mathamatics xam idea

Mathematics
STANDARD
As per the latest syllabus released by the CBSE dated
31st March, 2021 vide Circular No. Acad–26/2021 and the changes
in Assessment Practices (released on 22nd April, 2021 vide
Circular No. Acad–31/2021) for the session 2021-22

Compiled by:

Abhay Ji Dubey
Ajay Kumar Choubey

Neha Kalra

n Model Question Papers
n Competency-based Questions/
Objective Type Questions
n Important NCERT & Exemplar Problems
n CBSE Sample Question Paper-2021 (Solved)

Class X

Printing History: Edition: 2021-22

Syllabus Covered: Central Board of Secondary Education, Delhi

Price: Six Hundred Three Rupees (` 603/-)

ISBN: 978-93-91003-41-8

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be borne by the purchaser.

Contents

PART–A

UNIT I: Number Systems 3
1. Real Numbers
20
UNIT II: Algebra 42
2. Polynomials 77
3. Pair of Linear Equations in Two Variables 106
4. Quadratic Equations
5. Arithmetic Progressions 132

UNIT III: Coordinate Geometry 164
6. Coordinate Geometry 200
230
UNIT IV: Geometry
7. Triangles 246
8. Circles 275
9. Constructions
310
UNIT V: Trigonometry 346
10. Introduction to Trigonometry
11. Heights and Distances 390
419
UNIT VI: Mensuration
12. Areas Related to Circles
13. Surface Areas and Volumes

UNIT VII: Statistics & Probability
14. Statistics
15. Probability

PART–B 447

n Periodic Test – 1 453
l Pen Paper Test
l Multiple Assessment 459

n Periodic Test – 2 467
l Pen Paper Test 487
l Multiple Assessment 489
527
n Periodic Test – 3
l Pen Paper Test
l Multiple Assessment

n CBSE Sample Question Paper (Standard)–2021 (Solved)

n Blue Prints

n Model Question Papers 1 to 5 (Unsolved)

n APPENDIX
Competency-Based Questions
(Case Study Based Questions)

Syllabus

Unit No. Course Structure Class – X Marks: 80
I (Annual Examination) 06
II 20
III Units Name 06
IV Number Systems 15
V Algebra 12
VI Coordinate Geometry 10
VII Geometry 11
Trigonometry
Mensuration 80
Statistics & Probability

Total

UNIT I: NUMBER SYSTEMS

1. REAL NUMBERS (15) Periods

Euclid’s division lemma, Fundamental Theorem of Arithmetic - statements after
reviewing work done earlier and after illustrating and motivating through examples,
Proofs of irrationality of 2, 3, 5 Decimal representation of rational numbers in
terms of terminating/non-terminating recurring decimals

UNIT II: ALGEBRA

1. POLYNOMIALS (7) Periods

Zeros of a polynomial. Relationship between zeros and coefficients of quadratic
polynomials. Statement and simple problems on division algorithm for polynomials
with real coefficients.

2. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES (15) Periods

Pair of linear equations in two variables and graphical method of their solution,
consistency/inconsistency.

Algebraic conditions for number of solutions. Solution of a pair of linear equations
in two variables algebraically - by substitution, by elimination and by cross
multiplication method. Simple situational problems. Simple problems on equations
reducible to linear equations.

3. QUADRATIC EQUATIONS (15) Periods

Standard form of a quadratic equation ax2 + bx + c = 0, (a ≠ 0). Solutions of

quadratic equations (only real roots) by factorization, and by using quadratic

formula. Relationship between discriminant and nature of roots.

Situational problems based on quadratic equations related to day to day activities
to be incorporated.

4. ARITHMETIC PROGRESSIONS (8) Periods

Motivation for studying Arithmetic Progression Derivation of the nth term and sum of

the first n terms of A.P. and their application in solving daily life problems.

UNIT III: COORDINATE GEOMETRY

1. LINES (In two-dimensions) (14) Periods

Review: Concepts of coordinate geometry, graphs of linear equations. Distance
formula. Section formula (internal division). Area of a triangle.

UNIT IV: GEOMETRY

1. TRIANGLES (15) Periods

Definitions, examples, counter examples of similar triangles.

1. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other
two sides in distinct points, the other two sides are divided in the same ratio.

2. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is
parallel to the third side.

3. (Motivate) If in two triangles, the corresponding angles are equal, their
corresponding sides are proportional and the triangles are similar.

4. (Motivate) If the corresponding sides of two triangles are proportional, their
corresponding angles are equal and the two triangles are similar.

5. (Motivate) If one angle of a triangle is equal to one angle of another triangle
and the sides including these angles are proportional, the two triangles are
similar.

6. (Motivate) If a perpendicular is drawn from the vertex of the right angle of a
right triangle to the hypotenuse, the triangles on each side of the perpendicular
are similar to the whole triangle and to each other.

7. (Prove) The ratio of the areas of two similar triangles is equal to the ratio of the
squares of their corresponding sides.

8. (Prove) In a right triangle, the square on the hypotenuse is equal to the sum of
the squares on the other two sides.

9. (Prove) In a triangle, if the square on one side is equal to sum of the squares
on the other two sides, the angles opposite to the first side is a right angle.

2. CIRCLES (8) Periods

Tangent to a circle at, point of contact

1. (Prove) The tangent at any point of a circle is perpendicular to the radius
through the point of contact.

2. (Prove) The lengths of tangents drawn from an external point to a circle are
equal.

3. (Motivate) Alternative Segment theorem: If a chord is drawn through the point
of contact of a tangent to a circle, then the angles made by the chord with the
tangent are respectively equal to the angles subtended by the chord in the
alternate segments.

3. CONSTRUCTIONS (8) Periods
1. Division of a line segment in a given ratio (internally).
2. Tangents to a circle from a point outside it.
3. Construction of a triangle similar to a given triangle.

UNIT V: TRIGONOMETRY

1. INTRODUCTION TO TRIGONOMETRY (10) Periods

Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their
existence (well defined); motivate the ratios whichever are defined at 0° and 90°.
Values of the trigonometric ratios of 30°, 45° and 60°. Relationships between the
ratios

2. TRIGONOMETRIC IDENTITIES (15) Periods

Proof and applications of the identity sin2A + cos2A = 1. Only simple identities to

be given.Trigonometric ratios of complementary angles.

3. HEIGHTS AND DISTANCES: Angle of elevation, Angle of Depression.

(8) Periods

Simple problems on heights and distances. Problems should not involve more
than two right triangles. Angles of elevation / depression should be only 30°, 45°,
and 60°.

UNIT VI: MENSURATION

1. AREAS RELATED TO CIRCLES (12) Periods

Motivate the area of a circle; area of sectors and segments of a circle. Problems
based on areas and perimeter / circumference of the above said plane figures.
(In calculating area of segment of a circle, problems should be restricted to
central angle of 60°, 90° and 120° only. Plane figures involving triangles, simple
quadrilaterals and circle should be taken.)

2. SURFACE AREAS AND VOLUMES (12) Periods

1. S urface areas and volumes of combinations of any two of the following: cubes,
cuboids, spheres, hemispheres and right circular cylinders/cones. Frustum of
a cone.

2. Problems involving converting one type of metallic solid into another and other
mixed problems. (Problems with combination of not more than two different
solids be taken).

UNIT VII: STATISTICS AND PROBABILITY

1. STATISTICS (18) Periods

Mean, median and mode of grouped data (bimodal situation to be avoided).
Cumulative frequency graph.

2. PROBABILITY (10) Periods

Classical definition of probability. Simple problems on finding the probability of an
event.

The changes for classes IX-X (2021-22)internal year-end/
Board Examination are as under:

Year-end Classes IX-X (2021-22)
Examination/ Modified
Board Examination (2020-21)
Existing
(Theory)

Composition  Objective type  Competency Based Questions would
Questions including be minimum 30%
Multiple Choice
Question-20%   T hese can be in the form of Multiple-
Choice Questions, Case- Based
 Case-based/Source- Questions, Source Based Integrated
based Integrated Questions or any other types
Questions-20%
 Objective Questions will be 20 %
 S hort Answer/ Long
Answer Questions-  Remaining 50% Short Answer/ Long
Remaining 60% Answer Questions- (as per existing
pattern)

Design of Question Paper

MATHEMATICS-Standard
CLASS – X (2021-22)

Time: 3 hours Max. Marks: 80

S. No. Typology of Questions Total %
Marks Weightage
(approx.)

Remembering: Exhibit memory of 43 54
previously learned material by recalling
facts, terms, basic concepts, and
answers.

1. Understanding: Demonstrate
understanding of facts and ideas by
organizing, comparing, translating,
interpreting, giving descriptions, and
stating main ideas

Applying: Solve problems to new
situations by applying acquired
2. knowledge, facts, techniques and rules in 19 24

a different way.

Analysing: Examine and break
information into parts by identifying
motives or causes. Make inferences and
find evidence to support generalizations

Evaluating: Present and defend opinions
by making judgments about information,
3. validity of ideas, or quality of work based 18 22
80 100
on a set of criteria.

Creating: Compile information together
in a different way by combining elements
in a new pattern or proposing alternative
solutions

Total

INTERNAL ASSESSMENT 20 Marks
 Pen Paper Test and Multiple Assessment (5+5) 10 Marks
 Portfolio 05 Marks
 Lab Practical
05 Marks
(Lab activities to be done from the prescribed books)

PART–A

SS BASIC CONCEPTS—A FLOW CHART
SS MORE POINTS TO REMEMBER
SS OBJECTIVE TYPE QUESTIONS

OO Multiple Choice Questions
OO Fill in the Blanks
OO Very Short Answer Questions

SS SHORT ANSWER QUESTIONS
SS LONG ANSWER QUESTIONS
SS HOTS [HIGHER ORDER THINKING SKILLS]
SS PROFICIENCY EXERCISE
SS SELF-ASSESSMENT TEST

MATHEMATICS



Real Numbers 1

BASIC CONCEPTS – A FLOW CHART

rn bn

Real Numbers 3

MORE POINTS TO REMEMBER

 Algorithm: An algorithm means a series of well-defined steps which gives a procedure for
solving a type of problem.

 Lemma: A lemma is a proven statement used for proving another statement.
A positive integer n is prime, if it is not divisible by any prime less than or equal to n .
For example, 5, 7 are prime as they are not divisible by prime less than 5 , 7 i.e., by 2.
9 is composite as 9 is divisible by 3 = 9 .
If x is a positive prime, then x is an irrational number.
For example, 7 is a positive prime ⇒ 7 is an irrational number.

Multiple Choice Questions [1 mark]

Choose and write the correct option in the following questions.

1. 7 is (b) an irrational number
(a) an integer (d) none of these
(c) a rational number

2. The decimal expansion of the rational number 47 will terminate after
23 52

(a) one decimal place (b) two decimal places

(c) three decimal places (d) more than three decimal places

3. For some integer q, every odd integer is of the form [NCERT Exemplar]

(a) q (b) q + 1 (c) 2q (d) 2q + 1

4. The product of three consecutive integers is divisible by

(a) 5 (b) 6 (c) 7 (d) none of these

5. The product of two consecutive integers is divisible by

(a) 2 (b) 3 (c) 5 (d) 7

6. The largest number which divides 615 and 963 leaving remainder 6 in each case is

(a) 82 (b) 95 (c) 87 (d) 93

7. n2 – 1 is divisible by 8 if n is

(a) an integer (b) a natural number (c) an odd integer (d) an even integer

8. The largest number which divides 70 and 125 leaving remainders 5 and 8 respectively is
[NCERT Exemplar]

(a) 13 (b) 65 (c) 875 (d) 1750

9. If two positive integers a and b are written as a = x3y2 and b = xy3; x, y are prime numbers, then

LCM (a, b) is [NCERT Exemplar]

(a) xy (b) xy2 (c) x3y3 (d) x2y2

10. The product of a non zero rational and an irrational number is [NCERT Exemplar]

(a) always irrational (b) always rational

(c) rational or irrational (d) one

4 Xam idea Mathematics–X

11. The product of two irrational numbers is

(a) always irrational (b) always rational

(c) rational or irrational (d) one

12. The decimal expansion of the rational number 14587 will terminate after [NCERT Exemplar]
1250

(a) one decimal place (b) two decimal places

(c) three decimal places (d) four decimal places

13. The decimal expansion of number 46 is
22 ×5×3
(a) terminating (b) non-terminating repeating

(c) non-terminating non-repeating (d) none of these

14. If two positive integers a and b are written as a = x4 y2 and , b = x2 y3; x, y are prime numbers,
then HCF (a, b) is

(a) x4 y3 (b) xy (c) x2 y3 (d) x2 y2

15. The exponent of 2 in prime factorisation of 144 is

(a) 4 (b) 5 (c) 6 (d) 3

16. The LCM of two numbers is 1200. Which of the following cannot be their HCF?

(a) 600 (b) 500 (c) 400 (d) 200

17. If 3 is the least prime factor of number a and 7 is the least prime factor of number b, then the
least prime factor of (a + b) is

(a) 2 (b) 3 (c) 5 (d) 10

18. If HCF (26, 169) = 13 then LCM (26, 169) is

(a) 26 (b) 52 (c) 338 (d) 13

19. 3.27 is (b) a rational number (c) a natural number (d) an irrational number
(a) an integer

20. If n is any natural number then 6n – 5n always end with

(a) 1 (b) 3 (c) 5 (d) 7

Answers

1. (b) 2. (c) 3. (d) 4. (b) 5. (a) 6. (c)
7. (c) 8. (a) 9. (c) 10. (a) 11. (c) 12. (d)
13. (b) 14. (d) 15. (a) 16. (b) 17. (a) 18. (c)
19. (b) 20. (a)

Fill in the Blanks [1 mark]

Complete the following statements with appropriate word(s) in the blank space(s).

1. Euclid’s Division Lemma is a restatement of _________________ .
2. If every positive even integer is of the form 2q, then every positive odd integer is of the form

_________________, where q is some integer.
3. 2, 3, 7 , etc are _________________ numbers.
4. Every point on the number line corresponds to a _________________ number.

Real Numbers 5

5. Every real number is either a _________________ number or an _________________ number.
6. The product of three numbers is _________________ to the product of their HCF and LCM.
7. If p is a prime number and it divides a2 then it also divides _________________, where a is a

positive integer.
8. If a = bq + r then least value of r is _________________.
9. Numbers having non-terminating non-repeating decimal expansion are known as ____________.
10. An algorithm which is used to find HCF of two positive numbers is _________________ .
11. HCF of two numbers is always a factor of their _________________ .
12. 0 is _________________ (a rational/an irrational) number.
13. 1 is neither _________________ nor _________________ number.
14. LCM of the smallest composite number and the smallest prime number is _________________.

15. Co-prime number is a set of numbers which have 1 as their _________________ .

Answers

1. long division process 2. 2q + 1 3. irrational 4. real
8. zero
5. rational, irrational 6. not equal 7. a 12. a rational

9. irrational numbers 10. Euclid’s division algorithm 11. LCM
13. prime; composite 14. 4 15. HCF

Very Short Answer Questions [1 mark]

1. What is the HCF of the smallest composite number and the smallest prime number?

[CBSE 2018]

Sol. Smallest composite number = 4

Smallest prime number = 2

So, HCF (4, 2) = 2 6
1250
2. The decimal representation of will terminate after how many places of decimal?

Sol. 6 = 3 = 3 # 24 = 48 = 48 = 0 . 0048
1250 625 54 24 ^5 # 2h4 104

This representation will terminate after 4 decimal places.

3. The HCF of two numbers a and b is 5 and their LCM is 200. Find the product ab.

[CBSE 2019 (30/5/2)]

Sol. We know that

Product of two numbers = product of their LCM and HCF

ab = 200 × 5

= 1000

4. What is the HCF of 33 × 5 and 32 × 52?

Sol. HCF of 33 × 5 and 32 × 52 = 32 × 5 = 45

5. If a is an odd number, b is not divisible by 3 and LCM of a and b is P, what is the LCM of
3a and 2b?

Sol. Since a is odd, it is not divisible by 2 (so we need to multiply (from 2b) with LCM of a and b) since
b is not divisible by 3 we need to multiply 3 from 3a with LCM of a, b.

Hence, LCM (3a and 2b) = 6P
6. If P is prime number then, what is the LCM of P, P2, P3?
Sol. P3

6 Xam idea Mathematics–X

7. Two positive integers p and q can be expressed as p = 2 and q = a2b, a and b are prime
numbers. What is the LCM of p and q?

Sol. a2b2

8. A number N when divided by 14 gives the remainder 5. What is the remainder when the same
number is divided by 7?

Sol. 5, because 14 is multiple of 7.

Therefore, remainder in both cases are same.

9. If HCF (336, 54) = 6. Find LCM (336, 54). [CBSE 2019 (30/2/1)]

Sol. LCM × HCF = Product of numbers

LCM × 6 = 336 × 54

LCM = 336 × 54
6

= 3024

10. What are the possible values of remainder r, when a positive integer a is divided by 3?

Sol. According to Euclid’s division lemma

a = 3q + r, where 0 # r 1 3 and r is an integer.
Therefore, the values of r can be 0, 1 or 2.

11. A rational number in its decimal expansion tihse1f.o7r3m51.qpW? hGaitvecarneaysooun. say about the prime
factors of q when this number is expressed in

Sol. As 1.7351 is a terminating decimal number, so q must be of the form 2n 5m, where m, n are natural

numbers.

12. Find after how many places the decimal form of the number 27 will terminate?
23 .54 .32
[CBSE 2019 (30/3/3)]
Sol. 27 33 ×2 3×2
23 ×54 ×32 = 23 × 54 ×32 × 2 = (2×5) 4

So, the decimal form will end after four decimal places.

Short Answer Questions-I [2 marks]

1. Express 429 as product of its prime factors. [CBSE 2019 (30/3/3)]

Sol. 3 429

13 143

11

\ 429 = 3 × 13 × 11

2. Write whether the square of any positive integer can be of the form 3m + 2, where m is a

natural number. Justify your answer. [NCERT Exemplar]

Sol. No, since any positive integer can be written as 3q, 3q + 1, 3q + 2, then, square will be
(3q)2 = 9q2 = 3.(3q2) = 3m, (3q + 1)2 = 9q2 + 6q + 1 = 3 (3q2 + 2q) + 1 = 3m + 1,

(3q + 2)2 = 9q2 + 12q + 4 = 3(3q2 + 4q +1) + 1 = 3m + 1.

3. Can two numbers have 18 as their HCF and 380 as their LCM? Give reason. [NCERT Exemplar]
Sol. No, because here HCF (18) does not divide LCM (380).

4. Write a rational number between 2 and 3 . 17 [CBSE 2019 (30/1/2)]
10
Sol. A rational number between 2 and 3 is 2.89 = 1.7 = . Real Numbers 7

5. The product of two consecutive integers is divisible by 2. Is this statement true or false? Give
reason.

Sol. True, because n(n + 1) will always be even, as one out of the n or n+1 must be even.

6. Explain why 3 × 5 × 7 + 7 is a composite number.
Sol. 3 × 5 ×7 + 7= 7 (3 × 5 +1) = 7 × 16, which has more than two factors.

7. What is the least number that is divisible by all the numbers from 1 to 10?
Sol. Required number = LCM of 1, 2, 3, … 10 = 2520.

8. Find the sum of 0.68 + 0.73 .

Sol. 0.68 + 0.73 = 68 + 73 = 141 = 1.42
99 99 99

9. “The product of three consecutive positive integers is divisible by 6”. Is this statement true or
false? Justify your answer.

Sol. True, because n(n + 1) (n + 2) will always be divisible by 6, as at least one of the factors will be
divisible by 2 and at least one of the factors will be divisible by 3.

10. Find the HCF of 612 and 1314 using prime factorisation. [CBSE 2019 (30/5/3)]

Sol. 2 612 2 1314

2 306 3 657

3 153 3 219

3 51 73

17

612 = 2 × 2 × 3 × 3 × 17

1314 = 2 × 3 × 3 × 73

HCF = 2 × 3 × 3 = 18

11. Find the HCF of 1260 and 7344 using Euclid’s algorithm. [NCERT, CBSE 2019 (30/1/1)]

OR [CBSE 2019 (30/4/1)]
Use Euclid’s division algorithm to find the HCF of 255 and 867.

Sol. Since 7344 > 1260 and from Euclid’s algorithm

a = bq + r

⇒ 7344 = 1260 × 5 + 1044 ⇒ 1260 = 1044 × 1 + 216

⇒ 1044 = 216 × 4 + 180 ⇒ 216 = 180 × 1 + 36
⇒ 180 = 36 × 5 + 0 ⇒ HCF = 36

OR

Solution is similar as above.

Only values are changed

Ans: 51

Short Answer Questions-II [3 marks]

1. An army contingent of 616 members is to march behind an army band of 32 members in a

parade. The two groups are to march in the same number of columns. What is the maximum

number of columns in which they can march? [NCERT]

Sol. For the maximum number of columns, we have to find the HCF of 616 and 32.

Now, since 616 > 32, we apply division lemma to 616 and 32.

8 Xam idea Mathematics–X

We have, 616 = 32 × 19 + 8

Here, remainder 8 ≠ 0. So, we again apply division lemma to 32 and 8.

We have, 32 = 8 × 4 + 0

Here, remainder is zero. So, HCF (616, 32) = 8

Hence, maximum number of columns is 8.

2. Find the LCM and HCF of 12, 15 and 21 by applying the prime factorisation method. [NCERT]

Sol. The prime factors of 12, 15 and 21 are
12 = 22 × 3, 15 = 3 × 5 and   21 = 3 × 7

Therefore, the HCF of these integers is 3.
22, 31, 51 and 71 are the greatest powers involved in the prime factorisation of 12, 15 and 21.
So, LCM (12, 15, 21) = 22 × 31 × 51 × 71 = 420.

3. If HCF of 65 and 117 is expressible in the form 65n – 117, find the value of n.
[CBSE 2019 (30/3/3)]

Sol. HCF (65, 117) is given by

65 = 5 × 13

117 = 13 × 3 × 3

⇒ HCF = 13

According to question,

65n – 117 = 13

⇒ 65n = 130

⇒ n=2

4. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the

field, while Ravi takes 12 minutes for the same. Suppose they both start from the same point

and at the same time, and go in the same direction. After how many minutes will they meet

again at the starting point? [NCERT]

Sol. To find the time after which they meet again at the starting point, we have to find LCM of 18 and

12 minutes. We have 2 18 2 12

18 = 2 × 32 3 9 2 6
and 12 = 22 × 3 3 3 3 3

Therefore, LCM of 18 and 12 = 22 × 32 = 36 1 1

So, they will meet again at the starting point after 36 minutes.

5. Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form

9m, 9m +1 or 9m +8 for some integer m. [NCERT]

Sol. Let x be any positive integer. Then, it is of the form 3q or 3q + 1 or 3q + 2 So, we have the
following cases:

Case I: When x = 3q

In this case, we have
x3 = (3q)3 = 27q3 = 9(3q3) = 9m, where m = 3q3

Case II: When x = 3q + 1

In this case, we have

x3 = (3q + 1)3

⇒ x3 = 27q3 + 27q2 + 9q + 1

⇒ x3 = 9q(3q2 + 3q + 1) + 1

⇒ x3 = 9m +1, where m = q(3q2 + 3q +1)

Real Numbers 9

Case III: When x = 3q + 2

In this case, we have
x3 = (3q + 2)3
⇒ x3 = 27q3 + 54q2 + 36q +8
⇒ x3 = 9q(3q2 + 6q + 4) + 8

⇒ x3 = 9m + 8, where m = q(3q2 + 6q + 4)

Hence, x3 is either of the form 9m or 9m + 1 or 9m +8.



6. Express the number 0.3178 in the form of rational number a .
b

Sol. Let x = 0.3178

Then x = 0.3178178178... ... (i)

10x = 3.178178178... ... (ii)

10000x = 3178.178178... ... (iii)

On subtracting (ii) from (iii), we get

9990x = 3175 ⇒ x= 3175 = 635
\ 9990 1998
635
0.3178 = 1998

7. If n is an odd positive integer, show that (n2 – 1) is divisible by 8. [NCERT Exemplar]

Sol. We know that an odd positive integer n is of the form (4q + 1) or (4q + 3) for some integer q.

Case-I: When n = (4q + 1)

In this case n2 – 1 = (4q + 1)2 – 1 = 16q2 + 8q = 8q (2q + 1)

which is clearly divisible by 8.

Case-II: When n = (4q + 3)

In this case, we have

n2 – 1 = (4q + 3)2 –1 = 16q2 + 24q + 8 = 8 (2q2 + 3q +1)

which is clearly divisible by 8.

Hence (n2 – 1) is divisible by 8.

8. Find the value of x, y and z in the given factor tree. Can the value of ‘x’ be
found without finding the value of ‘y’ and ‘z’? If yes, explain.

Sol. z = 2 × 17 = 34; y = 34 × 2 = 68 and x = 2 × 68 = 136

Yes, value of x can be found without finding value of y or z as
x = 2 × 2 × 2 × 17 which are prime factors of x.

9. Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some

integer. [NCERT]

OR

Show that any positive odd integer is of the form 6m + 1 or 6m + 3 or 6m + 5, where m is some

integer. [CBSE 2019(30/5/2)]

Sol. Let a be any positive odd integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some
integer q ≥ 0 and 0 ≤ r < 6.

i.e., the possible remainders are 0, 1, 2, 3, 4, 5.

Thus, a can be of the form 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where q is some
integer.

Since a is odd integer, so a cannot be of the form 6q or 6q + 2 or 6q + 4 (since they are even).

Thus, a is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.

10 Xam idea Mathematics–X

Hence, any odd positive integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.

OR
Let a be any positive odd integer and b = 6. Then, by Euclid’s algorithm, a = 6m + r, for some

integer m ≥ 0 and 0 ≤ r < 6.

i.e., the possible remainders are 0, 1, 2, 3, 4, 5.

Thus, a can be of the form 6m, or 6m + 1 or 6m + 2 or 6m + 3 or 6m + 4 or 6m + 5, where m
is some integer.

Since a is odd integer, so a cannot be of the form 6m or 6m + 2 or 6m + 4 (since they are even).

Thus, a is of the form 6m + 1, 6m + 3 or 6m + 5, where m is some integer.

Hence, any odd positive integer is of the form 6m + 1 or 6m + 3 or 6m + 5, where m is some integer.

10. Write the smallest number which is divisible by both 306 and 657. [CBSE 2019 (30/2/1)]

Sol. Here, to find the required smallest number we will find LCM of 306 and 657.

306 = 2 × 32 × 17 2 306 3 657
657 = 32 × 73 3 153 3 219
LCM = 2 × 32 × 17 × 73 = 22338 3 51 73
17

Long Answer Questions [5 marks]

1. Use Euclid’s division lemma to show that the square of any positive integer is either of the

form 3m or 3m + 1 for some integer m. [NCERT]

Sol. Let a be an arbitrary positive integer.

Then by Euclid’s division algorithm, corresponding to the positive integers a and 3 there exist
non-negative integers q and r such that

a = 3q + r where 0 ≤ r < 3

⇒ a2 = 9q2 + 6qr + r2 .... (i)

Case-I: When r = 0

a2 = 9q2 = 3(3q2) = 3m where m = 3q2 [From (i)]

Case-II: When r = 1

a2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1 where m = 3q2 + 2q [From (i)]

Case-III: When r = 2

a2 = 9q2 + 12q + 4 = 3(3q2 + 4q + 1) + 1 = 3m + 1 where m = (3q2 + 4q + 1)

[From (i)]

Hence, square of any positive integer is either of the form 3m or 3m +1 for some integer m.

2. Show that one and only one out of n, n + 2, n + 4 is divisible by 3, where n is any positive

integer. [NCERT Exemplar]

Sol. Let q be the quotient and r be the remainder when n is divided by 3.

Therefore, n = 3q + r, where r = 0, 1, 2

⇒ n = 3q or n = 3q + 1 or n = 3q + 2

Case I: if n = 3q, then n is divisible by 3, but n + 2 and n + 4 are not divisible by 3.

Case II: if n = 3q + 1 then n + 2 = 3q + 3 = 3(q + 1), which is divisible by 3 and

n + 4 = 3q + 5, which is not divisible by 3.

n = 3q + 1, then n is not divisible by 3

So, only (n + 2) is divisible by 3.

Real Numbers 11

Case III: if n = 3q + 2, then n + 2 = 3q + 4, which is not divisible by 3 and
 (n + 4) = 3q + 6 = 3(q + 2), which is divisible by 3.

n = 3q + 2 then n is not divisible by 3.

So, only (n + 4) is divisible by 3.

Hence, one and only one out of n, (n + 2), (n + 4) is divisible by 3.

3. Use Euclid’s division algorithm to find the HCF of: [NCERT]

(i) 960 and 432

(ii) 4052 and 12576. [CBSE 2019 (C)(30/1/1)]

Sol. (i) Since 960 > 432, we apply the division lemma to 960 and 432.

We have, 960 = 432 × 2 + 96

Since the remainder 96 ≠ 0, so we apply the division lemma to 432 and 96.

We have, 432 = 96 × 4 + 48

Again remainder 48 ≠ 0 so we again apply division lemma to 96 and 48.

We have, 96 = 48 × 2 + 0

The remainder has now become zero. So our procedure stops.

Since the divisor at this stage is 48.
Hence, HCF of 960 and 432 is 48.
i.e., HCF (960, 432) = 48

(ii) Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get

12576 = 4052 × 3 + 420

Since the remainder 420 ≠ 0, we apply the division lemma to 4052 and 420, to get

4052 = 420 × 9 + 272

We consider the new divisor 420 and the new remainder 272, and apply the division lemma
to get

420 = 272 × 1 + 148

We consider the new divisor 272 and the new remainder 148, and apply the division lemma
to get

272 = 148 × 1 + 124

We consider the new divisor 148 and the new remainder 124, and apply the division lemma
to get

148 = 124 × 1 + 24

We consider the new divisor 124 and the new remainder 24, and apply the division lemma
to get

124 = 24 × 5 + 4

We consider the new divisor 24 and the new remainder 4, and apply the division lemma to
get

24 = 4 × 6 + 0

The remainder has now become zero, so our procedure stops. Since the divisor at this stage
is 4, the HCF of 12576 and 4052 is 4.

4. Using prime factorisation method, find the HCF and LCM of 30, 72 and 432. Also show that
HCF × LCM ≠ product of the three numbers.

Sol. Given numbers = 30, 72, 432

30 = 2 × 3 × 5; 72 = 23 × 32 and 432 = 24 × 33

Here, 21 and 31 are the smallest powers of the common factors 2 and 3 respectively.

So, HCF (30, 72, 432) = 21 × 31 = 2 × 3 = 6

Again, 24, 33 and 51 are the greatest powers of the prime factors 2, 3 and 5 respectively.

12 Xam idea Mathematics–X

So, LCM (30, 72, 432) = 24 × 33 × 51 = 2160

HCF × LCM = 6 × 2160 = 12960

Product of numbers = 30 × 72 × 432 = 933120

Therefore, HCF × LCM ≠ Product of the numbers.

5. Prove that 5 is an irrational number. [CBSE 2019 (30/5/1)]

Sol. Let us assume, to the contrary, that 5 is a rational number.

Then, there exist co-prime positive integers a and b such that

5 = a , (b ≠ 0)
b

So, a = 5 b

Squaring both sides, we have

a2 = 5b2 …(i)

⇒ 5 divides a2 ⇒ 5 divides a

So, we can write

a = 5c (where c is any integer)

Putting the value of a = 5c in (i), we have

25c2 = 5b2 ⇒ 5c2 = b2

It means 5 divides b2 and so 5 divides b.

So, 5 is a common factor of both a and b which is a contradiction.

So, our assumption that 5 is a rational number is wrong. [NCERT]
Hence, we conclude that 5 is an irrational number.
6. Show that 5 – 3 is an irrational number.

Sol. Let us assume that 5 – 3 is a rational number.
So, 5 – 3 may be written as

   5 – 3 = p , where p and q are integers, having no common factor except 1 and q ≠ 0.
q

⇒  5 – p = 3   ⇒  3 = 5q – p
q q

Since 5q – p is a rational number as p and q are integers.
q

∴ 3 is also a rational number which is a contradiction.

Thus, our assumption is wrong.

Hence, 5 – 3 is an irrational number. [NCERT]
7. Check whether 6n can end with the digit 0 for any natural number n.

Sol. If the number 6n, for any n, were to end with the digit zero, then it would be divisible by 5. That
is, the prime factorisation of 6n would contain the prime 5. But 6n = (2 × 3)n = 2n × 3n so the
primes in factorisation of 6n are 2 and 3. So the uniqueness of the Fundamental Theorem of
Arithmetic guarantees that there are no other primes except 2 and 3 in the factorisation of 6n.
So there is no natural number n for which 6n ends with digit zero.

Real Numbers 13

HOTS [Higher Order Thinking Skills]

1. Show that there is no positive integer n for which n – 1 + n + 1 is rational.

Sol. Let there be a positive integer n for which n – 1 + n + 1 be rational number.

n–1 + n+1= p ; where p, q are integers and q ≠ 0 ...(i)
q

⇒ 1 n+1 = q
n–1 + p

⇒ _ n–1 + n–1 – n+1 n + 1i = q
p
n + 1i×_ n – 1 –

⇒ ^ n –1 – ^ n+1 = q
n – 1h – n + 1h p

⇒ n–1 – n+1 = q
n –1– n –1 p

⇒ n+1– n –1 = q
2 p

⇒ n+1 – n –1 = 2q ...(ii)
p

Adding (i) and (ii), we get p 2q
q p
n–1 + n+1 + n+1 – n–1 = +

⇒ 2 n+1 = p2 + 2q2
pq

⇒ n+1 = p2 + 2q2
⇒ 2pq

n+1 is rational number as p2 + 2q2 is rational.
2pq

⇒ n + 1 is perfect square of positive integer. ...(A)

Again subtracting (ii) from (i), we get

n–1 + n+1 – n+1+ n–1 = p – 2q ⇒2 n–1 = p2 – 2q2
q p pq

⇒ n –1 is rational number as p2 – 2q2 is rational.
2pq

⇒ n – 1 is also perfect square of positive integer. ...(B)

From (A) and (B)

n + 1 and n – 1 are perfect squares of positive integer. It contradict the fact that two perfect

squares differ at least by 3.
Hence, there is no positive integer n for which n –1 + n + 1 is rational.
2. Let a, b, c, k be rational numbers such that k is not a perfect cube.

If a + bk1/3 + ck2/3 then prove that a = b = c = 0.

14 Xam idea Mathematics–X

Sol. Given, a + bk1/3 + ck2/3 = 0 ...(i)

Multiplying both sides by k1/3, we have

ak1/3 + bk2/3 + ck = 0 ...(ii)

Multiplying (i) by b and (ii) by c and then subtracting, we have

⇒ (ab + b2k1/3 + bck2/3) – (ack1/3 + bck2/3 + c2k) = 0

⇒ (b2 – ac)k1/3 + ab – c2k = 0

⇒ b2 – ac = 0 and ab – c2k = 0 [Since k1/3 is irrational]

⇒ b2 = ac and ab = c2k

⇒ b2 = ac and a2b2 = c4k2

⇒ a2(ac) = c4k2 [By putting b2 = ac in a2b2 = c4k2]

⇒ a3c – k2c4 = 0

⇒ (a3 – k2c3)c = 0

⇒ a3 – k2c3 = 0, or c = 0

Now, a3 – k2c3 = 0

⇒ k2 = a3
c3

⇒ ^ k2h1/3 = d a3 1/3
c3
n

⇒ k2/3 = a
c

This is impossible as k2/3 is irrational and a is rational.
c
∴ a3 – k2c3 ≠ 0

Hence, c = 0
Substituting c = 0 in b2 – ac = 0, we get b = 0
Substituting b = 0 and c = 0 in a + bk1/3 + ck2/3= 0, we get a = 0

Hence, a = b = c = 0

3. Find the largest number which on dividing 1251, 9377 and 15628 leaves remainders 1, 2 and

3 respectively. [NCERT Exemplar, CBSE 2019 (30/3/3)]

Sol. Given numbers are 1251, 9377 and 15628.

Now as per question, subtract 1, 2 and 3 from respective numbers.

1251 – 1 = 1250

9377 – 2 = 9375

15628 – 3 = 15625

Now HCF of 1250, 9375 and 15625 is

1250 = 5 × 5 × 5 × 5 × 2

9375 = 5 × 5 × 5 × 5 × 5 × 3

15625 = 5 × 5 × 5 × 5 × 5 × 5

HCF = 54 = 625

∴ Required largest number is 625.

Real Numbers 15

PROFICIENCY EXERCISE

QQ Objective Type Questions: [1 mark each]

1. Choose and write the correct option in each of the following questions.

(i) The 3.27 is

(a) an integer (b) a rational (c) a natural number (d) an irrational number

(ii) If a and b are co-prime numbers then a2 and b2 are

(a) co-prime (b) not co-prime (c) even number (d) odd number
(b) non-terminating
(iii) The decimal form of 129 is
22 .57 .75
(a) terminating

(c) non-terminating non-repeating (d) none of them

(iv) For some integer a, every odd integer is of the form

(a) 2 a + 1 (b) 2 a (c) a + 1 (d) a

(v) If the LCM of p and 18 is 36 and the HCF of p and 18 is 2 then p =

(a) 2 (b) 3 (c) 4 (d) 1

2. Fill in the blanks.

(i) The ratio between the LCM and HCF of 5, 15, 20 is _______________ .

(ii) If p is positive prime, then p is an _______________ number.

(iii) HCF of 168 and 126 is _______________ .

(iv) The decimal representation of 15 will terminate after _______________ decimal places.
(v) If two positive integers a and b 400 written as a = x3y2 and b = xy3; x, y are prime number
are

then LCM is _______________ .

QQ Very Short Answer Questions : [1 mark each]

3. Write whethe r 2 4 52+53 2 0 on s implification gives an irrational or a rational number.
[CBSE 2018 (C)
(30/1)]

4. Find the rational number between 2 and 7 . [CBSE 2019 (30/4/2)]

5. Write the number of zeros in the end of a number whose prime factorization is

22 × 53 × 32 × 17. [CBSE 2019 (30/4/2)]

6. The LCM of two numbers is 9 times their HCF. The sum of LCM and HCF is 500. Find the HCF

of two numbers. [CBSE 2019 (C) (30/1/1)]

7. What can you say about the prime factorisation of the denominators of the rational number 0.134 ?

8. A rational number in its decimal expansion is 1.7112. What can you say about the prime factors
p
of q, when this number is expressed in the form q ?

9. State wtyhpeethoefrde2c06i0mahl aesxtpearnmsiinonatidnogesor2n2o×n2-59t×erm7 inhaatvine?g repeating decimal expansion.
10. What

11. If two positive integers a and b are written as a = x4y2 and b = x3y, where x, y are prime numbers,

then find HCF (a, b).

12. If two positive integers a and b are written as a = xy2 and b = x3y, where x, y are prime numbers,

then find LCM (a, b). 47
23 52
13. After how many decimal places will the decimal expansion of the rational number

terminate? 189
125
14. After how many places will the decimal expansion of terminate?

16 Xam idea Mathematics–X

QQ Short Answer Questions-I: [2 marks each]

15. Given that 2 is irrational, prove that ^5 + 3 2h is an irrational number [CBSE 2018 (30/1)]

16. On a morning walk, three persons step out together and their steps measure 30 cm, 36 cm and

40 cm respectively. What is the minimum distance each should walk so that each can cover the

same distance in complete steps? [CBSE 2019 (30/3/1)]

17. Prove that n2 + n is divisible by 2 for any positive integer n. [CBSE 2019 (C) (30/1/1)]

18. The numbers 525 and 3000 are both divisible only by 3, 5, 15, 25 and 75. What is HCF
(525, 3000)? Justify your answer.

19. Can the numbers 6n, n being a natural number end with the digit 5? Give reasons.

20. Given that LCM (26, 169) = 338, find HCF (26, 169).

21. Write whether every positive integer can be of the form 4q + 2, where q is an integer. Justify your
answer.

22. A positive integer is of the form 3q + 1, q being a natural number. Can you write its square in any
form other than 3m + 1 i.e., 3m or 3m + 2 for some integer m? Justify your answer.

23. Find the LCM of x2 – 4 and x4 – 16.

QQ Short Answer Questions-II: [3 marks each]

24. Using Euclid’s division algorithm find the HCF of the numbers 867 and 255.

[CBSE 2018 (C)]

25. Prove that 2 + 5 3 is an irrational number, given that 3 is an irrational number.

[CBSE 2019 (30/2/1)]

26. Prove that 3 is an irrational number. [CBSE 2019 (30/3/1)]

27. Prove that 2 + 3 3 is an irrational number when it is given that 3 is an irrational number.

[CBSE 2019 (30/4/2)]

28. Use Euclid’s algorithm to find the HCF of 4052 and 12576. [CBSE 2019 (C) (30/1/1)]
29. Show that 12n cannot end with the digit 0 for any natural number n. [CBSE 2020 (30/5/1)]

30. If the HCF (210, 55) is expressible in the form 210 × 5 – 55y, find y.

31. Find the greatest number that will divide 445, 572 and 699 leaving remainders 4, 5 and 6 respectively.
257
32. Write the denominator of the rational number 5000 in the form 2m × 5n, where m, n are

non-negative integers. Hence, write its decimal expansion, without actual division.

33. Using prime factorisation method, find the LCM of 21, 28, 36, 45.

34. Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.

35. Rahul takes 24 minutes to complete one round of circular track, while Sahil takes 18 minutes

for the same. Suppose they both start at the same time from the same point and go in the same

direction. After how many minutes will they meet again at the starting point?
36. Show that 9n cannot end with the digit 2 for any n ∈ N.

37. Express 3825 as product of its prime factors using factor tree.

38. Show that the square of any odd integer is of the form 4q + 1, for some integer q.

39. Show that 2 3 is an irrational number.

40. The length, breadth and height of a room are 8 m 25 cm, 6 m 75 cm and 4 m 50 cm respectively.

Determine the longest rod which can measure the three dimensions of the room exactly.

41. On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and

45 cm respectively. What is the minimum distance each should walk so that each can cover the

same distance in complete steps? [NCERT Exemplar]

42. Find the LCM and HCF of 336 and 54 and verify that LCM × HCF = Product of the two numbers.

43. Find the largest number which divides 318 and 739 leaving remainder 3 and 4 respectively.

Real Numbers 17

QQ Long Answer Questions: [5 marks each]

44. Show that cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is
also of the form 6m + r.

45. Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n
is any positive integer.

(Hint: Any positive integer can be written in the form of 5q, 5q + 1, 5q + 2, 5q + 3, 5q + 4)

46. Use Euclid’s division lemma to show that the square of any positive integer is either of the form
3m or 3m + 1 for some integer m.

47. Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some

integer q. [NCERT Exemplar]

48. Prove that 2+ 5 is an irrational number.

49. Show that 3 – 5 is an irrational number.

50. Show that p + q is an irrational number, where p, q are primes.

51. Prove that 5 is an irrational number and hence show that 3 + 5 is also an irrational number.
52. For any positive integer n, prove that n3 – n is divisible by 6.

Answers

1. (i) (b) (ii) (a) (iii) (c) (iv) (a) (v) (c)

2. (i) 12 : 1 (ii) irrational (iii) 42 (iv) 4 (v) x3y3

3. Rational 4. Any rational number between 1.4 and 2.6. 5. two 6. HCF = 50

7. Since 0.134 has non-terminating repeating decimal expansion, its denominator has factors other than 2 or 5.

8. q has the factors of the form 2n × 5m for whole numbers n and m. 9. Terminating

10. Non-terminating repeating 11. x3y 12. x3y3 13. 3 14. 3

16. 360 cm 18. HCF = 75, as HCF is the highest common factor.

19. No, because 6n = (2 × 3)n = 2n × 3n, so the only primes in the factorisation of 6n are 2 and 3, and not 5.

20. 13 21. No, because an integer can be written in the form 4q, 4q + 1, 4q + 2, 4q + 3.

22. No, (3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1 23. (x2 + 4) (x2 – 4) 24. HCF = 51

28. 4 30. 19 31. 63 32. 23×54, 0.0514 33. 1260 34. 9,99,720

35. 72 minutes 37. 40. 75 cm

41. 2520 cm 42. H CF=6, LCM=3024 43. 105

SELF-ASSESSMENT TEST
Time allowed: 1 hour
Max. marks: 40

Section A

1. Choose and write the correct option in the following questions. (4 × 1 = 4)

(i) If two positive integers a and b are written as a = xy2 and b = x3y, where x, y are prime
numbers, then LCM (a, b) is

(a) x2y2 (b) xy (c) x3y2 (d) none of these

(ii) If the HCF of 65 and 117 is expressible in the form 65m –117, then the value of m is

(a) 4 (b) 2 (c) 11 (d) 3

18 Xam idea Mathematics–X

(iii) The product of a non-zero rational and an irrational number is

(a) always rational (b) always irrational (c) one (d) rational or irrational

(iv) Euclid’s division lemma states that for two positive integers a and b, there exist unique

integers q and r such that a = bq + r, where r must satisfy [NCERT Exemplar]

(a) 1< r <b (b) 0< r ≤ b (c) 0≤ r < b (d) 0 <r < b

2. Fill in the blanks. (3 × 1 = 3)

(i) 2 + 2 is a/an _______________ .

(ii) If HCF (a, b) = 2 and LCM (a, b) = 27, the value of ab is _______________ .

(iii) A rational number can be expressed as terminating decimal when the factors of the
denominator are in the form _______________ .

3. Solve the following questions. (3 × 1 = 3)

(i) If the LCM of x and 18 is 36 and the HCF of x and 18 is 2 then find out the value of x.

(ii) Find the least number which is a perfect square and is divisible by each of 16, 20 and 24.

(iii) What is the least value of r in division lemma a = bq + r?

Section B

QQ Solve the following questions. (3 × 2 = 6)
4. Prove that 2 is an irrational number. [CBSE 2019 (30/1/2)]
5. Using Euclid’s division algorithm, find HCF of 2048 and 960. [CBSE 2019 (30/2/2)]
6. Show that 9n cannot end with digit 0 for any n∈ N.

QQ Solve the following questions. (3 × 3 = 9)
7. Find HCF of (x2 – 3x +2) and (x2 – 4x + 3).

8. Find the least number that is divisible by first five even numbers.

9. Show that 2+ 3 is an irrational number. [CBSE 2019 (30/1/2)]
2

QQ Solve the following questions. (3 × 5 = 15)

10. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m,
9m + 1 or 9m + 8.

OR

Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given

numbers. [CBSE 2018]

11. Show that one and only one out of n, n + 2, n + 4 is divisible by 3, where n is any positive integer.

12. Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any
integer q.

Answers

1. (i) (c) (ii) (b) (iii) (b) (iv) (c)

2. (i) irrational number (ii) 54 (iii) 2n5m

3. (i) x = 4 (ii) 3600 (iii) 0

5. 64 7. x – 1 8. 120 10. OR HCF = 4, LCM = 9696

zzz

Real Numbers 19

2 Polynomials

BASIC CONCEPTS – A FLOW CHART

e

20 Xam idea Mathematics–X

Polynomials 21

MORE POINTS TO REMEMBER

 Value of a Polynomial: If p(x) is a polynomial and ‘a’ is any real number, then the real
number obtained by replacing x by ‘a’ in
p(x), is called the value of p(x) at x = a and is
represented by p(a).

For example, Let p(x) ≡ x2 + 2x + 5 be a
polynomial. The value of p(x) at 3 denoted by
p(3) is given by p(3) = 32 + 2 × 3 + 5 = 20.

 Graph of a Polynomial: The graph of a
polynomial p(x) is a free handed curve passing
through points (x1, y1), (x2, y2), (x3, y3) .........
etc., where y1, y2, y3 ....... are the values of
polynomial p(x) at x1, x2, x3 ....... respectively.

For example, Graph of p(x) ≡ 2x +1 (Fig 2.1)

 Geometrical meaning of the zeros of a Fig. 2.1
Polynomial: The zeros of a polynomial are
the x-coordinates of the points where the
graph of polynomial intersects or touches
x-axis.

For example, In the following polynomial p(x), 1, 4, 7 are zeros of p(x). (Fig 2.2)

p(x)

Fig. 2.2

Multiple Choice Questions [1 mark]

Choose and write the correct option in the following questions.

1. If 5 is a zero of the quadratic polynomial, x2 – kx – 15 then the value of k is

(a) 2 (b) –2 (c) 4 (d) – 4

2. The zeros of the quadratic polynomial x2 + ax + b, a, b > 0 are

(a) both positive (b) both negative

(c) one positive one negative (d) can’t say

3. If the zeros of the quadratic polynomial x2 + (a + 1) x + b are 2 and –3, then [NCERT Exemplar]

(a) a = – 7, b = – 1 (b) a = 5, b = – 1

(c) a = 2, b = – 6, (d) a = 0, b = – 6

22 Xam idea Mathematics–X

4. Given that one of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is zero, the product of

the other two zeros is [NCERT Exemplar]
c c b
(a) – a (b) a (c) 0 (d) – a

5. If one of the zeros of a quadratic polynomial of the form x2 + ax + b is the negative of the
other, then it

(a) has no linear term and the constant term is negative.

(b) has no linear term and the constant term is positive.

(c) can have a linear term but the constant term is negative.

(d) can have a linear term but the constant term is positive.

6. If one of the zeros of the quadratic polynomial (k – 1) x2 + k x + 1 is – 3, then the value of k

is [NCERT Exemplar]

(a) 4 (b) –4 (c) 2 (d) –2
3 3 3 3

7. The number of polynomials having zeros as – 2 and 5 is [NCERT Exemplar]

(a) 1 (b) 2 (c) 3 (d) more than 3

8. If one of the zeros of the cubic polynomial x3 +ax2 + bx + c is – 1, then the product of the other

two zeroes is [NCERT Exemplar]

(a) b – a + 1 (b) b – a – 1 (c) a – b + 1 (d) a – b – 1

9. If one root of the polynomial p(y) = 5y2 + 13y + m is reciprocal of other, then the value of m is

(a) 6 (b) 0 (c) 5 (d) 1
5

10. Given that two of the zeros of the cubic polynomial ax3 + bx2 + cx + d are 0, the value of c is

(a) less than 0 (b) greater than 0 (c) equal to 0 (d) can’t say

11. A quadratic polynomial with 3 and 2 as the sum and product of its zeros respectively is

(a) x2 + 3x – 2 (b) x2 – 3x + 2 (c) x2 – 2x + 3 (d) x2 – 2x – 3

12. A quadratic polynomial, whose zeros are 5 and – 8 is

(a) x2 + 13x – 40 (b) x2 + 4x – 3 (c) x2 – 3x + 40 (d) x2 + 3x – 40

13. The number of polynomials having zeros 1 and –2 is

(a) 1 (b) 2 (c) 3 (d) more than 3

14. The zeros of the quadratic polynomial x2 + kx + k, k ! 0

(a) both cannot be positive (b) both cannot be negative

(c) are always equal (d) are always unequal

15. The degree of the remainder r(x) when p (x) = bx3 + cx + d is divided by a polynomial of
degree 4 is

(a) less than 4 (b) less than 3

(c) equal to 3 (d) less than or equal to 3

16. If the graph of a polynomial intersects the x-axis at exactly two points, then it
(a) cannot be a linear or a cubic polynomial
(b) can be a quadratic polynomial only
(c) can be a cubic or a quadratic polynomial
(d) can be a linear or a quadratic polynomial

Polynomials 23

17. If one zero of the quadratic polynomial x2–5x + k is – 4 , then the value of k is

(a) 36 (b) –36 (c) 18 (d)–18

18. If a polynomial of degree 6 is divided by a polynomial of degree 2, then the degree of the
quotient is

(a) less than 4 (b) less than 2 (c) equal to 2 (d) equal to 4

19. Which of the following is not the graph of a quadratic polynomial? [NCERT Exemplar]

(a) (b)

(c) (d)

Fig. 2.3 (a to d)

20. A quadratic polynomial with sum and product of its zeros as 8 and –9 respectively is
(a) x2 –8x + 9 (b) x2 – 8x – 9 (c) x2 + 8x – 9 (d) x2 + 8x + 9

ANSWERS

1. (a) 2. (b) 3. (d) 4. (b) 5. (a) 6. (a)
7. (d) 8. (a) 9. (c) 10. (c) 11. (b) 12. (d)
13. (d) 14. (a) 15. (c) 16. (c) 17. (b) 18. (d)
19. (d) 20. (b)

Fill in the Blanks [1 mark]

Complete the following statements with appropriate word(s) in the blank space(s).

1. _________________ is not equal to zero when the divisor is not a factor of dividend.
2. The algebraic expression in which the variable has non-negative integral exponents only is

called _________________ .
3. A _________________ is a polynomial of degree 0.
4. We get the original polynomial if we multiply the _________________ together.
5. The highest power of a variable in a polynomial is called its _________________.

24 Xam idea Mathematics–X

6. A _________________ polynomial is of degree one.
7. A cubic polynomial is of degree _________________ .
8. _________________ equation is valid for all values of its variables.
9. A linear polynomial is represented by a _________________ .
10. Degree of remainder is always _________________ than degree of divisor.
11. Degree of a _________________ polynomial is not defined.
12. Graph of a quadratic polynomial is a _________________ .
13. Sum of zeros of a quadratic polynomial bx2 + cx + a is _________________ .
14. A quadratic polynomial can have atmost _________________ zeros.
15. Number of zeros that a polynomial f(x) = (x – 2)2 +4 can have is _________________ .

Answers

1. Remainder 2. Polynomial 3. Constant 4. Factors 5. Degree 6. Linear
11. zero
7. Three 8. Identity 9. straight line 10. smaller / less
–c [1 mark]
12. Parabola 13. b 14. 2 15. two

Very Short Answer Questions

The graphs of y = p(x) for some polynomials (for questions 1 to 4) are given below. Find the number

of zeros in each case. [NCERT]

1. 2.

Fig. 2.4 Fig. 2.5
3. 4.

Fig. 2.6 Fig. 2.7

Polynomials 25

Sol. 1. There is no zero as the graph does not intersect the x-axis.
2. The number of zeros is four as the graph intersects the x-axis at four points.
3. The number of zeros is three as the graph intersects the x-axis at three points.
4. The number of zeros is three as the graph intersects the x-axis at three points.

Answer the following questions and justify your answer: (Q. 5 — Q. 7)

5. What will be the quotient and remainder on division of ax2 + bx + c by px3 + qx2 + rx + 5, p ≠ 0?
Sol. 0, ax2 + bx + c; as degree of divisor is greater than degree of dividend.

6. If on division of a polynomial p(x) by a polynomial g(x), the quotient is zero, what is the
relation between the degrees of p(x) and g(x)?

Sol. Since the quotient is zero, therefore

deg p(x) < deg g(x)

7. Can x – 2 be the remainder on division of a polynomial p(x) by x + 3?

Sol. No, as degree (x – 2) = degree (x + 3)

8. Find the quadratic polynomial whose zeros are –3 and 4. [NCERT Exemplar]

Sol. Sum of zeros = –3 + 4 = 1,

Product of zeros = – 3 × 4 = –12

\ Required polynomial = x2 – x – 12

9. If one zero of the quadratic polynomial x2 – 5x – 6 is 6 then find the other zero.

Sol. Let a, 6 be the zeros of given polynomial.

Then a + 6 = 5 ⇒ a = –1 .

10. If both the zeros of the quadratic polynomial ax2 + bx + c are equal and opposite in sign, then
find the value of b.

Sol. Let a and –a be the roots of given polynomial.

Then a + (– a) = 0 ⇒ – b =0 ⇒ b = 0.
a

11. What number should be added to the polynomial x2 – 5x + 4, so that 3 is the zero of the

polynomial?

Sol. Let f(x) = x2 – 5x + 4
Then f(3) = 32 – 5 × 3 + 4 = –2

For f(3) to be zero, 2 must be added to f(x).

12. Can a quadratic polynomial x2 + kx + k have equal zeros for some odd integer k > 1?

Sol. No, for equal zeros, k2 – 4k = 0 or k = 0, 4

⇒ k should be even.

13. If the zeros of a quadratic polynomial ax2 + bx + c are both negative, then can we say a, b and

c all have the same sign? Justify your answer. c
a
Sol. Yes, because – b = sum of zeros < 0, so that b 2 0. Also the product of the zeros = 2 0.
a a

14. If the graph of a polynomial intersects the x-axis at only one point, can it be a quadratic

polynomial?

Sol. Yes, because every quadratic polynomial has at the most two zeros.

15. If the graph of a polynomial intersects the x-axis at exactly two points, is it necessarily a
quadratic polynomial?

Sol. No, x4 – 1 is a polynomial intersecting the x-axis at exactly two points.

26 Xam idea Mathematics–X

Short Answer Questions-I [2 marks]

1. If one of the zeros of the quadratic polynomial f(x) = 4x2 – 8kx – 9 is equal in magnitude but
opposite in sign of the other, find the value of k.

Sol. Let one root of the given polynomial be a.

Then the other root = – a

\ Sum of the roots = (– a) + a = 0

⇒ –b =0 or 8k =0 or k=0
a 4

2. If one of the zeros of the quadratic polynomial (k + 2) x2 + kx + 4 is – 4, then find the value of

k.

Sol. Since – 4 is a zero of the given polynomial

\ (k + 2) (– 4)2 + k(– 4) + 4 = 0

⇒ 16 k + 32 – 4k + 4 = 0

12k = – 36

k=–3

3. If 1 is a zero of the polynomial p(x) = ax2 – 3(a – 1)x –1, then find the value of a.

Sol. Put x = 1 in p(x)
\ p(1) = a(1)2 – 3(a – 1) × 1 – 1 = 0

⇒ a – 3a + 3 – 1 = 0 ⇒ –2a = –2 ⇒ a = 1

4. If a and b are zeros of polynomial p(x) = x2 – 5x + 6, then find the value of a + b – 3ab.

Sol. Here, a + b = 5, ab = 6

⇒ a + b –3ab = 5 – 3 × 6 = –13

5. Find the zeros of the polynomial p(x) = 4x2 – 12x + 9.

Sol. p(x) = 4x2 – 12x + 9 = (2x – 3)2

For zeros, p(x) = 0

⇒ (2x – 3)(2x – 3) = 0
⇒ 3 3
x= 2 , 2 .

6. If one root of the polynomial p(y) = 7y2 + 14y + m is reciprocal of other, then find the value

of m. 1 1 m m
a a 7 7
Sol. Let the roots be a and . Then aa k= or 1= or m = 7

7. If a and b are zeros of p(x) = x2 + x – 1, then find 1 + 1 .
a b

Sol. Here, a + b = –1, ab = –1, so 1 + 1 = b+a = –1 =1
a b ab –1

8. If α and β are zeros of the polynomial f(x) = ax2 +bx + c, then find 1 + 1 .
Sol. Given : f(x) = ax2 + bx + c a2 b2

a + b = –b and ab = c
a a

a12 + 1 = b2 + a2 = ^a + bh2 – 2ab
b2 ^a bh2 ^abh2

= > ab22 – 2c H× a2 = b2 – 2ac × a2 = b2 – 2ac
a c2 a2 c2 c2

Polynomials 27

9. If the product of two zeros of the polynomial p(x) = 2x3 + 6x2 – 4x + 9 is 3, then find its third
zero.

Sol. Let a, b, γ be the roots of the given polynomial and ab = 3

Then abc = – 9 ⇒ 3 # c = –9 or γ = −3
2 2 2

10. Find the quadratic polynomial sum and product of whose zeros are –1 and –20 respectively.

Also find the zeros of the polynominal so obtained. [CBSE 2019, (30/4/2)]

Sol. Let a and b be the zeros of the quadratic polynomial.

∴ Sum of zeros, a + b = – 1

and product of zeros, a . b = – 20

Now, quadratic polynomial be

x2 – (a + b) . x + ab

= x2 – (–1) x – 20

= x2 + x – 20

Now, for zeros of this polynominal

x2 + x – 20 = 0 ⇒ x2 + 5x – 4x – 20 = 0

⇒ x (x + 5) – 4 (x + 5) = 0 ⇒ (x + 5) (x – 4) = 0

⇒ x = – 5, 4

∴ zeros are – 5 and 4

Short Answer Questions-II [3 marks]

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the
coefficients (Q. 1 - Q. 2).

1. 6x2 – 3 – 7x [NCERT]

Sol. We have, p(x) = 6x2 – 3 – 7x

p(x) = 6x2 – 7x – 3 (In general form)

= 6x2 – 9x + 2x – 3 = 3x (2x – 3) + 1 (2x – 3)

= (2x – 3) (3x + 1)

The zeros of polynomial p(x) is given by

p(x) = 0 ⇒ (2x – 3) (3x + 1) = 0 ⇒   x = 3 , – 1
2 3
3 1
Thus, the zeros of 6x2 – 7x – 3 are a = 2 and b = – 3

Now, sum of the zeros = a +b = 3 – 1 = 9 – 2 = 7
2 3 6
6

and (Coefficient of x) = – (–7) = 7
Coefficient of x2 6 6
– (Coefficient of x)
Therefore, sum of the zeros = Coefficient of x2

Again, product of zeros = a . b = 3 # d – 1 n = – 1
2 3 2

and Constant term = –3 = – 1
Coefficient of x2 6 2

28 Xam idea Mathematics–X

Therefore, product of zeros = Constant term Hence verified.
Coefficient of x2 [NCERT]
2. 4u2 + 8u

Sol. We have, p(u) = 4u2 + 8u ⇒ p(u) = 4u (u + 2)

The zeros of polynomial p(u) is given by

p(u) = 0 ⇒ 4u (u + 2) = 0

\ u = 0, –2

Thus, the zeros of 4u2 + 8u are a = 0 and b = –2

Now, sum of the zeros = a + b = 0 – 2 = – 2

and – (Coefficient of u) = –8 = –2
Coefficient of u2 4

Therefore, sum of the zeros = – (Coefficient of u)
Coefficient of u2

Again, product of the zeros = a . b = 0 × (–2) = 0

and Constant term = 0 = 0
Coefficient of u2 4

Therefore, product of zeros = Constant term
Coefficient of u2

Hence verified

3. Check whether the first polynomial is a factor of the second polynomial by dividing the second

polynomial by the first polynomial:

(i) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2 (ii) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12 [NCERT]

Sol. (i) We have, 3x2 – 4x + 2

gx2 + 3x + 1 3x4 + 5x3 – 7x2 + 2x + 2

3x4 ! 9x3 ! 3x2

–4x3 – 10x2 + 2x

" 4x3 " 12x2 " 4x
2x2 + 6x + 2

2x2 ! 6x ! 2

0
Clearly, remainder is zero, so x2 + 3x + 1 is a factor of polynomial 3x4 + 5x3 – 7x2 + 2x + 2.

(ii) We have, 2t2 + 3t + 4


gt2 – 3 2t4 + 3t3 – 2t2 – 9t – 12

2t4 " 6t2
3t3 + 4t2 – 9t

3t3 " 9t
4t2 – 12

4t2 "12
0

  Clearly, remainder is zero, so t2 – 3 is a factor of polynomial 2t4 + 3t3 – 2t2 – 9t – 12.

Polynomials 29

4. If a and b are the zeros of the quadratic polynomial f(x) = 2x2 – 5x + 7, find a polynomial
whose zeros are 2a + 3b and 3a + 2b.

Sol. Since a and b are the zeros of the quadratic polynomial f(x) = 2x2 – 5x + 7

\ a+ b = – (–5) = 5 and ab = 7
2 2 2
Let S and P denote respectively the sum and product of the zeros of the required polynomial.

Then, S = (2a + 3b) + (3a + 2b) = 5(a + b) = 5# 5 = 25
2 2
and P = (2a + 3b) (3a + 2b)

⇒ P = 6a2 + 6b2 + 13ab = 6a2 + 6b2 + 12ab + ab

= 6(a2 + b2 + 2ab) + ab = 6(a + b)2 + ab

⇒ P = 6 # c 5 2 + 7 = 75 + 7 = 41
2 m 2 2 2

Hence, the required polynomial g(x) is given by

g(x) = k(x2 – Sx + P)

or g(x) = kc x2 – 25 x + 41m , where k is any non-zero real number.
2

5. What must be subtracted from p(x) = 8x4 + 14x3 – 2x2 + 7x – 8 so that the resulting polynomial
is exactly divisible by g(x) = 4x2 + 3x – 2?

Sol. Let y be subtracted from polynomial p(x)

\ 8x4 + 14x3 – 2x2 + 7x – 8 – y is exactly divisible by g(x)

Now, 2x2 + 2x – 1

g4x2 + 3x – 2 8x4 + 14x3 – 2x2 + 7x – 8 – y

8x4 ! 6x3 " 4x2
8x3 + 2x2 + 7x – 8 – y

8x3 ! 6x2 " 4x

– 4x2 + 11x – 8 – y

" 4x2 " 3x ! 2
14x – 10 – y

 Remainder should be 0.

\ 14x – 10 – y = 0

or 14x – 10 = y or y = 14x – 10

\ Hence (14x – 10) should be subtracted from p(x) so that it will be exactly divisible by g(x).
6. What must be added to f(x) = 4x4 + 2x3 – 2x2 + x – 1 so that the resulting polynomial is

divisible by g(x) = x2 + 2x – 3?

Sol. By division algorithm, we have

f(x) = g(x) × q(x) + r(x)

⇒ f(x) – r(x) = g(x) × q(x)   ⇒ f(x) + {–r(x)} = g(x) × q(x)

Clearly, RHS is divisible by g(x). Therefore, LHS is also divisible by g(x). Thus, if we add –r(x) to
f(x), then the resulting polynomial is divisible by g(x). Let us now find the remainder when f(x) is
divided by g(x).

30 Xam idea Mathematics–X

4x2 – 6x + 22
gx2 + 2x – 3 4x4 + 2x3 – 2x2 + x – 1

4x4 ! 8x3 " 12x2

–6x3 + 10x2 + x – 1

" 6x3 "12x2 ! 18x
22x2 – 17x – 1

22x2 ! 44x " 66

\ –61x + 65
r(x) = –61x + 65 or –r(x) = 61x – 65

Hence, we should add –r(x) = 61x – 65 to f(x) so that the resulting polynomial is divisible by g(x).

7. Find the zeros of the quadratic polynomial 7y2 – 11 y – 2 and verify the relationship between
zeros and the cofficients. 3 3 [CBSE 2019, (30/2/1)]

Sol. Given: p(x) = 7y2 – 11 y – 2
3 3
1 1
= 3 (21y2 –11y –2) = 3 (21y2 – 14y + 3y – 2)

= 1 [7y (3y – 2) + 1 (3y – 2)] = 1 (7y + 1) (3y – 2)
3 3

Equating p(x) = 0

⇒ 1 (7y + 1) (3y – 2) = 0 ⇒ y = –1 and y = 2
Now, 3 7 3
–11
–b – c 3 m 11
a 21
sum of zeros = = 7 =

= 2 – 1 = 14 – 3 = 11
3 7 21 21

and product of zeros = c = – 2 = –2
a 3 21

7

= –1 × 2 = –2 Hence verified.
7 3 21

8. If a and b are the zeros of the quadratic polynomial f(x) = x2 – 4x +3, find the value of

a4 b2 + a2 b4 . [CBSE 2019, (30/4/1)]

Sol. Given quadratic polynomial is

f(x) = x2 – 4x + 3 – (–4)
1
∴ Sum of zeros =a+b = = 4

⇒ α + β = 4 ab = 3 = 3
and product of zeros, 1

 a2 + b2 = (a + b)2 –2ab

⇒ a2 + b2 = (4)2 –2×3 = 16 – 6 = 10

Now, a4 b2 + a2 b4 = a2 b2 (a2 + b2) = (3)2 ×10 = 9 ×10 = 90

∴ a4 b2 + a2 b4 = 90

Polynomials 31

9. If one zero of the polynomial 3x2 – 8x + 2k + 1 is seven times the other, find the value of k.

Sol. Let a and b be the zeros of the polynomial. Then as per question b = 7a

Now sum of zeros = a + b = a + 7a = –d –8 n
3
8 1
⇒ 8 a = 3 or a = 3
and
a#b = a # 7a = 2k +1
3
2
⇒ 7a2 = 2k +1 ⇒ 7c 1 m = 2k +1 ca a = 1 m
3 3 3 3

⇒ 7 = 2k +1 ⇒ 7 = 2k +1
9 3 3

⇒ 7 – 1= 2k ⇒ k = 2
3 3

10. For what value of k, is the polynomial f(x) = 3x4– 9x3 + x2+15x + k completely divisible by
3x2 – 5?

Sol. f (x) = 3x4 –9x3 + x2 + 15x + k

g (x) = 3x2 – 5

Since f(x) is completely divisible by g(x), r (x) = 0 therefore by long division

3x2 – 5 3x4 – 9x3 + x2 +15x + k x2 – 3x – 2
–3x4 +– 5x2

–9x3+6x2+15x + k
+–9x3
+– 15x

6x2 + k
–6x2 +– 10

But r(x) = 0 k + 10 = r(x)

∴ k+10 = 0 ⇒ k = –10

11. If one zero of polynomial (a2 + 9)x2 + 13x + 6a is reciprocal of the other, find the value of a.

Sol. Let one zero of the given polynomial be a.

Then, the other zero is 1
a

\ Product of zeros = a # 1 =1
a

But, as per the given polynomial product of zeros = 6a
a2 +9
6a
\ a2 + 9 =1 ⇒ a2 + 9 = 6a

⇒ a2 – 6a + 9 = 0 ⇒ (a – 3)2 = 0

⇒ a – 3 = 0 ⇒ a = 3

Hence, a = 3.

12. If the polynomial (x4 + 2x3 + 8x2 + 12x + 18) is divided by another polynomial (x2 + 5), the
remainder comes out to be (px + q). Find values of p and q.

Sol. Let f(x) = (x4 + 2x3 + 8x2 + 12x + 18) and g(x) = (x2 + 5)

On dividing f(x) by g(x), we get



32 Xam idea Mathematics–X

x2 + 2x + 3

gx2 + 5 x4 + 2x3 + 8x2 + 12x + 18

x4 ! 5x2

2x3 + 3x2 + 12x

2x3 ! 10x

3x2 + 2x + 18

3x2 ! 15
2x + 3

Now, px + q = 2x + 3 ⇒ p = 2, q = 3 (By comparing the coefficient of x and constant term).

Long Answer Questions [5 marks]

1. Verify that the numbers given alongside the cubic polynomial below are its zeros. Also verify

the relationship between the zeros and the coefficients. [NCERT]

x3 – 4x2 + 5x – 2; 2, 1, 1

Sol. Let p(x) = x3 – 4x2 + 5x – 2

On comparing with general polynomial p(x) = ax3 + bx2 + cx + d, we get a = 1, b = –4, c = 5 and

d = –2.

Given zeros 2, 1, 1.
p(2) = (2)3 – 4(2)2 + 5(2) – 2 = 8 – 16 + 10 – 2 = 0
\

and p(1) = (1)3 – 4(1)2 + 5(1) – 2 = 1 – 4 + 5 – 2 = 0.

Hence, 2, 1 and 1 are the zeros of the given cubic polynomial.
Again, consider a = 2, b =1, g = 1

\ a + b + g = 2 + 1 + 1 = 4
and – (Coefficient of x2
a+b+c= Coefficient of x3 ) = –b = – (–4) = 4
a 1

ab + bg + ga = (2) (1) + (1) (1) + (1) (2) = 2 + 1 + 2 = 5

and ab + bg + ga = Coefficient of x = c = 5 =5
Coefficient of x3 a 1

abg = (2) (1) (1) = 2

and abc = – (Constant term) = –d = – (–2) =2
Coefficient of x3 a 1

2. Find a cubic polynomial with the sum of the zeros, sum of the products of its zeros taken two

at a time, and the product of its zeros as 2, –7, –14 respectively. [NCERT]

Sol. Let the cubic polynomial be p(x) = ax3 + bx2 + cx + d. Then

Sum of zeros = –b =2 c
a a
Sum of the products of zeros taken two at a time = = –7

and product of the zeros = –d = –14
a

⇒ b = –2, c = –7, – d = –14 or d =14
\ a a a a
b c d
p(x) = ax3 + bx2 + cx + d   ⇒ p (x)= a:x3 + a x2 + a x + a D

Polynomials 33

p(x) = a[x3 + (–2)x2 + (–7)x + 14] ⇒ p(x) = a[x3 – 2x2 – 7x + 14]

For real value of a = 1, p(x) = x3 – 2x2 – 7x + 14

3. Find the zeros of the polynomial f(x) = x3 – 5x2 – 2x + 24, if it is given that the product of its
two zeros is 12.

Sol. Let a, b and g be the zeros of polynomial f(x) such that ab = 12.

We have, a+b+c= –b = – (– 5) =5
a 1

ab +bc + ca = c = –2 = –2 and abc = –d = –24 = –24
a 1 a 1

Putting ab = 12 in abg = –24, we get

12g = –24 ⇒ c=– 24 =– 2
12

Now, a + b + g = 5 ⇒ a + b – 2 = 5

⇒ a + b = 7 ⇒ a=7–b
⇒ 7b – b2 = 12
(7 – b) b = 12 [ ab = 12] ⇒

⇒ b2 – 7b + 12 = 0 ⇒ b2 – 3b – 4b + 12 = 0

⇒ b(b – 3) – 4 (b – 3) = 0 ⇒ (b – 4) (b – 3) = 0 ⇒ b = 4 or b = 3

\ a = 3 or a = 4

So, zeroes of f(x) are 3, 4, – 2

4. If the remainder on division of x3 – kx2 + 13x – 21 by 2x – 1 is –21, find the quotient and the
value of k. Hence, find the zeros of the cubic polynomial x3 – kx2 + 13x.

Sol. Let f(x) = x3 – kx2 + 13x – 21

T hen, fc 1 m = –21 ⇒ c 1 3 kc 1 2 + 13c 1 m – 21= – 21
2 2 2 2
m– m

or 1 – 1 k + 13 – 21+ 21= 0   or   k = 53 & k = 53 1 x2 – 13x
8 4 2 4 8 2 2
g2x –1
\ f (x)= x3 – 53 x2 +13x – 21 x3 – 53 x2 + 13x
2 2

Now, f(x) = q(x)(2x – 1) – 21 x3 " 1 x2
2
⇒ x3 – 53 x2 +13x – 21= q (x) (2x – 1) – 21
2 –26x2 + 13x

⇒ c x3 – 53 x2 +13xm ' (2x – 1)= q (x) " 26x2 ! 13x
2 0

i.e., x3 – 53 x2 +13x =(2x – 1) c 1 x2 – 13xm= 1 x (2x – 1) (x – 26)
2 2 2

For zeros, x3 – 53 x2 +13x = 0 ⇒ 1 x (2x – 1) (x – 26) = 0 ⇒ x = 0, 1 , 26
2 2 2

5. Obtain all other zeros of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeros are 5 and – 5 .
3 3

5 5 5 [NCERT]
5 5 3 3 x2 3 is a factor of the
Sol. Since two zeros are 3 and – , so cx – m c x + m= –
given polynomial. 3

Now, we divide the given polynomial by c x2 – 5 m to obtain other zeros.
3

34 Xam idea Mathematics–X

gx2 –5 3x2 + 6x + 3
3 3x4 + 6x3 –2x2 – 10x – 5

3x4 " 5x2

6x3 + 3x2 – 10x
6x3 " 10x

3x2 – 5
3x2 "5

0 5
3
So, 3x4 + 6x3 – 2x2 – 10x – 5 = c x2 – m (3x2 + 6x + 3)

Now, 3x2 + 6x + 3 = 3(x2 + 2x + 1) = 3(x + 1)2 = 3(x + 1)(x + 1)

So its zeros are –1, –1.

Thus, all the zeros of given polynomial are 5 / 3 , – 5 / 3 , –1 and –1.

6. Given that 2 is a zero of the cubic polynomial 6x3 + 2 x2–10x – 4 2 , find its other zeros.

[NCERT Exemplar]

Sol. The given polynomial is f(x) = (6x3 + 2 x2 – 10x – 4 2) . Since 2 is the zero of f(x), it follows

that (x – 2) is a factor of f(x).

On dividing f(x) by (x – 2) , we get

6x2 + 7 2 x + 4

gx – 2 6x3 + 2 x2 – 10x – 4 2
6x3 +– 6 2 x2
7 2 x2 – 10x

7 2 x2 +– 14x

4x – 4 2

\ +– 4x +– 4 2
0

f (x)= 0 & (x – 2) (6x2 +7 2 x + 4)=0 & (x – 2) (3 2 x + 4) ( 2 x +1)=0

x – 2 = 0, 3 2 x + 4 = 0, 2 x +1= 0

Hence, x = 2, x=– 2 2 , x= –2 and all zeros of f(x) are 2, –2 2 , –2 .
3 2 3 2

HOTS [Higher Order Thinking Skills]

1. If a, b, g be zeros of polynomial 6x3 + 3x2 – 5x + 1, then find the value of a–1 + b–1 + g–1.

Sol. p(x) = 6x3 + 3x2 – 5x + 1 so a = 6, b = 3, c = –5, d = 1

 a, b and g are zeros of the polynomial p(x).
\
a+b+c= –b = –3 = –1
a 6 2

ab + ac +bc = c = –5 and abc = –d = –1
a 6 a 6

Now a–1 + b–1 + c–1 = 1 + 1 + 1 = bc + ac + ab = – 5/6 =5
a b c abc – 1/6

2. Find the zeros of the polynomial f(x) = x3 – 12x2 + 39x – 28, if it is given that the zeros are in AP.

Sol. Let a, b, g are the zeros of f(x). If a, b, g are in AP, then,

Polynomials 35

b – a = g – b ⇒ 2b = a + g …(i)

a +b + c = –b = – (–12) =12 ⇒ a + g = 12 – b …(ii)
a 1
From (i) and (ii)

2b = 12 – b or 3b = 12 or b = 4

Putting the value of b in (i), we have

8 = a + g …(iii)
– (– 28) …(iv)
abc = – d = 1 = 28
a
7
(a g) 4 = 28 or a g = 7 or c= a

Putting the value of c = 7 in (iii), we get
a
7
⇒ 8=a+ a ⇒ 8a = a2 + 7

⇒ a2 – 8a + 7 = 0 ⇒ a2 – 7a – 1a + 7 = 0

⇒ a(a – 7) –1 (a – 7) = 0 ⇒ (a – 1) (a – 7) = 0

⇒ a = 1 or a = 7

Putting a = 1 in (iv), we get Putting a = 7 in (iv), we get

c= 7 c = 7
or 1 7
g = 7 or g = 1

and b = 4 and b = 4

\ zeros are 1, 7, 4. \ zeros are 7, 4, 1.

Hence zeros are 1, 4, 7 or 7, 4, 1

3. If the polynomial f(x) = x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial
x2 – 2x + k, the remainder comes out to be x + a. Find k and a.
[NCERT]

Sol. By division algorithm, we have Dividend = Divisor × Quotient + Remainder
⇒ Dividend – Remainder = Divisor × Quotient
⇒ Dividend – Remainder is always divisible by the divisor.
When f(x)=x4 – 6x3+16x2 – 25x+10 is divided by x2 – 2x+k the remainder comes out to be x + a.

\ f(x) – (x + a) = x4 – 6x3 + 16x2 – 25x + 10 – (x + a)

= x4 – 6x3 + 16x2 – 25x + 10 – x – a = x4 – 6x3 + 16x2 – 26x + 10 – a

is exactly divisible by x2 – 2x + k

Let us now divide x4 – 6x3 + 16x2 – 26x + 10 – a by x2 – 2x + k.

x2 – 4x + (8 – k)

gx2 – 2x + k x4 – 6x3 + 16x2 – 26x + 10 – a
x4 " 2x3 ! kx2
– 4x3 + (16 – k) x2 – 26x + 10 – a

" 4x3 ! 8x2 " 4kx

(8 – k) x2 – (26 – 4k) x + 10 – a

! (8 – k) x2 " (16 – 2k) x ! (8k – k2)

(– 10 + 2k) x + (10 – a – 8k + k2)
For f(x) – (x + a) = x4 – 6x3 + 16x2 – 26x + 10 – a to be exactly divisible by x2 – 2x + k, we must

have (–10 + 2k) x + (10 – a – 8k + k2) = 0 for all x
⇒   –10 + 2k = 0 and 10 – a – 8k + k2 = 0
⇒  k = 5 and 10 – a – 40 + 25 = 0
⇒  k = 5 and a = –5

36 Xam idea Mathematics–X

PROFICIENCY EXERCISE

QQ Objective Type Questions: [1 mark each]

1. Choose and write the correct option in each of the following questions.

(i) If 2 and α are zeros of 2x2 – 6x + 2 then the value of α is

(a) 2 (b) 3 (c) 1 (d) 5

(ii) If two of the zeros of the cubic polynomial ax3 + bx2 + cx + d are 0, then the third zero is

(a) b (b) c (c) –d (d) –b
a a a a

(iii) The product of the zeros of the polynomial 4x2 + 3x + 7 is

(a) 3 (b) – 3 (c) 7 (d) –7
4 4 4 4

(iv) A quadratic polynomial with sum and product of its zeros as 8 and – 9 respectively is

(a) x2 – 8x + 9 (b) x2 – 8x – 9 (c) x2 + 8x – 9 (d) x2 + 8x + 9

(v) If one of the zeros of the quadratic polynomial (k – 1) x2 + kx + 1 is – 3, then the value of k

is 4 (b) –4 (c) 2 (d) –2
(a) 3 3 3 3

2. Fill in the blanks.

(i) If the product of zeros of the polynomial x2 – 9x + a is 8, then its zeros are _____ and _____.

(ii) If the polynomial p(x) = x3 – kx2 + 3x + 2 is exactly divisible by (x + 1) then the value of k
is ___________ .

(iii) Zeros of the quadratic polynomial t2 – 15 are ___________ and ___________.

(iv) If β is a zero of f(x) then __________ is one of the factors of f(x).

(v) Graph of a cubic polynomial cuts x-axis at atmost ______________ points.

QQ Very Short Answer Questions: [1 mark each]
3. If one zero of the quadratic polynomial x2 + x – 2 is –2, find the other zero.
4. Find the other zero of the quadratic polynomial y2 + 7y – 60 if one zero is –12.

5. Find the quadratic polynomial whose zeros are – 3 and – 5.

6. Find the quadratic polynomial whose zeros are 2 and – 6.

7. What number should be added to the polynomial x2 + 7x – 35 so that 3 is the zero of the
polynomial?

The graph of y = p(x) for some polynomials (Q 8 to 9) are given below. Find the number of zeros in
each case.
8.

Polynomials 37

9.


10. Can y + 5 be the remainder on division of a polynomial f(y) by y – 2?
11. Can x2 – 1 be the quotient on division of x6 + 2x3 + x – 1 by a polynomial in x of degree 5?
12. If on division of a polynomial p(x) by a polynomial g(x), the quotient is zero, what is the relation

between the degrees of p(x) and g(x)?
13. If one zero of the quadratic polynomial p(x) = x2 + 4kx – 25 is negative of the other, find the value of k.

QQ Short Answer Questions–I: 1 1 [2 marks each]
a b
14. If a, b are the zeros of the polynomial f(x) = x2 – 3x + 2, then find + .

15. If a and 1 are the zeros of polynomial 4x2 – 2x + (k–4), then find the value of k.
a

16. If the sum of the zeros of the polynomial f(x) = 2x3 – 3kx2 + 4x – 5 is 6, then find the value of k.

17. If 1 is the zero of the quadratic polynomial x2 + kx – 5, then find the value of k.

18. Find the zeros of the polynomial 5y2 – 11y + 2.

19. If one of the zeros of the quadratic polynomial (k – 2)x2 – 2x – (k + 5) is 4, find the value of k.
1 1
20. If a, b are the zeros of the polynomial x2 + x – 6, find the value of a2 + b2 .

21. If the remainder on division of x3 + 2x2 + kx + 3 by x – 3 is 21, find the quotient and the value of k.

22. If a, b are the two zeros of the polynomial f(y) = y2 – 8y + a and a2 + b2 = 40, find the value of a.

23. If a and b are zeros of p(x) = x2 + x – 1, then find a2b + ab2.

QQ Short Answer Questions–II: [3 marks each]

24. If the sum of the zeros of the quadratic polynomial f(x) = kx2 + 2x + 3k is equal to their product,

find the value of k.

25. Find a quadratic polynomial each with the given numbers as the sum and product of the zeros

respectively.

(i) 2 , – 1 (ii) 0, – 4 3 (iii) –3 , –1 (iv) 21 , 5
3 3 25 2 8 16

Also find the zeros of those polynomials by factorisation.

26. Find the zeros of the following polynomials and verify the relationship between the zeros and

the coefficients of the polynomials. 11 2
(i) 3x2 + 4x – 4 (ii) 7y2 3 3
– y– (iii) p2 – 30

(iv) 3 x2 – 11x + 6 3 (v) a(x2 + 1) –x(a2 +1) (vi) 6x2 + x – 2

27. Check whether g(x) is a factor of p(x) by dividing the first polynomial by the second polynomial:
(i) p(x) = 4x3 + 8x + 8x2 + 7, g(x) = 2x3 – x + 1

38 Xam idea Mathematics–X

(ii) p(x) = x4 – 5x + 6, g(x) = 2 – x2
(iii) p(x) = 13x3 – 19x2 + 12x + 14, g(x) = 2 – 2x + x2
28. If (x – 2) is a factor of x3 + ax2 + bx + 16 and b = 4a, find the values of a and b.
29. If a and b are the zeros of the quadratic polynomial f(x) = 3x2 – 5x – 2, then evaluate a2 + b2.
30. If a and b are the zeros of the quadratic polynomial f(t) = t2 – p(t + 1) – c, show that

(a + 1) (b + 1) = 1 – c.
31. What must be subtracted from x3 – 6x2 + 13x – 6 so that the resulting polynomial is exactly

divisible by x2 + x + 1?

32. What must be added to f(x) = x4 + 2x3 – 2x2 + x – 1, so that the resulting polynomial is divisible

by g(x) = x2 + 2x – 3?

33. If the zeros of the quadratic polynomial x2 + (a + 1)x + b are 2 and – 3, then find a and b.
1
34. Find the zeros of the polynomial x2 + 6 x –2 and verify the relation between the coefficients

and the zeros of the polynomial. [NCERT Exemplar]

35. Obtain all zeros of 3x4 – 15x3 + 13x2 + 25x – 30 if two of its zeros are 5 and – 5 .
3 3

[CBSE 2018 (C) 30/1]

36. Find the value of k such that the polynomial x2 – (k + 6) x + 2 (2 k – 1) has sum of its zeros equal

to half of their product. [CBSE 2019 (C) 30/1/1]

37. Check whether g(x) is a factor of p(x) by dividing polynomial p(x) by polynomial g(x),

where p(x) = x5 – 4x3 + x2 + 3x +1, g(x) = x3 – 3x + 1. [CBSE 2019 (C) 30/3/1]

38. Find all the zeros of the polynomial x4 + x3 – 14x2 – 2x + 24 if two of its zeroes are 2 and – 2 .

[CBSE 2019 (C) 30/3/1]
39. Apply division algorithm to check if g(x) = x2 – 3x + 2 is a factor of the polynomial

f(x) = x4 – 2x3 – x + 2. [CBSE 2019 (C) 30/3/2]

40. Find all the zeroes of 2x4 – 13x3 + 19x2 + 7x – 3, if you know that two of its zeroes are

2 + 3 and 2 – 3 . [CBSE 2019 (C) 30/1/1]

QQ Long Answer Questions: [5 marks each]

41. Given that x – 5 is a factor of the cubic polynomial x3 – 3 5 x2 + 13x – 3 5, find all the zeros

of the polynomial. [NCERT Exemplar]

42. If a and b are zeros of polynomial f(x) = 2x2 + 11x + 5, then find

(i) a4 + b4 (ii) 1 + 1 – 2ab
a b

43. If a and b are the zeros of the polynomial f(x) = 4x2 – 5x + 1, find a quadratic polynomial whose

zeros are α2 and β2 .
βα

44. On dividing the polynomial f(x) = x3 – 5x2 + 6x – 4 by a polynomial g(x), the quotient and
remainder are x – 3 and –3x + 5 respectively. Find the polynomial g(x).

45. Verify that the numbers given alongside the cubic polynomials below are their zeros. Also verify

the relationship between the zeros and the coefficients. 1
2
(i) x3 – 2x2 – 5x + 6; –2, 1, 3 (ii) 2x3 + 7x2 + 2x – 3; –3, –1,
46. (i) Obtain all other zeros of 2x4 + 7x3 – 19x2 – 14x
+ 30, if two of its zeros are 2 and – 2 .

(ii) Obtain all other zeros of 2x3 + x2 – 6x – 3, if two of its zeros are – 3 and 3 .

Polynomials 39

Answers

1. (i) (c) (ii) (d) (iii) (c) (iv) (b) (v) (a)

2. (i) 1 ; 8 (ii) – 2 (iii) – 15; 15 (iv) (x – β) (v) three

3. 1 4. 5 5. x2 + 8x + 15 6. x2 + 4x – 12 7. 5

8. 1 149.. 032 10. No 11. No 12. deg p(x) < deg g(x)
19. k = 3
13. k = 0 15. k = 8 16. 4 17. 4
1 13
18. 5 , 2 20. 36 21. k = –9 and quotient = x2 + 5x + 6 22. a = 12

23. 1 24. k= –2 25. (i) 1 (3x2 – 2x – 1); –1 , 1 (ii) x2 − 4 11
3 3 3
3; 2(3)4 , −2(3)4

(iii) 21 5 (2 5 x2 +3x – 5); –5 , 1 (iv) 1 (16x2 - 42x + 5); 1 , 5
2 5 16 8 2

26. (i) –2, 2 (ii) –1 , 2 (iii) 30 , – 30 (iv) 2 , 3 3 (v) a, 1 (vi) –2 , 1
3 7 3 3 a 3 2

27. (i) No (ii) No (iii) Yes 28. a = –2, b = –8 29. 37 4 –3
31. 19x + 1 32. x – 2 33. a = 0, b = –6 9 3 2
34. zeroes are ,

35. 53, − 5 , 2 and 3 36. k = 7
3
37. Since remainder ≠ 0 ∴ g(x) is a factor of p(x), remainder = 2, Quotient = x2 – 1

38. All zeroes are – 4, 3, 2, – 2

39. Quotient = x2 + x + 1, Remainder = 0, yes g(x) is factor of f(x)

40. 2 + 3, 2 – 3, – 1 and 3. 41. 5, 5+ 2, 5, – 2 42. (i) 10001 (ii) –36
2 16 5

43. 1 (16x2 – 65x + 4) 44. x2 – 2x + 3 46. (i) 2, − 2, −5 and 3 (ii) 3, − 3 and 1
16 2 2

SELF-ASSESSMENT TEST

Time allowed: 1 hour Max. marks: 40

Section A

1. Choose and write the correct option in the following questions. (4 × 1 = 4)

(i) A quadratic polynomial, whose zeros are – 3 and 4, is
x2
(a) x2 – x + 12 (b) x2 + x + 12 (c) 2 – x – 6 (d) 2x2 + 2x – 24
2
(ii) The zeros of the quadratic polynomial x2 + 99x + 127 are

(a) both positive (b) both negative

(c) one positive and one negative (d) both equal
(iii) The zeros of the quadratic polynomial ax2 + bx + c, c ≠ 0 are equal, then

(a) c and a have opposite signs (b) c and b have opposite signs

(c) c and a have the same sign (d) c and b have the same sign

(iv) Product of zeros of a cubic polynomial is
–d c d –b
(a) a (b) b (c) b (d) a

2. Fill in the blanks. (3 × 1 = 3)

(i) Graph of a quadratic polynomial is shape of a _______________ if a > 0.
(ii) Sum of zeros of a cubic polynomial is _______________ .

40 Xam idea Mathematics–X

(iii) The value of polynomial p(x) at x = k is denoted as _______________ .

3. Solve the following questions. (3 × 1 = 3)

(i) If α and β are zeros of the quadratic polynomial f (x) = x2 – x –4, then find the value of

1 + 1 – ab .
a b

(ii) If the square of difference of the zeros of the quadratic polynomial x2 +px + 45 is equal

to 144. Find out the value of p. 1 + 1 – 1 is zero, then what is the product
x+a x+b c
(iii) If the sum of zeros of the polynomial
of zeros of the given polynomial?

Section B

QQ Solve the following questions. (3 × 2 = 6)

4. If a and b are the zeros of the quadratic polynomial f(x) = 3x2 – 5x – 2, then evaluate
b2
(i) a3 + b3 (ii) a2 + a
b

5. If the polynomial f(x) = ax3 + bx – c is divisible by the polynomial g(x) = x2 + bx + c, then find

the value of ab.
6. If one root of the polynomial f(x) = x2 + 5x + k is reciprocal of the other, find the value of k.

QQ Solve the following questions. (3 × 3 = 9)

7. If 2 and – 3 are the zeros of the polynomial ax2 + 7x + b, then find the values of a and b. – q .
8. If 3 + a) is a factor of two polynomials x2 + px + q and x2 + mx + n then prove that a= n – p
m
(x

9. Given that 3 is a zero of the polynomial x3 + x2 – 3x – 3, find its other two zeros.

QQ Solve the following questions. (3 × 5 = 15)

10. If a and b are the zeros of the quadratic polynomial f(x) = 3x2 – 7x – 6, find a polynomial whose
zeros are (i) a2 and b2 (ii) 2a + 3b and 3a + 2b

11. Find all zeros of the polynomial ( 2x4 – 9x3 + 5x2 + 3x – 1) if two of its zeros are (2 + 3) and

(2– 3) . [CBSE 2018 (30/1)]

12. If the zeros of the polynomial f(x) = x3 – 3x2 – 6x + 8 are of the form a – b, a, a + b, find all the zeros.

Answers

1. (i) (c) (ii) (b) (iii) (c) (iv) (a)

2. (i) parabola (ii) –b (iii) p (k)
a
– (a2 + b2)
3. (i) 15 (ii) ±18 (iii) 2
4

4. (i) 215 (ii) – 215 5. 1 6. k = 1 7. a = 3, b = – 6
27 18

9. – 3, –1

10. (i) 1 (9x2 –85x + 36) (ii) 1 (3x2 –35x + 92)
9 3
–1
11. (2 + 3), (2 – 3), 2 and 1 12. –2, 1 and 4 or 4, 1 and – 2

zzz

Polynomials 41

3 Pair of Linear
EquationVsariniabTwleos
BASIC CONCEPTS – A FLOW CHART

42 Xam idea Mathematics–X