a4) are all factors of g(x).

So, g(x) = (x − a1)(x − a2)(x − a3)(x − a4) × h(x)
where h(x) is polynomial with integer coefficients.
g (b) = f (b) − 1991
= 1993 − 1991 = 2 (by our choice of b)
But, g(b) = (b − a1)(b − a2)(b − a3)(b − a4) h(b) = 2
Thus, (b − a1)(b − a2)(b − a3)(b − a4) are all divisors of 2 and are distinct.
∴ (b − a1)(b − a2)(b − a3)(b − a4) are 1, −1, 2, −2 in some order, and h(b) is an
integer.

∴  g(b) = 4 . h(b) ≠ 2.
Hence, such b does not exist.

1.4  Fundamental Theorem of Algebra
Every polynomial function of degree ≥ 1 has at least one zero in the complex numbers.
In other words, if we have

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1.4  Chapter 1
f ( x ) = an x n + an −1 x n −1 +

+ a1 x + a0

with n ≥ 1, then there exists atleast one h ∈, such that
an hn + an −1 hn −1 +

+ a1h + a0 = 0.

From this, it is easy to deduce that a polynomial function of degree ‘n’ has exactly n
zeroes.

i.e., f(x) = a(x - r1)(x - r2)…(x - rn)
Notes:

1. Some of the zeroes of a polynomial may repeat.
2. If a root a is repeated m times, then m is called multiplicity of the root ‘a’ or a is

called m fold root.
3. The real numbers of the form 3,
‘quadratic surds’. In general,

5,

12,

27, …,

5 + 3 , etc. are called,

a , b , and a + b , etc. are quadratic surds, if a,

b are not perfect squares. In a polynomial with integral coefficients (or rational
coefficients), if one of the zeroes is a quadratic surd, then it has the conjugate of
the quadratic surd also as a zero.
Illustration  f (x) = x2 + 2x + 1 = (x + 1)2 and the zeroes of f (x) are −1 and −1. Here,
it can be said that f (x) has a zero −1 with multiplicity two.
Similarly, f (x) = (x + 2)3(x − 1) has zeroes −2, −2, −2, 1, i.e., the zeroes of f (x) are
−2 with multiplicity 3 and 1.
Example 5  Find the polynomial function of lowest degree with integral coefficients

with 5 as one of its zeroes.
Solution:  Since the order of the surd 5 is 2, you may expect that the polynomial of

the lowest degree to be a polynomial of degree 2.
Let,

P(x) = ax2 + bx + c; a, b, c ∈

P ( 5 ) = 5a + 5b + c = 0 ⇒ (5a + c) + 5b = 0

But,
So,

5 is irrational.
5a + c = 0 and b = 0
⇒  c = −5a  and  b = 0.

So, the required polynomial function is P(x) = ax2 - 5a, a ∈ » \ {0}
You can find the other zero of this polynomial to be − 5.
Aliter: You know that any polynomial function having, say, n zeroes α1, α2, …, αn
can be written as P(x) = (x − α1)(x − α2) … (x − αn) and clearly, this function is of nth
degree. Here, the coefficients may be rational, real or complex depending upon the
zeroes α1, α2, …, αn.
If the zero of a polynomial is 5, then P0(x) = (x - 5) or a(x − 5).
But, we want a polynomial with rational coefficients.
So, here we multiply (x − 5) by the conjugate of x − 5 , i.e., x +
get the polynomial P(x) = (x −

M01_Polynomials_C01.indd 4

5) (x +

5. Thus, we

5), where the other zero of P(x) is − 5.

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Polynomials  1.5

Now, P1(x) = x2 − 5, with coefficient of x2 = 1, x = 0 and constant term −5, and all
these coefficients are rational numbers.
Now, we can write the required polynomial as P(x) = ax2 − 5a where a is a non-zero
integer.
Example 6  Obtain a polynomial of lowest degree with integral coefficient, whose one

of the zeroes is 5 + 2 .
Solution:  Let, P (x ) = x − ( 5 + 2 ) = [( x − 5 ) − 2 ].

Now, following the method used in the previous example, using the conjugate, we
get:
P1(x) = [( x − 5 ) − 2 ][( x − 5 ) + 2 ]


= (x2 − 2 5x + 5) − 2



= (x2 + 3 − 2 5x)

P2(x) = [( x 2 + 3) − 2 5 x ][( x 2 + 3) + 2 5 x ]


= (x2 + 3)2 − 20x2




= x4 + 6x2 + 9 − 20x2


= x4 − 14x2 + 9
P(x) = ax4- 14ax2 + 9a, where a ∈ , a ≠ 0.
The other zeroes of this polynomial are

5 − 2 , − 5 + 2 , − 5 − 2.

1.4.1  Identity Theorem
A polynomials f(x) of degree n is identically zero if it vanishes for atleast n + 1 distinct
values of ‘x’.
Proof: Let a1, a2, … an be n distinct values of x at which f(x) becomes zero.
Then we have


f(x) = a(x - x1)(x - x2)…(x - xn)

Let an+1 be the n+1th value of x at which f(x) vanishes. Then
f(an+1) = a(an+1 - a1)(an+1 - a2)…(an+1 - an) = 0
As an+1 is different from a1, a2 … an none of the number an+1 - ai vanishes for i =
1, 2, 3, … n. Hence a = 0 ⇒ f(x) ≡ 0.
Using above result we can say that,
If two polynomials f(x) and g(x) of degree m, n respectively with m ≤ n have equal
values at n + 1 distinct values of x, then they must be equal.
Proof: Let P(x) = f(x) - g(x), now degree of P(x) is at most ‘n’ and it vanishes for at
least n + 1 distinct values of x ⇒ P(x) ≡ 0 ⇒ f(x) ≡ g(x).
Corollary: The only periodic polynomial function is constant function.
i.e., if f(x) is polynomials with f (x + T) = f(x) ∀ x ∈ for some constant T then f(x) =
constant = c (say)

Proof: Let f(0) = c
⇒ f(0) = f(T) = f(2T) = … = c
⇒ Polynomial f(x) and constant polynomial g(x) = c take same values at an infinite
number of points. Hence they must be identical.

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1.6  Chapter 1
Example 7  Let P(x) be a polynomial such that x ⋅ P(x - 1) = (x - 4) P(x) ∀ x ∈.
Find all such P(x).
Solution:  Put x = 0, 0 = -4 P(0)


⇒ P(0) = 0
Put x = 1, 1 ⋅ P(0) = -3 P(1)

⇒ P(1) = 0
Put x = 2, 2 ⋅ P(1) = -2 P(2)

⇒ P(2) = 0
Put x = 3, 3 ⋅ P(2) = -P(3)

⇒ P(3) = 0
Let us assume P(x) = x(x - 1) (x - 2) (x - 3) Q(x), where Q(x) is some polynomial.
Now using given relation we have
x( x − 1)( x − 2)( x − 3)( x − 4)Q( x − 1) = x( x − 1)( x − 2)( x − 3)( x − 4)Q( x )
⇒ Q( x − 1) = Q( x ) ∀x ∈ » − {0,1, 2, 3, 4}

⇒ Q( x − 1) = Q( x ) ∀x ∈ » (From identity theoorem)
⇒ Q( x ) is periodic
⇒ Q( x ) = c
⇒ P ( x ) = cx( x − 1)( x − 2)( x − 3)

Example 8  Let P(x) be a monic cubic equation such that P(1) = 1, P(2) = 2, P(3) = 3,

then find P(4).
Solution:  as P(x) is a monic, coefficient of highest degree will be ‘1’.

Let Q(x) = P(x) - x, where Q(x) is also monic cubic polynomial.

Q(1) = P (1) − 1 = 0; Q( 2) = P ( 2) − 2 = 0; Q(3) = P (3) − 3 = 0
⇒ Q( x ) = ( x − 1)( x − 2)( x − 3)
⇒ P ( x ) = Q( x ) + x = ( x − 1)( x − 2)( x − 3) + x
⇒ P ( x ) = ( 4 − 1)( 4 − 2)( 4 − 3) + 4 = 10

Build-up Your Understanding 1
1. Find a fourth degree equation with rational coefficients, one of whose roots is,
3 + 7.
2. Find a polynomial equation of the lowest degree with rational coefficients whose
one root is 3 2 + 3 3 4 .
3. Form the equation of the lowest degree with rational coefficients which has
2 + 3 and 3 + 2 as two of its roots.
4. Show that (x – 1)2 is a factor of xn – nx + n – 1.
5.If a, b, c, d, e are all zeroes of the polynomial (6x5 + 5x4 + 4x3 + 3x2 + 2x + 1), find
the value of (1 + a) (1 + b) (1 + c) (1 + d) (1 + e).
6. If 1, a1, a2, …, an−1 be the roots of the equation xn - 1 = 0, n ∈, n ≥ 2 show that
n = (1 − a1) (l − a2)(1 − a3) … (1 − an-1).
7.If α, β, γ, δ be the roots of the equation x4 + px3 + qx2 + rx + s = 0, show that


(1 + α2) (1 + β2) (1 + γ 2) (1 + δ 2) = (1 – q + s)2 + (p – r)2.

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Polynomials  1.7

8. If f(x) = x4 + ax3 + bx2 + cx + d is a polynomial such that f(1) =10, f(2) = 20, f(3)
= 30, find the value of f (12) + f ( −8) . 
[CMO, 1984]
10
9.The polynomial x2k + 1 + (x + 1)2k is not divisible by x2 + x + 1. Find the value
of k ∈ .
10. Find all polynomials P(x) with real coefficients such that
(x - 8)P(2x) = 8(x - 1)P(x).
3

11. Let (x - 1) divides (p(x) + 1) and (x + 1)3 divides (p(x)-1). Find the polynomial
p(x) of degree 5.

1.5  Polynomial Equations
Let, P(x) = anxn + an−1 xn−1 + … + a1x + a0; an ≠ 0, n ≥ 1 be a polynomial function.
Then, P(x) = an x n + an −1 x n −1 +

+ a1 x + a0 = 0 is called a polynomial equation in

x of degree n. Thus,

1. Every polynomial equation of degree n has n roots counting repetition.
2.If anxn + an−1xn−1 + … + a1x + a0 = 0(1)
an ≠ 0 and ai, (i = 0, 1, 2, 3, …, n) are all real numbers and if, α + iβ is a zero of (1),
then α - iβ is also a root. For real polynomial, complex roots occur in conjugate pairs.
However, if the coefficients of Eq. (1) are complex numbers, it is not necessary that
the roots occur in conjugate pairs.
Example 9  Form a polynomial equation of the lowest degree with 3 + 2i and 2 + 3i as

two of its roots, with rational coefficients.
Solution:  Since, 3 + 2i and 2 + 3i are roots of polynomial equation with rational coef-

ficients, 3 − 2i and 2 - 3i are also the roots of the polynomial equation. Thus, we have
identified four roots so that there are 2 pairs of roots and their conjugates. So, the lowest degree of the polynomial equation should be 4. The polynomial equation should be

P(x) = a [x − (3 − 2i)][x − (3 + 2i)][x − (2 + 3i)] [x − (2 - 3i)] = 0
⇒  a [(x − 3)2 + 4][(x − 2)2 + 9] = 0
⇒  a ((x − 3)2(x − 2)2 + 9(x − 3)2 + 4(x − 2)2 + 36) = 0
⇒  a ((x2 − 5x + 6)2 + 9(x2 − 6x + 9) + 4(x2 − 4x + 4) + 36) = 0
⇒  a (x4 − 10x3 + 50x2 − 130x + 169) = 0,  a ∈\{0}

1.5.1  Rational Root Theorem
An important theorem regarding the rational roots of polynomial equations:
p
If the rational number , where p, q ∈ », q ≠ 0, gcd(p, q) = 1, i.e., p and q are
q
relatively prime, is a root of the equation
anxn + an−1xn−1 + … + a1x + a0 = 0
where a0, a1, a2,…, an are integers and an ≠ 0, then p is a divisor of a0 and q that of an.
Proof: Since


M01_Polynomials_C01.indd 7

p
is a root, we have
q

8/11/2017 1:36:21 PM


1.8  Chapter 1
n

 p
 p
an   + an −1  
q
q

n −1

+

⇒ an p n + an −1qp n −1 +

+ a1

p
+ a0 = 0
q


+ a1q n −1 p + a0 q n = 0 

⇒ an −1 p n −1 + an − 2 p n − 2 q +

+ a1q n − 2 p + a0 q n −1 = −

(1)

an p n
(2)
q

Since the coefficients an−1, an−2,…, a0 and p, q are all integers, hence the left-hand side
is an integer, so the right- hand side is also an integer. But, p and q are relatively prime
to each other, therefore q should divide an.
Again,
n
n −1
  an p + an −1qp +

+ a1q n −1 p = a0 q n

⇒ an p n −1 + an −1qp n − 2 +
⇒

+ a1q n −1 =

p | a0

a0 q n

p 

(3)

As a consequence of the above theorem, we have the following corollary.

1.5.2  Corollary (Integer Root Theorem)
Every rational root of x n + an −1 x n −1 + + a0 ;0 ≤ i ≤ n − 1 is an integer, where ai(i = 0,
1, 2, …, n − 1) is an integer, and each of these roots is a divisor of a0.
4

3

2

Example 10  Find the roots of the equation x + x − 19x − 49x − 30, given that the

roots are all rational numbers.
Solution:  Since all the roots are rational by the above corollary, they are the divisors

of -30.
The divisors of -30 are ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30.
By applying the remainder theorem, we find that -1, -2, -3, and 5 are the roots.
Hence, the roots are -1, -2, -3 and +5.
3

2

Example 11  Find the rational roots of 2x - 3x - 11x + 6 = 0.
Solution:  Let the roots be of the form


p
,  where (p, q)= 1 and q > 0.
q

Then, since q | 2, q must be 1 or 2
and p | 6  ⇒  p = ±1, ±2, ±3, ±6
By applying the remainder theorem,
1
 −2 
3
f   = f   = f   = 0.
2
 1 
1
(Corresponding to q = 2 and p = 1; q = 1, p = -2; q = 1, p = 3, respectively.)
So, the three roots of the equation are

M01_Polynomials_C01.indd 8

1
, -2, and 3.
2

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Polynomials
3


1.9

2

Example 12 Solve: x − 3x + 5x − 15 = 0.
3

2

Solution: x − 3x + 5x − 15 = 0

⇒

(x2 + 5)(x − 3) = 0

⇒ x = ± 5 i, 3.

5 i , − 5 i.

So the solution are 3,

1000

− x500 + x100 + x + 1 = 0 has no rational roots.
p
Solution: If there exists a rational root, let it be where (p, q) = 1, q ≠ 0. Then, q
q
should divide the coefficient of the leading term and p should divide the constant term.
Example 13 Show that f (x) = x


Thus, q | 1

⇒

And p | 1 ⇒
Thus,

q = ± 1,
p = ±1

p
= ±1
q

If the root

p
= 1,
q

Then, f (1) = 1 − 1 + 1 + 1 + 1 = 3 ≠ 0,
so, 1 is not a root.
If

p
= −l,
q

Then, f (−1) = 1 − 1 + 1 − 1 + 1 = 1 ≠ 0
And hence, (−1) is not a root.

Thus, there exists no rational roots for the given polynomial.
Francois Viète

1.6 vieTa’s RelaTions
If a1, a2, a3, …, an are the roots of the polynomial equation
anxn + an−1xn−1 + an−2xn−2 + … + a0 = 0 (an ≠ 0),

∑ αi = −

then,

1≤i ≤ n

∑

1≤i < j < k ≤ n

an −1
a
; ∑ αi ⋅ α j = n−2
an 1≤i < j ≤ n
an

α iα j α k = −

an −3
,
an

; α1α 2α 3


α n = ( −1) n

a0
an

If we represent the sum ∑ai, ∑aiaj, …, ∑aiaj… an, respectively, as σ1, σ2, σ3, …, σn,
(Read it ‘sigma 1’, ‘sigma 2’, etc.)
then,

σ1 = −

1540–23 Feb 1603
Nationality: French

an −1
a
, σ 2 = n − 2 , ...
an
an

σ r = ( −1) r

an − r
a
,… , σ n = ( −1) n 0
an
an

These relations are known as Vieta’s relations.

Let us consider the following quadratic, cubic and biquadratic equations and see how
we can relate σ 1 , σ 2 , σ 3 ,..., with the coefficients.

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1.10  Chapter 1
1.ax2 + bx + c = 0, where α and β are its roots. Thus,
c
−b
 and  σ 2 = αβ =
a
a
2.ax3 + bx2 + cx + d = 0, where α, β and γ are its roots. Thus,

σ1 = α + β =

σ1 = α + β + γ = −

b
a

σ 2 = αβ + βγ + γα =

σ 3 = αβγ =
Here, expressing σ2 = α(β + γ) + βγ =
erty in computations.


c
a

−d
a
c
will be helpful when we apply this propa

3.ax4 + bx3 + cx2 + dx + e = 0, where α, β, γ, δ are its roots. Thus,

σ1 = α + β + γ + δ =

−b
a

σ 2 = αβ + αγ + αδ + βγ + βδ + γδ =
σ 3 = αβγ + αβδ + αγδ + βγδ =
σ 4 = αβγδ =

c
a

−d
,
a

e
a

Here, again, σ2 can be written as (α + β)(γ + δ ) + αβ + γδ and σ3 can be written

as αβ (γ + δ ) + γδ (α + β).
2

Example 14  If x + ax + b + 1 = 0, where a, b ∈  and b ≠ -1, has a root in integers
2

then prove that a + b2 is a composite.

Solution:  Let, α and β be the two roots of the equation where, α ∈ . Then,

α + β = -a
(1)
α ⋅ β = b + 1(2)
∴  β = -a - α is an integer. Also, since b + 1 ≠ 0, β ≠ 0.
From Eqs. (1) and (2), we get
a2 + b2 = (α + β)2 + (αβ - l)2
= α2 + β2 + α2β2 + 1
= (1 + α2)(l + β2)
Now, as α ∈  and β is a non-zero integer, 1 + α2 > 1 and 1 + β2 > 1.
Hence, a2 + b2 is composite number.
Example 15  For what value of p will the sum of the squares of the roots of

x2 - px = 1 - p be minimum?

2

Solution:  If r1 and r2 are the roots of x - px + (p - 1) = 0,

then rl + r2 = p, r1r2 = p - 1


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Polynomials  1.11

r21 + r22 = (r1 + r2)2 - 2r1r2 = p2 - 2p + 2 = (p - l)2 + 1
and r21 + r22 is minimum when (p - l)2 is minimum, then p = 1.
Thus, for p = 1, the sum of the squares of the roots is minimum.
Example 16  Let u, v be two real numbers none equql to -1, such that u, v and uv are
the roots of a cubic polynomial with rational coefficients. Prove or disprove that uv is
rational.
3

2

Solution:  Let, x + ax + bx + c = 0 be the cubic polynomial of which u, v, and uv are

the roots and a, b, and c are all rationals.
u + v + uv = -a
⇒ u + v = -a - uv,
uv + uv2 + u2v = b
u2v2 = - c

and

2

(1)

(2)
(3)

2

From (2)b = uv + uv + u v = uv(1 + v + u)
= uv(1 - a - uv)  (From (1))
= (1 - a) uv - u2v2
= (1 - a)uv + c
⇒ (1 - a) uv = b - c
As a ≠ 1, uv =

(b − c)
and since, a, b, c are rational, uv is ­rational.
1− a

Note that a = 1 ⇒ 1 + u + v + uv = 0 ⇒ (1 + u)(1 + v) = 0 ⇒ u = -1 or v = -1, which is
not the case.
3

2

Example 17  Solve the cubic equation 9x - 27x + 26x - 8 = 0, given that one of the

root of this equation is double the other.
Solution:  Let the roots be α, 2α and β.

−27
=3
9

⇒  β = 3(1 - α)(1)
26
(2)
2α 2 + 3αβ =
9
3α + β = −

Now,

2α 2 β =

8
(3)
9

From Eqs. (1) and (2), we get
2α 2 + 3α × 3(1 − α ) =

26
9

⇒  63α2 - 81α + 26 = 0
⇒  (21α - 13)(3α - 2) = 0
So, α =
If α =

13
2
or
21 3


13
21

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1.12  Chapter 1
 13  24 8
=
∴ β = 3 1 −  =
 21  21 7
This leads to 2α 2 β = 2 ×
So, taking α =

169 8 8
× ≠ (a contradiction)
441 7 9

1
2
 2
, β = 3 1 −  = 3 × = 1
3
3
 3

Hence, α + 2α + β =


2 4
+ + 1 = 3,
3 3

4 3× 2
26
2α 2 + 3αβ = 2 × +
×1 =
,
9
3
9


and

4
8
2α 2 β = 2 × × 1 =
9
9

Thus, the roots are

2 4
, , and 1.
3 3
3


2

Example 18  Solve the equation 6x - 11x + 6x - 1 - 0, given that the roots are in

harmonic progression.
Solution:  Let the roots be α, β and γ.

Since they are in HP,
∴

β=

2αγ
(1)
α +γ
11
(2)
6

Now,

σ1 = α + β + γ =



σ 2 = β (α + γ ) + αγ = 1 (3)



σ 3 = αβγ =


1
(4)
6

Using Eqs. (1) and (3), we get
2αγ
× (α + γ ) + αγ = 1
(α + γ )
⇒ 3αγ = 1
⇒  αγ =

1
(5)
3

From Eqs. (4) and (5), we get

β=

1 1 1
÷ = (6)
6 3 2

From Eqs. (2) and (6), we get

α +γ =
∴

M01_Polynomials_C01.indd 12


α=

11 1 8 4
− = =
6 2 6 3

4
−γ
3

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