NCERT Solutions Class 9 Maths Chapter 4 Exploring Algebraic Identities

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 9 · M AT H S

NCERT Solutions

Chapter 4: Exploring Algebraic
Identities

NCERT Textbook — Ganita Manjari

BOOK PAGES SECTIONS QUESTIONS MEDIUM

68 – 91 20 45 English

Solutions, notes, sample papers & more at 58 pages

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

CLASS 9 · MATHS · GANITA MANJARI

NCERT Solutions — Chapter 4: Exploring Algebraic
Identities
An identity is an equation that stays true for every value of the letters in it. This chapter builds the standard
identities twice over — once by cutting up squares, rectangles and cubes so you can see why each one holds,
and once by plain distributivity — and then turns them round to factorise expressions, square numbers
quickly and simplify rational expressions.

TEXTBOOK BOOK PAGES

Ganita Manjari (Class 9) 68 – 91

SECTIONS QUESTIONS

20 45

MEDIUM

English

Think and Reflect — Page 69
Section 4.1 Introduction

Q1 Try and find other patterns like this one. For example, you could consider 4
consecutive squares and see if you can find a pattern.

Take four consecutive squares and compare the outer pair with the inner pair. The answer is
always 4.

FOUR CONSECUTIVE SQUARES (OUTER SUM) − (INNER SUM) RESULT

1, 4, 9, 16 (1 + 16) − (4 + 9) = 17 − 13 4

4, 9, 16, 25 (4 + 25) − (9 + 16) = 29 − 25 4

25, 36, 49, 64 (25 + 64) − (36 + 49) = 89 − 85 4

Now prove it. Four consecutive numbers can be written as (n − 1), n, (n + 1), (n + 2).

Page 1 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

(n − 1)² + (n + 2)² − [ n² + (n + 1)² ]

= (n² − 2n + 1) + (n² + 4n + 4) − n² − (n² + 2n + 1)

= 2n² + 2n + 5 − 2n² − 2n − 1
= 4, whatever n is

Every term containing n cancels, so the answer cannot depend on which four squares you
picked.

Why it happens: The gap between consecutive squares grows by a fixed amount.
Since (k + 1)² − k² = 2k + 1, consecutive gaps are 2n − 1, 2n + 1, 2n + 3 — each 2 more
than the one before. The chapter’s own pattern with three squares is exactly this
constant difference of gaps, which is 2. With four squares you are comparing the last
gap with the first, and they differ by two steps of 2, that is 4.

Try This: Do the same with cubes. Take four consecutive cubes and compute (fourth
− third) − 2(third − second) + (second − first). You will always get 6 — and 6 = 3! is not
a coincidence.

Think and Reflect — Page 71
Section 4.2 Visualising Identities

Q1 What can you say about a and b if (a + b)² < a² + b²?

Everything is decided by one term. Subtract:

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

So (a + b)² < a² + b² exactly when 2ab < 0, that is when

ab < 0 — a and b have opposite signs (one positive, one negative, neither zero)

Check: a = 10, b = −2 gives (a + b)² = 8² = 64 while a² + b² = 100 + 4 = 104, and indeed 64 < 104.

Page 2 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Why it happens: The two expressions differ by the cross term 2ab and by nothing
else. Both a² and b² are never negative, so they can never decide the comparison —
only the sign of the product ab can. This is exactly the hint the book gives: use the
expansion to find the deciding term.

Q2 What can you say about a and b if (a + b)² > a² + b²?

Same difference, opposite sign.

(a + b)² − (a² + b²) = 2ab > 0 ⇒ ab > 0

⇒ a and b have the same sign — both positive, or both negative

Check both cases: a = 10, b = 2 gives 144 > 104. And a = −2, b = −3 gives (−5)² = 25 while a² + b² =
4 + 9 = 13, so 25 > 13.

Tip: The book’s example used two positive numbers, which is why (a + b)² came out
larger. Two negative numbers behave the same way, because their product is still
positive.

Q3 When will (a + b)² be equal to a² + b²?

(a + b)² = a² + b² ⇔ 2ab = 0 ⇔ ab = 0

⇔ a = 0 or b = 0 (or both)

For instance a = 7, b = 0: (7 + 0)² = 49 and 7² + 0² = 49.

Page 3 of 58

as e
Class 9 Maths Chapter 4 Exploring Algebraic Identities
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: A product of two real numbers is zero only when at least one factor

m l a se
is zero — there are no other possibilities. So the equation (a + b)² = a² + b² is not an
o g = 0 and b = 0. This
identity: it.cis an ordinary equation, satisfied only on the two linesaa
m
se the distinction the section is making. An identity such as (a + b)² = a² +
l a
g + b² holds everywhere; an equation such as (a + b)² = a² + b² holds only at special
is precisely
a2ab
values.
co m
e m . ag
g l as
Exercise Set 4.1 — Pages 71–72
a

co m
m.
Section 4.2 Visualising Identities

o m l a se
.c the identity (a + b)² = a² + 2ab + b², expand the following:a g (i) (7x + 4y)² (ii) (7x/5
m
Using
se + 3y/2)² (iii) (2.5p + 1.5q)² (iv) (3s/4 + 8t)² (v) (x + 1/2y)² (vi) (1/x + 1/y)²
Q1

l a
ag
s
m a
m .co
In every part, name a and b first, then write down a² + 2ab + b² without multiplying anything out
agl
by hand.
l a se
(i) (7x + 4y)² — here a = 7x, b = 4y. a g

com
(7x)² + 2(7x)(4y) + (4y)²
m .
m as e
.co
= 49x² + 56xy + 16y²
a g l
a s em
agl (ii) (7x/5 + 3y/2)² — a = 7x/5, b = 3y/2.
se m
com g l a
. a
em
(7x/5)² + 2(7x/5)(3y/2) + (3y/2)²
a s
agl
= 49x²/25 + (42/10)xy + 9y²/4

= 49x²/25 + 21xy/5 + 9y²/4

co m
m .
m as e
l
(iii) (2.5p + 1.5q)² — a = 2.5p, b = 1.5q.

m .co a g
l a se
ag (2.5p)² + 2(2.5p)(1.5q) + (1.5q)²

.c
= 6.25p² + 7.5pq + 2.25q²
s e m
m a
(as fractions: 25p²/4 + 15pq/2 + 9q²/4)
m . co agl
l a se
ag b = 8t.
(iv) (3s/4 + 8t)² — a = 3s/4,

co m
m .
m ase
.co


a g l Page 4 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

(3s/4)² + 2(3s/4)(8t) + (8t)²

= 9s²/16 + 12st + 64t²

(v) (x + 1/2y)² — a = x, b = 1/(2y).

x² + 2(x)(1/2y) + (1/2y)²

= x² + x/y + 1/(4y²)

(vi) (1/x + 1/y)² — a = 1/x, b = 1/y.

(1/x)² + 2(1/x)(1/y) + (1/y)²

= 1/x² + 2/(xy) + 1/y²

Why the identity may be used on fractions and on 1/x: an identity is true for all
values of its letters. So a and b need not be whole numbers, or even numbers
written without a variable in the denominator — any expression at all may be
substituted, provided it is defined. In part (v) that means y ≠ 0, and in part (vi), x ≠ 0
and y ≠ 0.

Check it yourself: Put x = 1, y = 1 in (vi). The left side is (1 + 1)² = 4 and the right side
is 1 + 2 + 1 = 4. A single well-chosen substitution catches most slips.

Q2 Using the same identity, find the values of the following: (i) (64)² (ii) (105)² (iii)
(205)²

Split each number so that one part is a round number and the other is small.
(i) 64² = (60 + 4)²

= 60² + 2 × 60 × 4 + 4²
= 3600 + 480 + 16

= 4096

(ii) 105² = (100 + 5)²

Page 5 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

= 100² + 2 × 100 × 5 + 5²

= 10000 + 1000 + 25

= 11025

(iii) 205² = (200 + 5)²

= 200² + 2 × 200 × 5 + 5²

= 40000 + 2000 + 25

= 42025

Why this is quicker than long multiplication: the three pieces a², 2ab and b² are all
easy when a is a multiple of 10 — a² and 2ab end in zeros, so the only real arithmetic
is the small square b². You are trading one hard multiplication for three easy ones.

Tip: Choose the split that makes b small. For 64 you could also write (70 − 6)², but 60
+ 4 keeps b at 4 and is easier.

Think and Reflect — Page 73
Section 4.3 Factorisation of Algebraic Expressions Using Identities

Q1 What if we replace b by − b in (a + b)² = a² + 2ab + b²?

You get the second standard identity, free of charge.

(a + (−b))² = a² + 2a(−b) + (−b)²

↓ a + (−b) = a − b, 2a(−b) = −2ab, (−b)² = +b²

(a − b)² = a² − 2ab + b²

Only the middle term changes sign, because it is the only term in which b appears an odd
number of times.

Page 6 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Why the substitution is allowed: this is the whole point of the word identity. (a + b)²
= a² + 2ab + b² is true for every value of b — so it is in particular true when the
number standing in the place of b happens to be −b. Nothing is being assumed; we
are simply reading the identity at a different value.

The book also gives the picture. Start with a square of side a and split the side into (a − b) and b:

a−b b

a−b (a − b)² b(a−b)

b ab

A square of side a. Cutting off the bottom strip ab and the side rectangle b(a − b) leaves the square (a
− b)².

Page 7 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Area of the whole square = a²

Area of the shaded strip along the bottom = ab

Area of the strip on the right = b(a − b)
So (a − b)² = a² − ab − b(a − b)

= a² − ab − ba + b²

= a² − 2ab + b²

Did you know? The two strips overlap in a b × b corner, which is why subtracting ab
and b(a − b) — rather than ab and ab — is what puts the +b² back. The picture is
doing the same bookkeeping as the algebra.

Exercise Set 4.2 — Pages 74–75
Section 4.3 Factorisation of Algebraic Expressions Using Identities

Q1 Factor completely: (i) 9x² + 24xy + 16y² (ii) 4s² + 20st + 25t² (iii) 49x² + 28xy + 4y² (iv)
64p² + (32/3)pq + (4/9)q² *(v) 3a² + 4ab + (4/3)b² *(vi) (9/5)s² + 6sv + 5v² (Hint: 2 was
taken out as a common factor in Example 7. Is it possible to do something similar in
Exercises (v) and (vi) above?)

In each part, ask: is the first term a square, is the last term a square, and is the middle term
twice the product of those two square roots? If yes, the expression is a perfect square.
(i) 9x² + 24xy + 16y²

9x² = (3x)², 16y² = (4y)², 2(3x)(4y) = 24xy ✓

= (3x + 4y)²

(ii) 4s² + 20st + 25t²

4s² = (2s)², 25t² = (5t)², 2(2s)(5t) = 20st ✓

= (2s + 5t)²

(iii) 49x² + 28xy + 4y²

Page 8 of 58

as e
Class 9 Maths Chapter 4 Exploring Algebraic Identities
a g l AglaSem · NCERT Solutions

co m
e m.
49x² = (7x)², 4y² = (2y)², 2(7x)(2y) = 28xy ✓
m l as
= (7x + 2y)²
m .co a g
l a se
a g
(iv) 64p² + (32/3)pq + (4/9)q²

co m
64p² = (8p)², (4/9)q² = (2q/3)², 2(8p)(2q/3) = 32pq/3 ✓
e m . ag
g l as
a
= (8p + 2q/3)² — or, clearing fractions, (4/9)(12p + q)²

*(v) 3a² + 4ab + (4/3)b² — here 3a² is not a square of anything neat, so follow the hint and pull
co m
em.
as
out a common factor first. Take out 1/3:
m l
.co a g
a s e+m4ab + (4/3)b² = (1/3)(9a² + 12ab + 4b²)
gl
3a²
a 9a² = (3a)², 4b² = (2b)², 2(3a)(2b) = 12ab ✓

m a s
.co agl
= (1/3)(3a + 2b)²

se m
l a
ag 1/5:
*(vi) (9/5)s² + 6sv + 5v² — take out

co m
(9/5)s² + 6sv + 5v² = (1/5)(9s² + 30sv + 25v²)
m .
m as e
.co
9s² = (3s)², 25v² = (5v)², 2(3s)(5v) = 30sv ✓
a g l
se m
l a
= (1/5)(3s + 5v)²
ag
se m
com
Why the common factor has to come out first: in Example 7 the book met 50p²,
g l a
m . a
e
whose square root is √50 p — a surd, which the identity a² + 2ab + b² cannot use
as into 25p², a genuine square. Parts (v) and (vi) are the
tidily. Taking out 2 turned l50p²
a g
same trap in reverse: 3a² and (9/5)s² are not squares of rational expressions, but
multiplying inside by 3 and by 5 respectively makes them so. The factor you pull out
co m
is exactly the one that clears the denominators and leaves square coefficients.
m .
m as e
.co a g l
se m Check it yourself: Put a = b = 1 in (v). The original gives 3 + 4 + 4/3 = 25/3; the
g l a
a answer gives (1/3)(3 + 2)² = 25/3. ✓
c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 9 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Q2 Find the values of the following using the identity (a − b)² = a² − 2ab + b². (i) (79)² (ii)
(193)² (iii) (299)²

Each number is just below a round number, so subtract rather than add.
(i) 79² = (80 − 1)²

= 80² − 2 × 80 × 1 + 1²

= 6400 − 160 + 1 = 6241

(ii) 193² = (200 − 7)²

= 200² − 2 × 200 × 7 + 7²

= 40000 − 2800 + 49 = 37249

(iii) 299² = (300 − 1)²

= 300² − 2 × 300 × 1 + 1²

= 90000 − 600 + 1 = 89401

Why choose subtraction here: 79 = 80 − 1 keeps b = 1, while 79 = 70 + 9 would force
b = 9 and a much larger b². The rule of thumb is the same for both identities — go to
the nearest round number, and let the sign look after itself.

Think and Reflect — Page 76

Page 10 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Section 4.4 More Identities

Q1 Label the squares and rectangles in Fig. 4.4 so that it represents the identity (a + b +
c)² = a² + b² + c² + 2ab + 2bc + 2ca.

a b c

a

b

c

Fig. 4.4, page 76 — a square of side (a + b + c) cut into nine parts, printed without labels.

The side of the big square is split into a, b and c both horizontally and vertically, so the nine
pieces have areas (row height) × (column width):

Page 11 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

a b c

a a² ab ac

b ab b² bc

c ac bc c²

A square of side a + b + c cut by both splittings. The three squares on the diagonal give a², b² and c²;
the six rectangles pair up into 2ab, 2bc and 2ca.

COLUMN A COLUMN B COLUMN C

row a a² ab ac

row b ab b² bc

row c ac bc c²

Total area = a² + b² + c² + (ab + ab) + (bc + bc) + (ac + ac)

= a² + b² + c² + 2ab + 2bc + 2ca

and that same area is (a + b + c)², being a square of side a + b + c.

Page 12 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Why the rectangles come in pairs: the piece in row a, column b and the piece in
row b, column a are different rectangles in the picture but have the same area, ab —
one is a tall ab, the other a wide ab. The grid is symmetric about its main diagonal,
so every off-diagonal area appears exactly twice. That symmetry is the geometric
reason each cross term carries a factor 2, and it is why the identity has 2ab, 2bc, 2ca
and not ab, bc, ca.

Tip: The same picture in n strips gives (a₁ + a₂ + … + aₙ)² = sum of all aᵢ² + twice the
sum of all products aᵢaⱼ with i < j. Three letters is just the first interesting case.

Exercise Set 4.3 — Pages 76–77
Section 4.4 More Identities

Q1 Find the following squares using one of the above identities. Determine which of
these identities will make these calculations easier. (i) 117² (ii) 78² (iii) 198² (iv)
214² (v) 1104² (vi) 1120²

Pick the identity that keeps the pieces small: two-term identities when the number is close to a
round number, the three-term identity when it is not.
(i) 117² — use (a + b + c)² with 100 + 10 + 7.

= 100² + 10² + 7² + 2(100)(10) + 2(10)(7) + 2(7)(100)

= 10000 + 100 + 49 + 2000 + 140 + 1400

= 13689

(ii) 78² — use (a − b)² with 80 − 2.

= 80² − 2 × 80 × 2 + 2² = 6400 − 320 + 4 = 6084

(iii) 198² — use (a − b)² with 200 − 2.

= 40000 − 800 + 4 = 39204

(iv) 214² — use (a + b + c)² with 200 + 10 + 4.

Page 13 of 58

as e
Class 9 Maths Chapter 4 Exploring Algebraic Identities
a g l AglaSem · NCERT Solutions

co m
e m.
= 40000 + 100 + 16 + 2(200)(10) + 2(10)(4) + 2(4)(200)
m l as
.co
= 40000 + 100 + 16 + 4000 + 80 + 1600 = 45796
m a g
l a se
a g
(v) 1104² — use (a + b)² with 1100 + 4.

com
= 1100² + 2 × 1100 × 4 + 4² = 1210000 + 8800 + 16 = 1218816
e m . ag
g l as
(vi) 1120² — use (a + b)² with 1100 + 20. a

co m
em.
as
= 1210000 + 2 × 1100 × 20 + 400 = 1210000 + 44000 + 400 = 1254400
m l
.co a g
a s em
a gl Which identity, and why: a two-term split is enough when the number sits next to a
round number — 78, 198, 1104, 1120 all do. But 117 and 214 are two digits away

m a s
cohundreds agl
from anything round, so a two-term split would leave b = 17 or b = 14, and b² would

m .
e
be as hard as the original. Splitting into + tens + units makes every one of

g l as
the six pieces a one-digit multiplication. The number of terms you need is the
number of non-zero digits,aso the three-term identity is the natural tool for a three-

m
digit number.
. co
e m
m l as
.co a g
Check it yourself: 1120² can also be done as (112 × 10)² = 112² × 100 = 12544 × 100.
m answer, and a good reminder that pulling out powers of ten first often saves
a s eSame
a gl work.
se m
com g l a
m . a
ase
agl
Q2 Factor using suitable identities: (i) 16y² − 24y + 9 (ii) (9/4)s² + 6st + 4t² (iii) m²/9 +
mk/3 + k²/4 + 3nk + 2mn + 9n² (iv) p²/16 − 2 + 16/p² (v) 9a² + 4b² + c² − 12ab + 6ac −
4bc

co m
m .
s e
com− 24y + 9 — a square with a minus sign in the middle.agla

(i).16y²

a sem
agl c
16y² = (4y)², 9 = 3², 2(4y)(3) = 24y ✓
m .
m a s e
. co agl
= (4y − 3)²
e m
g l as
(ii) (9/4)s² + 6st + 4t² a

co m
m .
m ase
.co


a g l Page 14 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

(9/4)s² = ((3/2)s)², 4t² = (2t)², 2((3/2)s)(2t) = 6st ✓

= ((3/2)s + 2t)² — equivalently (1/4)(3s + 4t)²

(iii) m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n² — six terms, three of them squares, so try (a + b + c)².

m²/9 = (m/3)², k²/4 = (k/2)², 9n² = (3n)²

Check the three cross terms with a = m/3, b = k/2, c = 3n:

2ab = 2(m/3)(k/2) = mk/3 ✓
2bc = 2(k/2)(3n) = 3kn ✓

2ca = 2(3n)(m/3) = 2mn ✓

= (m/3 + k/2 + 3n)² — equivalently (1/36)(2m + 3k + 18n)²

(iv) p²/16 − 2 + 16/p² — the middle term is the clue: 2(p/4)(4/p) = 2.

p²/16 = (p/4)², 16/p² = (4/p)², 2(p/4)(4/p) = 2 ✓

= (p/4 − 4/p)² — equivalently (p² − 16)²/(16p²), i.e. (p − 4)²(p + 4)²/(16p²)

(v) 9a² + 4b² + c² − 12ab + 6ac − 4bc — two of the three cross terms are negative, so one of the
letters must carry a minus sign.

Try a′ = 3a, b′ = −2b, c′ = c:

2a′b′ = 2(3a)(−2b) = −12ab ✓

2b′c′ = 2(−2b)(c) = −4bc ✓

2c′a′ = 2(c)(3a) = 6ac ✓

= (3a − 2b + c)²

How to place the minus signs: the three cross terms of (a + b + c)² are 2ab, 2bc, 2ca.
Changing the sign of exactly one letter flips exactly two of them — the two in which
that letter appears. So a pattern of two minuses and one plus means one letter is
negative; three minuses is impossible; and all pluses means none (or all) are
negative. In (v) the negative terms are the ones containing b, so b is the letter to
negate.

Page 15 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Q3 Expand the following using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca: (i)
(p + 3q + 7r)² (ii) (3x − 2y + 4z)²

(i) (p + 3q + 7r)² — a = p, b = 3q, c = 7r.

= p² + (3q)² + (7r)² + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p)

= p² + 9q² + 49r² + 6pq + 42qr + 14pr

(ii) (3x − 2y + 4z)² — take a = 3x, b = −2y, c = 4z and keep the signs inside the brackets.

= (3x)² + (−2y)² + (4z)² + 2(3x)(−2y) + 2(−2y)(4z) + 2(4z)(3x)

= 9x² + 4y² + 16z² − 12xy − 16yz + 24zx

= 9x² + 4y² + 16z² − 12xy − 16yz + 24zx

Tip: Notice that the squared terms never lose their plus sign, however many
minuses are inside — (−2y)² = +4y². Only the cross terms can turn negative.

Q4 Is this an identity? (a + b − c)² + (a − b + c)² + (a − b − c)² = 2a² + 2b² + 2c²

No, it is not an identity. One substitution settles it, and the expansion shows exactly what has
gone wrong.

Take a = b = c = 1.

Left side = (1 + 1 − 1)² + (1 − 1 + 1)² + (1 − 1 − 1)² = 1 + 1 + 1 = 3

Right side = 2 + 2 + 2 = 6

3 ≠ 6, so the equation fails at a perfectly ordinary point.

Expanding all three squares shows the true value of the left side:

Page 16 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

(a + b − c)² = a² + b² + c² + 2ab − 2bc − 2ca

(a − b + c)² = a² + b² + c² − 2ab − 2bc + 2ca

(a − b − c)² = a² + b² + c² − 2ab + 2bc − 2ca

Adding: 3a² + 3b² + 3c² − 2ab − 2bc − 2ca

That is not 2a² + 2b² + 2c², so the equation holds only for those special a, b, c satisfying a² + b² +
c² = 2ab + 2bc + 2ca — an equation, not an identity.

The missing fourth term: the statement becomes true the moment you put (a + b +
c)² back in. Adding it contributes a² + b² + c² + 2ab + 2bc + 2ca, and now every cross
term cancels:

(a + b + c)² + (a + b − c)² + (a − b + c)² + (a − b − c)² = 4a² + 4b² + 4c²

The four sign patterns come in pairs that kill each cross term, and this is an identity — it is the
very one drawn in Fig. 4.6 on page 78.

Tip: To disprove a claimed identity, one counter-example is enough and it should be
the first thing you try. To prove one, you must expand.

In-text Questions — Page 77

Page 17 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Section 4.4 More Identities

Q1 Look at the following figure (Fig. 4.5). Justify the identity a² = (a + b)(a − b) + b² for
yourself.

a b

a

b

Fig. 4.5, page 77 — the curved arrow shows one piece being moved from where it is
drawn to the dotted position.

The figure is a cut-and-slide proof. Start with a square of side a and take a strip of width b off
the bottom.

Page 18 of 58

ase
Class 9 Maths Chapter 4 Exploring Algebraic Identities
a g l AglaSem · NCERT Solutions

co m
em.
com as
a a+b
. a g l
m
ase
agl

b(a−b)
a(a − b) a−b a(a − b)
o m
. c ag
a

a s em
cut & slide

ag l
b(a−b) b²
b²
m
square of side a
.co
(a + b)(a − b)e+mb²
c om g l as
. a
m of side a. The bottom strip is cut at width b: the piece b(a − b) swings round to the right-
ase
A square

agl
hand side, turning the shape into a rectangle (a + b) by (a − b), and the b × b corner is left over.

m a s
m.co agl
se
Before the cut: area = a²

g l a
a
The bottom strip is a × b, and it splits into b(a − b) and b².

co m
m .
e
Slide the piece b(a − b) round to the right-hand edge. It fits exactly, because the remaining
m l as
.co
block is (a − b) tall.
m a g
l a se
ag
m
After the slide: a rectangle of width a + b and height a − b, plus a leftover square b²

a se
area = (a + b)(a − b) + b²
.com a g l
m
ase
agl
Nothing was added or thrown away, so a² = (a + b)(a − b) + b²

co m
m .
Rearranged, this is the identity you met in Grade 8:
m as e
.co a g l
se m a² − b² = (a + b)(a − b)
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 19 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Why the pieces fit: the block left after removing the bottom strip is a wide and (a −
b) tall. The strip we cut off is b tall and (a − b) long, so standing it on end makes it (a
− b) tall and b wide — exactly the height of the block. Widths add: a + b. Heights
match: a − b. The identity is a statement that a rearrangement is possible, and the
picture is the rearrangement.

Did you know? In 750 CE Śhrīdharāchārya proposed this as a way to square
numbers mentally. Choose b so that a − b and a + b are easy:
55² = (55 + 5)(55 − 5) + 5² = 60 × 50 + 25 = 3025.
Try 98² = 100 × 96 + 4 = 9604, and 43² = 46 × 40 + 9 = 1849.

Think and Reflect — Page 78
Section 4.4 More Identities

Q1 Try to evaluate the following using a suitable identity: (i) 35² (ii) 65² (iii) 85² (iv)
105². Do you observe any interesting pattern?

Every number here ends in 5, so Śhrīdharāchārya’s form a² = (a + b)(a − b) + b² with b = 5 is the
right tool.

35² = (35 + 5)(35 − 5) + 5² = 40 × 30 + 25 = 1200 + 25 = 1225
65² = 70 × 60 + 25 = 4200 + 25 = 4225

85² = 90 × 80 + 25 = 7200 + 25 = 7225

105² = 110 × 100 + 25 = 11000 + 25 = 11025

The pattern. Every answer ends in 25, and the digits in front are the tens digit multiplied by the
next whole number:

Page 20 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

NUMBER N N(N + 1) SQUARE

35 3 3 × 4 = 12 1225

65 6 6 × 7 = 42 4225

85 8 8 × 9 = 72 7225

105 10 10 × 11 = 110 11025

Why the rule works — a proof, not a coincidence: any number ending in 5 is 10n +
5. Then

(10n + 5)² = (10n)² + 2(10n)(5) + 5²

= 100n² + 100n + 25

= 100·n(n + 1) + 25

The term 100·n(n + 1) simply writes n(n + 1) two places to the left, and the +25 fills the last two
places. So “n times n + 1, then write 25” is exactly the identity in disguise.

Try This: 995². Here n = 99, so n(n + 1) = 99 × 100 = 9900, and 995² = 990025.

Page 21 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Q2 Observe the two rows of figures below (Fig. 4.6). They represent an algebraic
identity. Try to identify it.

a+b+c

2c
a+b−c

a−b+c

2b
a−b−c

2a 2b 2c

Fig. 4.6, page 78 — the two rows of figures as printed with the question.

Read the labels off the two rows. The top row has four squares, of sides a + b + c, a + b − c, a − b
+ c and a − b − c. The bottom row has three squares, of sides 2a, 2b and 2c — and the pieces are
coloured to show that the top four cut up and reassemble into the bottom three.

Page 22 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

a+b+c a+b−c a−b+c a−b−c
sum of these four areas

2a 2b 2c
equals the sum of these three

The four squares of the top row have exactly the same total area as the three squares of the bottom
row.

(a + b + c)² + (a + b − c)² + (a − b + c)² + (a − b − c)² = (2a)² + (2b)² + (2c)²
that is, = 4a² + 4b² + 4c²

Proof. Expand the four squares and add:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

(a + b − c)² = a² + b² + c² + 2ab − 2bc − 2ca

(a − b + c)² = a² + b² + c² − 2ab − 2bc + 2ca

(a − b − c)² = a² + b² + c² − 2ab + 2bc − 2ca

Sum = 4a² + 4b² + 4c² — every cross term cancels

Page 23 of 58

as e
Class 9 Maths Chapter 4 Exploring Algebraic Identities
a g l AglaSem · NCERT Solutions

co m
m.
Why every cross term cancels: look down the 2ab column: the signs are +, +, −, −.

m l a se
Down the 2bc column: +, −, −, +. Down the 2ca column: +, −, +, −. In each column two
o g with each sign.
plus signs.cmeet two minus signs, so each cross term appears twice
a
Theas
m
e sign patterns are precisely the four ways of choosing signs for b and c,
l
g is what forces this balance.
four
awhich
o m
e
. c
m 6² + 4² + 2² + 0² = 36 + 16 + 4 + 0 = 56. Right: ag
s
Check it yourself: a = 3, b = 2, c = 1. Left:
a
agl
36 + 16 + 4 = 56. ✓

co m
se m.
o m
Think and Reflect — Page 79
g l a
m .c a
se
Section 4.5 Factorisation Using Algebra Tiles

l a
ag Q1 Suppose 7x is split as 2x + 5x; can a similar rectangular arrangement be formed?
m a s
.co agl
Consider other possibilities and check.

se m
g l a
a

No — the split 2x + 5x does not work. If two of the x-tiles go to the right of the x²-tile and five

co
go below it, then the unit tiles must fill a 2 by 5 corner, which holds only 10 units. We have 12, so
m
two tiles are left over and the shape is not a rectangle.
m .
m ase
.co
There are only three ways to split 7x into two whole-number pieces. Check each:
a g l
s m
eSPLIT
gl a
a
OF CORNER THE UNITS UNITS UNITS RECTANGLE?
7X MUST FILL NEEDED AVAILABLE

se m
com6 g l a
. a
1x + 6x 1 by 6 12 No

m
ase
agl
2x + 5x 2 by 5 10 12 No

3x + 4x 3 by 4 12 12 Yes

co m
m .
o m l a se
Why only one split can work: putting a x-tiles on one side and b on the other builds

m ag+ b) = x² + (a + b)x + ab.
.ca rectangle of sides (x + a) and (x + b). Its area is (x + a)(x
l a se
ag Matching this with x² + 7x + 12 forces two conditions at once: a + b = 7 and ab = 12.
Splitting 7x is easy — keeping the product at 12 at the same time is the real
.c
s e m
m a
constraint, and only 3 and 4 satisfy both.

e m . co agl
g l as
a

co m
m .
m as e
.co


a g l Page 24 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Tip: This is exactly why the ‘splitting the middle term’ method of Section 4.6 lists the
factor pairs of the constant term first, and only then checks which pair adds to the
middle coefficient. The tiles make the two conditions visible: one is a length, the
other is an area.

Think and Reflect — Page 79
Section 4.5 Factorisation Using Algebra Tiles

Q1 Figure out the product of x + 2 and x + 3 using algebra tiles.

Lay one x-tile and two unit tiles along the top edge, and one x-tile and three unit tiles down the
left edge. Fill the rectangle:

x+2

x² x x

x+3 x 1 1

x 1 1

x 1 1

Page 25 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

One x²-tile, 2 + 3 = 5 x-tiles and a 2 × 3 array of 6 unit tiles form a rectangle of sides x + 2 and x + 3.

1 x²-tile → x²

3 x-tiles to the right of it and 2 below it → 5x

a 2 × 3 block of unit tiles → 6

(x + 2)(x + 3) = x² + 5x + 6

Why the tiles give the answer: the big rectangle is measured two ways. Its
dimensions are x + 2 and x + 3, so its area is (x + 2)(x + 3). But it is also the sum of the
areas of the pieces, x² + 5x + 6. One region, two descriptions — and they must be
equal. That equality is the identity.

Q2 Lay out algebra tiles for x² + 11x + 30 in such a way that you will see its factors.

Split 11x so that the unit tiles form a rectangle of area 30. We need a + b = 11 and ab = 30, so a =
5 and b = 6.

Page 26 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

x+5

x² x x x x x

x 1 1 1 1 1

x 1 1 1 1 1
x+6
x 1 1 1 1 1

x 1 1 1 1 1

x 1 1 1 1 1

x 1 1 1 1 1

The x²-tile with 5 x-tiles beside it, 6 x-tiles below it and a 5 × 6 array of unit tiles. The finished rectangle
measures x + 5 by x + 6.

x² + 11x + 30 = x² + (5 + 6)x + (5)(6)

= (x + 5)(x + 6)

Other splits fail: 1 + 10 needs 10 units, 2 + 9 needs 18, 3 + 8 needs 24, 4 + 7 needs 28 — only 5 +
6 needs exactly 30.

Check it yourself: multiply back. (x + 5)(x + 6) = x² + 6x + 5x + 30 = x² + 11x + 30. ✓

Think and Reflect — Page 80

Page 27 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Section 4.5 Factorisation Using Algebra Tiles

Q1 We have seen that (x + 3)(x + 4) = x² + 7x + 12. Also (x + 6)(x + 7) = x² + 13x + 42.
Generalise the pattern to get an expression for (x + a)(x + b).

Look at where the numbers 7 and 12 came from, and where 13 and 42 came from:

PRODUCT COEFFICIENT OF X CONSTANT TERM

(x + 3)(x + 4) 7=3+4 12 = 3 × 4

(x + 6)(x + 7) 13 = 6 + 7 42 = 6 × 7

(x + a)(x + b) a+b ab

(x + a)(x + b) = x² + (a + b)x + ab

Proof by the distributive property:

(x + a)(x + b) = x(x + b) + a(x + b)

= x² + bx + ax + ab

= x² + (a + b)x + ab

Why the sum and the product both appear: multiplying out produces four terms.
The two ‘mixed’ terms bx and ax are the only ones carrying a single x, so they merge
into (a + b)x; the term with no x at all is ab. In the tile picture these are the two arms
of x-tiles and the corner block of unit tiles. The identity is the same statement as “a
rectangle’s area is the sum of its parts”.

Tip: Read the identity right to left and it becomes a factorising rule: to factor x² + px
+ q, hunt for two numbers whose sum is p and whose product is q. That is exactly
what Section 4.6 does.

In-text Questions — Page 80

Page 28 of 58

as e
Class 9 Maths Chapter 4 Exploring Algebraic Identities
a g l AglaSem · NCERT Solutions

Section 4.5 Factorisation Using Algebra Tiles
co m
em.
m l as
Q1
.co
Now consider the case where we have a rectangle of sidelengths 2x + 3 and 3x + 1,
m a g
l a se
as shown in Fig. 4.8. What can you say about its area (2x + 3)(3x + 1)?

ag

. com 2x + 3 ag
e m
g l as
a
o m
x² x² x xem.xc
com g l as
m . a
ase
agl
x² x² x x x
a s
3x + 1 .com agl
m
ase
agl
x² x² x x x
. com
m a sem
m . co x x ag1l 1 1
ase
agl
se m
com g l a
.
Fig. 4.8, page 80 — the rectangle of sidelengths 2x + 3 and 3x + 1, built from algebra tiles.
m a
ase
agl
m

. co
m
Count the tiles in Fig. 4.8. Two x-columns and three unit-columns across; three x-rows and one

m as e
.co l
unit-row down.
a g
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 29 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

2x + 3

x² x² x x x

x² x² x x x
3x + 1

x² x² x x x

x x 1 1 1

The rectangle (2x + 3) by (3x + 1) built from 6 x²-tiles, 11 x-tiles and 3 unit tiles.

TILE WHERE IT COMES FROM HOW MANY TOTAL AREA

x² 2 x-columns × 3 x-rows 6 6x²

x 3 unit-columns × 3 x-rows = 9, plus 2 x-columns × 1 unit-row = 2 11 11x

1 3 unit-columns × 1 unit-row 3 3

(2x + 3)(3x + 1) = 6x² + 11x + 3

Verify by the distributive property:

(2x + 3)(3x + 1) = 2x(3x + 1) + 3(3x + 1)

= 6x² + 2x + 9x + 3 = 6x² + 11x + 3 ✓

Page 30 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Why the middle coefficient is 11 and not 3 + 1: the two ends of the rectangle are
no longer plain x-tiles — there are 2 of them one way and 3 the other. So an x-tile
arises in two different ways, 3 × 3 = 9 of them from the wide side and 2 × 1 = 2 from
the tall side. This is why the general rule needs (pb + aq), not (a + b).

Q2 Fill in the blanks with the appropriate expressions to make the equation true: (px +
a)(qx + b) = (_____)x² + (_____)x + _____ . Also, verify your answer using the distributive
property.

(px + a)(qx + b) = (pq)x² + (pb + aq)x + ab

Verification by the distributive property:

(px + a)(qx + b) = px(qx + b) + a(qx + b)

= pq·x² + pb·x + aq·x + ab

= pq·x² + (pb + aq)x + ab

Check against Fig. 4.8, where p = 2, a = 3, q = 3, b = 1:

pq = 2 × 3 = 6 ✓

pb + aq = 2 × 1 + 3 × 3 = 2 + 9 = 11 ✓

ab = 3 × 1 = 3 ✓

giving 6x² + 11x + 3, exactly the tile count.

Why the three blanks are what they are: the product of two brackets has four
terms. Only one of them, px × qx, has two x’s — that is the x² coefficient pq. Two of
them have one x each — those combine into pb + aq. One has none — that is ab.
Setting p = q = 1 recovers the earlier rule (x + a)(x + b) = x² + (a + b)x + ab, so this is its
general form.

Page 31 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Tip: When you factor 6x² + 7x + 2 later, you are solving pq = 6, ab = 2, pb + aq = 7.
This identity tells you exactly which three conditions to satisfy.

Exercise Set 4.4 — Pages 81–82
Section 4.6 Factorisation Without Using Algebra Tiles

Q1 Fill in the blanks to complete the following identities: (i) s² − 11s + 24 = (________)
(________) (ii) (________) (x + 1) = (3x² − 4x −7) (iii) 10x² − 11x − 6 = (2x − ___) (___ + 2) (iv)
6x² + 7x + 2 = (____________) (___________)

(i) s² − 11s + 24 — need a + b = −11 and ab = +24, so both are negative. The factor pairs of 24 are
1×24, 2×12, 3×8, 4×6; only 3 + 8 = 11.

a = −3, b = −8

s² − 11s + 24 = s² − 3s − 8s + 24 = s(s − 3) − 8(s − 3)

= (s − 3)(s − 8)

(ii) (________)(x + 1) = 3x² − 4x − 7 — one factor is given, so divide it out. Since (x + 1) is a factor,
split −4x so that (x + 1) appears:

3x² − 4x − 7 = 3x² + 3x − 7x − 7

= 3x(x + 1) − 7(x + 1)

= (3x − 7)(x + 1)

(iii) 10x² − 11x − 6 = (2x − ___)(___ + 2) — the pattern (px + a)(qx + b) needs pq = 10, ab = −6 and
pb + aq = −11. With p = 2 and b = 2, we need q = 5 and a × 2 = −6, so a = −3.

10x² − 11x − 6 = 10x² − 15x + 4x − 6 (since −15 × 4 = −60 = 10 × −6 and −15 + 4 = −11)

= 5x(2x − 3) + 2(2x − 3)

= (2x − 3)(5x + 2)

So the blanks are 3 and 5x.

Page 32 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

(iv) 6x² + 7x + 2 — split 7x using two numbers with product 6 × 2 = 12 and sum 7: they are 3 and
4.

6x² + 7x + 2 = 6x² + 3x + 4x + 2

= 3x(2x + 1) + 2(2x + 1)

= (2x + 1)(3x + 2)

Why the product is ac and not just c when the leading coefficient is not 1: for (px
+ a)(qx + b), the constant is ab and the x-coefficient is pb + aq. Multiply the two split
pieces: (pb)(aq) = (pq)(ab) = (leading coefficient) × (constant term). So the two
numbers you look for must multiply to a·c, not to c. In (iv) that is 6 × 2 = 12, and in
(iii) it is 10 × (−6) = −60.

Check it yourself: Expand each answer. (2x − 3)(5x + 2) = 10x² + 4x − 15x − 6 = 10x² −
11x − 6. ✓

Q2 Select and use the identity that will help you to find the following products without
multiplying directly: (i) (41)² (ii) (27)² (iii) (23 × 17) (iv) (135)² (v) (97)² (vi) (18 × 29)
(vii) (34 × 43) (viii) (205)²

PRODUCT IDENTITY CHOSEN WORKING VALUE

(i) 41² (a + b)² (40 + 1)² = 1600 + 80 + 1 1681

(ii) 27² (a − b)² (30 − 3)² = 900 − 180 + 9 729

(iii) 23 × 17 (a + b)(a − b) (20 + 3)(20 − 3) = 400 − 9 391

(iv) 135² a² = (a + b)(a − b) + b² 140 × 130 + 25 = 18200 + 25 18225

(v) 97² (a − b)² (100 − 3)² = 10000 − 600 + 9 9409

(vi) 18 × 29 (x + a)(x + b) (20 − 2)(20 + 9) = 400 + 20(7) − 18 522

(vii) 34 × 43 (x + a)(x + b) (40 − 6)(40 + 3) = 1600 + 40(−3) − 18 1462

(viii) 205² (a + b)² (200 + 5)² = 40000 + 2000 + 25 42025

Page 33 of 58

as e
Class 9 Maths Chapter 4 Exploring Algebraic Identities
a g l AglaSem · NCERT Solutions

The two unequal products are worth writing out in full, since they use the third identity of
co m
Section 4.5.
e m.
m l as
m .co a g
l a se
(vi) 18 × 29 with x = 20, a = −2, b = 9:
g
a(x + a)(x + b) = x² + (a + b)x + ab

co m
ag
= 400 + (−2 + 9)(20) + (−2)(9)
m .
= 400 + 140 − 18 = 522
ase
a g l
(vii) 34 × 43 with x = 40, a = −6, b = 3:
co m
em.
m as
= 1600 + (−6 + 3)(40) + (−6)(3)

= 1600.c−o120 − 18 = 1462 a g l
a s em
ag l
s
How to pick the identity: if the two numbers are equally far from a round number,
m a
.co
they are a ± b and the difference-of-squares identity finishes in one step (that is (iii),
em agl
a s
23 and 17 about 20). If they are at different distances, use (x + a)(x + b) with x the
agl ending in 5, Śhrīdharāchārya’s form is fastest —
round number. And for a square
135² is 13 × 14 = 182 followed by 25.

co m
m .
o m l a se
m .cFactor the following: (i) 9a² + b² + 4c² − 6ab + 12ac − 4bcag(ii) 16s² + 25t² − 40st (iii) r²
e
las − r − 42 (iv) 49g² + 14gh + h² (v) 64u² + 121v² + 4w² − 176uv − 32uw + 44vw
Q3

ag
se m

com g l a
m . a
e
(i) 9a² + b² + 4c² − 6ab + 12ac − 4bc — three squares and three cross terms, so aim at (A + B +

g l as negative ones, so give b the minus sign.
C)². The terms containing b are the
a

com
A = 3a, B = −b, C = 2c
.
m a s em = 12ac ✓
2AB = 2(3a)(−b) = −6ab ✓ 2BC = 2(−b)(2c) = −4bc ✓ 2CA = 2(2c)(3a)
. co(3a − b + 2c)² a gl
as em =

agl c
m .
e
(ii) 16s² + 25t² − 40st

m a s
e m . co agl
as
= (4s)² − 2(4s)(5t) + (5t)²

= (4s − 5t)² a g l

com
m .
m ase
.co


a g l Page 34 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

(iii) r² − r − 42 — need a + b = −1 and ab = −42, so the two numbers have opposite signs and
differ by 1: they are −7 and 6.

r² − r − 42 = r² − 7r + 6r − 42

= r(r − 7) + 6(r − 7)

= (r − 7)(r + 6)

(iv) 49g² + 14gh + h²

= (7g)² + 2(7g)(h) + h²

= (7g + h)²

(v) 64u² + 121v² + 4w² − 176uv − 32uw + 44vw — the negative cross terms are the ones
containing u, so u should be the odd one out. Take A = 8u, B = −11v, C = −2w and check:

2AB = 2(8u)(−11v) = −176uv ✓

2BC = 2(−11v)(−2w) = +44vw ✓

2CA = 2(−2w)(8u) = −32uw ✓

= (8u − 11v − 2w)²

How the sign hunt works: in (A + B + C)² a cross term is negative exactly when its
two letters carry opposite signs. Here uv and uw are negative while vw is positive, so
v and w agree with each other and both disagree with u — hence u positive, v and w
negative (or the whole thing negated, which gives the same square). Reading the
sign pattern first saves you from trial and error.

Tip: A square never has a unique-looking answer: (8u − 11v − 2w)² and (−8u + 11v +
2w)² are the same expression. Either is a correct factorisation.

Think and Reflect — Page 82

Page 35 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Section 4.6 Factorisation Without Using Algebra Tiles

Q1 James and Reshma were talking about algebraic identities they learnt in school.
James: (a − b)²(a + b) = (a² − 2ab + b²)(a + b). Reshma: I have a different idea. (a − b)²(a
+ b) = (a − b)[(a − b)(a + b)] = (a − b)(a² − b²). I will find this product to get the answer.
According to you, who is correct and why?

Both are correct. They are two routes to the same expression, and both routes are legal.
Reshma’s is the shorter one.
James’s route — expand the square first:

(a² − 2ab + b²)(a + b)
= a³ + a²b − 2a²b − 2ab² + ab² + b³

= a³ − a²b − ab² + b³

Reshma’s route — regroup first, so that a² − b² appears:

(a − b)[(a − b)(a + b)] = (a − b)(a² − b²)

= a³ − ab² − a²b + b³

= a³ − a²b − ab² + b³

The two answers are identical, and both factor back as

a³ − a²b − ab² + b³ = a²(a − b) − b²(a − b) = (a − b)(a² − b²) = (a − b)²(a + b)

Why Reshma may regroup: multiplication is associative, so (a − b)²(a + b) may be
read as [(a − b)(a − b)](a + b) or as (a − b)[(a − b)(a + b)] — the brackets are ours, not
the expression’s. She then spots that one of the pairs, (a − b)(a + b), is a known
identity, so she replaces it by a² − b² and is left with a single easy multiplication.
James has to multiply a three-term expression by a two-term one, giving six terms,
two of which then cancel.

Try This: Combine identities the same way to get new results. For instance (a + b)³(a
− b)³ = [(a + b)(a − b)]³ = (a² − b²)³ — three lines of work replaced by one.

Page 36 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

In-text Questions — Page 85
Section 4.7 Finding New Identities

Q1 Try to multiply (x + y)(x² − xy + y²) using the distributive property. Predict what (x +
y)(x² − xy + y²) will be.

Prediction: x³ + y³. The chapter has just shown (x − y)(x² + xy + y²) = x³ − y³; replacing y by −y
turns the left side into (x + y)(x² − xy + y²) and the right side into x³ + y³. Now confirm it by
multiplying out.

(x + y)(x² − xy + y²)

= x(x² − xy + y²) + y(x² − xy + y²)

= (x³ − x²y + xy²) + (x²y − xy² + y³)

= x³ − x²y + x²y + xy² − xy² + y³

= x³ + y³

Why the middle terms cancel: the second bracket is arranged so that each of its
terms, multiplied by x, produces exactly what the next term produces when
multiplied by y — but with the opposite sign. The six terms therefore fall into two
cancelling pairs, leaving only the first and the last. That is why x² − xy + y² is the right
partner for x + y, while x² + xy + y² is the right partner for x − y.

Check it yourself: x = 2, y = 1. Left: (3)(4 − 2 + 1) = 3 × 3 = 9. Right: 8 + 1 = 9. ✓

Think and Reflect — Page 85
Section 4.7 Finding New Identities

Q1 We already know that x² − y² = (x − y)(x + y). Further, we have verified that x³ − y³ = (x
− y)(x² + xy + y²). Observe that x − y is a common factor of x² − y² and x³ − y³. Do you
think x − y is also a factor of x⁴ − y⁴? Note that x⁴ − y⁴ = (x²)² − (y²)² = (x² − y²)(x² + y²).
Can you see how x − y is a factor of x⁴ − y⁴?

Yes. The hint does almost all the work — the first bracket already contains x − y.

Page 37 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

x⁴ − y⁴ = (x²)² − (y²)²

= (x² − y²)(x² + y²) — difference of two squares

= (x − y)(x + y)(x² + y²) — difference of two squares again
So x⁴ − y⁴ = (x − y)(x + y)(x² + y²), and x − y is a factor.

Equally, the direct division gives

x⁴ − y⁴ = (x − y)(x³ + x²y + xy² + y³)

and the two answers agree, since (x + y)(x² + y²) = x³ + x²y + xy² + y³.

Why x − y must be a factor, without any algebra: put x = y in the expression x⁴ −
y⁴. You get 0. Any expression that vanishes whenever x = y has x − y as a factor —
that is the same reasoning that made x − y a factor of x² − y² and of x³ − y³. You will
meet this as the Factor Theorem in Grade 10.

Q2 How about x⁵ − y⁵? Does this also have x − y as a factor?

Yes. Here the trick of splitting into squares is not available (5 is odd), so multiply out directly and
watch the cancellation.

(x − y)(x⁴ + x³y + x²y² + xy³ + y⁴)

= (x⁵ + x⁴y + x³y² + x²y³ + xy⁴)

− (x⁴y + x³y² + x²y³ + xy⁴ + y⁵)

= x⁵ − y⁵ — every middle term appears once with each sign

The same pattern works for every power:

xⁿ − yⁿ = (x − y)(xⁿ⁻¹ + xⁿ⁻²y + xⁿ⁻³y² + … + xyⁿ⁻² + yⁿ⁻¹)

Page 38 of 58

as e
Class 9 Maths Chapter 4 Exploring Algebraic Identities
a g l AglaSem · NCERT Solutions

co m
m.
Why it always telescopes: multiplying the long bracket by x raises every exponent

m l a se
of x by one; multiplying it by −y raises every exponent of y by one. The list of terms
o
produced.cby x is the same as the list produced by y, shifted along a gby one place. So
m
seterm except the very first (xⁿ) and the very last (−yⁿ) is matched by an equal
l a
g of opposite sign. This is a telescoping sum, and it is why x − y is a factor of xⁿ −
every
aterm
yⁿ for every natural number n.
com
e m . ag
g l as
a
Did you know? This single fact explains why 2ⁿ − 1 can be prime only when n is
prime: if n = ab then 2ᵃ − 1 divides 2ⁿ − 1. Such primes are called Mersenne primes.

co m
em.
m
coReflect g l as
Think and
m . — Page 87 a
a e Simplifying Rational Expressions
s4.8
ag l
Section

a s
com t² + 2ts − 48s² and simplify the rational agl
Try to simplify the following rational expression: (36s² − 12st + t²)/(t² + 2ts − 48s²) =
(6s − t)²/[(___ + ___)(___ + ___)]. (Hint:.Factor
Q1

expressions assuming thatat²s+ em
a g l 2ts − 48s² ≠ 0.)

m

. co
Numerator. It is already a perfect square, which is why the book has written it for you:
e m
m l as
m .co a g
l a se 36s² − 12st + t² = (6s)² − 2(6s)(t) + t² = (6s − t)²

ag
Denominator. Treat it as a quadratic in t: find two terms whose sum is 2s and whose product is
se m
com g l a
. a
−48s². They are 8s and −6s.
m
ase
agl
t² + 2ts − 48s² = t² + 8ts − 6ts − 48s²

co m
= t(t + 8s) − 6s(t + 8s)

m .
= (t + 8s)(t − 6s)
m as e
.co a g l
a s emSo the blanks are filled by (t + 8s)(t − 6s), and now the fraction cancels:
agl
.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 39 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

(36s² − 12st + t²)/(t² + 2ts − 48s²) = (6s − t)²/[(t + 8s)(t − 6s)]

Note (6s − t)² = (t − 6s)², since squaring kills the sign

= (t − 6s)²/[(t + 8s)(t − 6s)]
= (t − 6s)/(t + 8s)

Why the sign flip is safe: (6s − t) and (t − 6s) are negatives of each other, and a
square does not notice the difference: (−u)² = u². Rewriting the numerator as (t − 6s)²
makes the common factor with the denominator visible. Cancelling it is allowed
because we are told t² + 2ts − 48s² ≠ 0, so in particular t − 6s ≠ 0.

Tip: When you cancel in a rational expression, always say why the factor is not zero.
That is what the condition in the question is for.

Exercise Set 4.5 — Page 87
Section 4.8 Simplifying Rational Expressions

Q1 Simplify the following rational expressions assuming that the expressions in the
denominators are not equal to zero: (i) (3p² − 3pq − 18q²)/(p² + 3pq − 10q²) (ii) (n³ −
3n²m + 3nm² − m³)/(5m² − 10mn + 5n²) (iii) (w³ − v³ + x³ + 3wvx)/(w² + v² + x² − 2wv −
2vx + 2wx) (iv) (4y² − 20yz + 25z²)/(25z² − 4y²) (v) [(x² + x − 6)(x² − 7x + 12)]/[(x² − 6x +
8)(x² − 9)] (vi) (p⁴ − 16)/(p² − 4p + 4)

The method is always the same: factor the top, factor the bottom, cancel whatever is common
and non-zero.
(i) (3p² − 3pq − 18q²)/(p² + 3pq − 10q²)

Numerator: 3p² − 3pq − 18q² = 3(p² − pq − 6q²) = 3(p − 3q)(p + 2q)

Denominator: p² + 3pq − 10q² = (p + 5q)(p − 2q)

= 3(p − 3q)(p + 2q) / [(p + 5q)(p − 2q)]

Page 40 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

The four factors are all different, so nothing cancels — the expression is already in its lowest
terms, and the fully factorised form above is the simplest answer. (A quick check: put p = 1, q = 0.
The original gives 3/1 = 3, and so does the factorised form.)
(ii) (n³ − 3n²m + 3nm² − m³)/(5m² − 10mn + 5n²) — the numerator is a cube of a difference.

Numerator = n³ − 3n²m + 3nm² − m³ = (n − m)³

Denominator = 5(m² − 2mn + n²) = 5(m − n)² = 5(n − m)²

= (n − m)³ / [5(n − m)²] = (n − m)/5

(iii) (w³ − v³ + x³ + 3wvx)/(w² + v² + x² − 2wv − 2vx + 2wx) — the numerator is the three-cubes
identity with v negated.

Numerator = w³ + (−v)³ + x³ − 3(w)(−v)(x)

= (w − v + x)(w² + v² + x² + wv + vx − wx)

Denominator = (w − v + x)²

= (w² + v² + x² + wv + vx − wx)/(w − v + x)

(iv) (4y² − 20yz + 25z²)/(25z² − 4y²)

Numerator = (2y − 5z)²

Denominator = (5z)² − (2y)² = (5z + 2y)(5z − 2y)

= (2y − 5z)² / [(5z + 2y)(5z − 2y)]

and (2y − 5z) = −(5z − 2y), so one factor cancels leaving a minus sign:

= −(2y − 5z)/(2y + 5z) = (5z − 2y)/(2y + 5z)

(v) [(x² + x − 6)(x² − 7x + 12)] / [(x² − 6x + 8)(x² − 9)]

Page 41 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

x² + x − 6 = (x + 3)(x − 2)

x² − 7x + 12 = (x − 3)(x − 4)

x² − 6x + 8 = (x − 2)(x − 4)
x² − 9 = (x − 3)(x + 3)

= [(x + 3)(x − 2)(x − 3)(x − 4)] / [(x − 2)(x − 4)(x − 3)(x + 3)]

Every factor on top appears below → = 1

(vi) (p⁴ − 16)/(p² − 4p + 4)

Numerator = (p²)² − 4² = (p² − 4)(p² + 4) = (p − 2)(p + 2)(p² + 4)

Denominator = (p − 2)²

= (p + 2)(p² + 4)/(p − 2)

Why cancelling needs a condition: writing (n − m)³/[5(n − m)²] = (n − m)/5 divides
top and bottom by (n − m)², which is legal only when n ≠ m. The question grants
exactly this by assuming the denominators are non-zero. Without that promise the
two expressions would not be equal everywhere — the left side would be undefined
at n = m while the right side is 0 there.

Tip: In part (iii) it pays to recognise the shape x³ + y³ + z³ − 3xyz before doing
anything else. Here y = −v, which turns −3xyz into +3wvx and −v³ into the −v³ you can
see. The denominator, meanwhile, is just (w − v + x)².

End-of-Chapter Exercises — Pages 88–90

Page 42 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Chapter 4 Exploring Algebraic Identities

Q1 Use suitable identities to find the following products: (i) (−3x + 4)² (ii) (2s + 7)(2s − 7)
(iii) (p² + 1/2)(p² − 1/2) (iv) (2n + 7)(2n − 7) (v) (s − 2t)(s² + 2st + 4t²) (vi) (1/2r − 4r)²
(vii) (−3m + 4k − l)² (viii) (x − y/3)³ (ix) (7k/2 − 2m/3)³

(i) (−3x + 4)² — read it as (4 − 3x)² and use (a − b)².

= 4² − 2(4)(3x) + (3x)² = 9x² − 24x + 16

(ii) (2s + 7)(2s − 7) — difference of two squares.

= (2s)² − 7² = 4s² − 49

(iii) (p² + 1/2)(p² − 1/2)

= (p²)² − (1/2)² = p⁴ − 1/4

(iv) (2n + 7)(2n − 7)

= (2n)² − 7² = 4n² − 49

(v) (s − 2t)(s² + 2st + 4t²) — this is (x − y)(x² + xy + y²) with x = s, y = 2t.

Check the partner bracket: x² = s², xy = s(2t) = 2st, y² = 4t² ✓

= s³ − (2t)³ = s³ − 8t³

(vi) (1/2r − 4r)² — with a = 1/(2r), b = 4r, note 2ab = 2 × (1/2r) × 4r = 4.

= (1/2r)² − 2(1/2r)(4r) + (4r)²

= 1/(4r²) − 4 + 16r²

(vii) (−3m + 4k − l)² — three-letter square with a = −3m, b = 4k, c = −l.

Page 43 of 58

as e
Class 9 Maths Chapter 4 Exploring Algebraic Identities
a g l AglaSem · NCERT Solutions

co m
e m.
= 9m² + 16k² + l² + 2(−3m)(4k) + 2(4k)(−l) + 2(−l)(−3m)
m l as
.co
= 9m² + 16k² + l² − 24km − 8kl + 6lm
m a g
l a se
a g
(viii) (x − y/3)³ — cube of a difference, a = x, b = y/3.

co m
= x³ − 3x²(y/3) + 3x(y/3)² − (y/3)³
e m . ag
g l as
a
= x³ − x²y + xy²/3 − y³/27

(ix) (7k/2 − 2m/3)³ — a = 7k/2, b = 2m/3.
co m
em.
com g l as
. a
em= 3 × (49k²/4) × (2m/3) = 49k²m/2
a³ = 343k³/8
a s
agl 3a²b

s
3ab² = 3 × (7k/2) × (4m²/9) = 14km²/3
m a
b³ = 8m³/27
m .co agl
l a se
a g
= 343k³/8 − 49k²m/2 + 14km²/3 − 8m³/27

co m
m .
m as e
.co g l
Tip: Parts (ii) and (iv) are the same expression with a different letter. Spotting that
a
a s em saves you the second calculation — identities do not care what the letters are called.

agl
se m
com (i) 17 × 21 (ii) 104 × 96 (iii) 24 × 16 (iv) g l a
.
Find the values using suitable identities:
a
em(−107)³ (viii) (−299)³
Q2

a s
147³ (v) 199³ (vi) 127³ (vii)

agl

co m
(i) 17 × 21 — both near 19, but at different distances, so use (x + a)(x + b) with x = 20.
m .
m as e
.=co(20 − 3)(20 + 1) = 400 + (−3 + 1)(20) + (−3)(1) a g l
se m
g l a
a = 400 − 40 − 3 = 357
c
m .
m a s e
. co agl
(ii) 104 × 96 — equally spaced about 100, so use a² − b².

se m
l a
= (100 + 4)(100 − 4) =ag
10000 − 16 = 9984

co m
m .
m ase
.co


a g l Page 44 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

(iii) 24 × 16 — equally spaced about 20.

= (20 + 4)(20 − 4) = 400 − 16 = 384

(iv) 147³ = (150 − 3)³

= 150³ − 3(150)²(3) + 3(150)(3)² − 3³

= 3375000 − 202500 + 4050 − 27

= 3176523

(v) 199³ = (200 − 1)³

= 8000000 − 3(40000)(1) + 3(200)(1) − 1
= 8000000 − 120000 + 600 − 1 = 7880599

(vi) 127³ = (130 − 3)³

= 2197000 − 3(16900)(3) + 3(130)(9) − 27

= 2197000 − 152100 + 3510 − 27 = 2048383

(vii) (−107)³ = −107³ = −(100 + 7)³

107³ = 1000000 + 3(10000)(7) + 3(100)(49) + 343

= 1000000 + 210000 + 14700 + 343 = 1225043

so (−107)³ = −1225043

(viii) (−299)³ = −299³ = −(300 − 1)³

299³ = 27000000 − 3(90000)(1) + 3(300)(1) − 1

= 27000000 − 270000 + 900 − 1 = 26730899

so (−299)³ = −26730899

Page 45 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Why an odd power keeps the sign: (−a)³ = (−1)³a³ = −a³. Cubing a negative number
gives a negative answer, so compute the cube of the positive number and put the
minus back at the end. (An even power would have absorbed the sign, as in (−3x +
4)² above.)

Q3 Factor the following algebraic expressions: (i) 4y² + 1 + 1/(16y²) (ii) 9m² − 1/(25n²)
(iii) 27b³ − 1/(64b³) (iv) x² + 5x/6 + 1/6 (v) 27u³ − 1/125 − 27u²/5 + 9u/25 (vi) 64y³ +
z³/125 (vii) p³ + 27q³ + r³ − 9pqr (viii) 9m² − 12m + 4 (ix) 9x³ − (8/3)y³ + z³/3 + 6xyz
(x) 4x² + 9y² + 36z² + 12xz + 36yz + 24xy (xi) 27u³ − 1/216 − 9u²/2 + u/4

(i) 4y² + 1 + 1/(16y²) — the middle term 1 is the giveaway: 2 × 2y × 1/(4y) = 1.

= (2y)² + 2(2y)(1/4y) + (1/4y)² = (2y + 1/(4y))²

(ii) 9m² − 1/(25n²) — difference of two squares.

= (3m)² − (1/5n)² = (3m + 1/(5n))(3m − 1/(5n))

(iii) 27b³ − 1/(64b³) — difference of two cubes, x = 3b, y = 1/(4b).

= (3b)³ − (1/4b)³

= (3b − 1/(4b))(9b² + 3/4 + 1/(16b²))

(the middle term is xy = 3b × 1/4b = 3/4)

(iv) x² + 5x/6 + 1/6 — need a + b = 5/6 and ab = 1/6; take a = 1/2, b = 1/3.

= (x + 1/2)(x + 1/3) — or, clearing fractions, (2x + 1)(3x + 1)/6

(v) 27u³ − 27u²/5 + 9u/25 − 1/125 — four terms with alternating signs: a cube of a difference.

a = 3u, b = 1/5:

a³ = 27u³, 3a²b = 3(9u²)(1/5) = 27u²/5, 3ab² = 3(3u)(1/25) = 9u/25, b³ = 1/125 ✓
= (3u − 1/5)³

Page 46 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

(vi) 64y³ + z³/125 — sum of two cubes, x = 4y, y′ = z/5.

= (4y)³ + (z/5)³

= (4y + z/5)(16y² − 4yz/5 + z²/25)

(vii) p³ + 27q³ + r³ − 9pqr — the three-cubes identity, with 3q in place of the second letter.

= p³ + (3q)³ + r³ − 3(p)(3q)(r)

= (p + 3q + r)(p² + 9q² + r² − 3pq − 3qr − rp)

(viii) 9m² − 12m + 4

= (3m)² − 2(3m)(2) + 2² = (3m − 2)²

(ix) 9x³ − (8/3)y³ + z³/3 + 6xyz — the coefficients are not cubes, so take out 1/3 first.

= (1/3)(27x³ − 8y³ + z³ + 18xyz)
= (1/3)[ (3x)³ + (−2y)³ + z³ − 3(3x)(−2y)(z) ]

= (1/3)(3x − 2y + z)(9x² + 4y² + z² + 6xy + 2yz − 3zx)

(x) 4x² + 9y² + 36z² + 12xz + 36yz + 24xy — the three squares point to 2x, 3y and 6z, and their
cross terms are 2(2x)(3y) = 12xy, 2(3y)(6z) = 36yz, 2(6z)(2x) = 24zx. So the expression that factors
is

4x² + 9y² + 36z² + 12xy + 36yz + 24zx = (2x + 3y + 6z)²

Note on the printed question: the English edition prints this part as “12xz + 36yz +
24xy” — the coefficients 12 and 24 attached to the wrong pairs. With those
coefficients the expression is not a perfect square and does not factor at all (put x = y
= z = 1: it gives 4 + 9 + 36 + 12 + 36 + 24 = 121 = 11², but x = 1, y = 1, z = 0 gives 4 + 9 +
24 = 37, which is not a square of 2 + 3 = 5). The Hindi edition prints the intended
form, 12xy + 36yz + 24zx, and that is the version factored above.

(xi) 27u³ − 9u²/2 + u/4 − 1/216 — again a cube of a difference.

Page 47 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

a = 3u, b = 1/6:

3a²b = 3(9u²)(1/6) = 9u²/2 ✓ 3ab² = 3(3u)(1/36) = u/4 ✓ b³ = 1/216 ✓

= (3u − 1/6)³

How to tell a cube from a square at a glance: a perfect-square trinomial has three
terms and the outer two are squares; a perfect-cube expansion has four terms, the
outer two are cubes, and the two middle coefficients are in the ratio 3a²b : 3ab² = a :
b. In (v), 27u²/5 : 9u/25 = 15u : 1, and indeed a : b = 3u : 1/5 = 15u : 1. ✓

Q4 Simplify the following: (i) (4x² + 4x + 1)/(4x² − 1) (ii) 9(3a³ − 24b³)/(9a² − 36b²) (iii) (s³
+ 125t³)/(s² − 2st − 35t²). Note: Assume that the denominators are not equal to 0.

(i)

Numerator = (2x)² + 2(2x)(1) + 1² = (2x + 1)²

Denominator = (2x)² − 1² = (2x + 1)(2x − 1)

= (2x + 1)² / [(2x + 1)(2x − 1)] = (2x + 1)/(2x − 1)

(ii) — take out the numerical factors first, then use the difference of cubes and the difference of
squares.

Numerator = 9 × 3(a³ − 8b³) = 27(a − 2b)(a² + 2ab + 4b²)

Denominator = 9(a² − 4b²) = 9(a − 2b)(a + 2b)

= 27(a − 2b)(a² + 2ab + 4b²) / [9(a − 2b)(a + 2b)]

= 3(a² + 2ab + 4b²)/(a + 2b)

(iii)

Page 48 of 58

as e
Class 9 Maths Chapter 4 Exploring Algebraic Identities
a g l AglaSem · NCERT Solutions

co m
e m.
Numerator = s³ + (5t)³ = (s + 5t)(s² − 5st + 25t²)
m l as
.co
Denominator = s² − 2st − 35t² = (s − 7t)(s + 5t)
m a g
l a se
g
a= (s² − 5st + 25t²)/(s − 7t)

co m
m . ag
l a se
Tip: In (ii) resist the urge to cancel the 9s straight away. Factor completely first — the
useful cancellation is (a − 2b), a g you would not have seen it if you had stopped at
and
3(a³ − 8b³)/(a² − 4b²).

co m
em.
m l as
.co a g
a s em
Find possible expressions for the length and breadth of each of the following

gl rectangles whose areas are given by the following expressions in square units. (i)
Q5

a 25a² − 30ab + 9b² (ii) 36s² − 49t²

m a s
em
.co agl
s

a
gl the area into two factors gives a possible pair of
a
Area = length × breadth, so factorising
dimensions.
(i) 25a² − 30ab + 9b²
co m
m .
m as e
.co a g l
m
= (5a)² − 2(5a)(3b) + (3b)² = (5a − 3b)²

a s eLength
agl = (5a − 3b) units, Breadth = (5a − 3b) units

se m
com
Both sides are equal, so this rectangle is in fact a square.
g l a
m . a
ase
(ii) 36s² − 49t²

agl
= (6s)² − (7t)² = (6s + 7t)(6s − 7t)

co m
Length = (6s + 7t) units, Breadth = (6s − 7t) units
m .
m as e
.co a g l
se m Why the answer says ‘possible’: a given area can be split in many ways — 2(3s +
g l a
a 3.5t)(6s − 7t) has the same product. The question asks for a factorisation into
c
m .
sensible algebraic expressions, and for that the identities give the natural one. For
m a s e
. co agl
(ii), taking the larger factor as the length is the usual convention, since 6s + 7t > 6s −
e m
as
7t whenever t > 0.

a g l

co m
m .
m ase
.co


a g l Page 49 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Q6 Find possible expressions for the length, breadth, and heights of each of the
following cuboids whose volumes are given by the following expressions in cubic
units. (i) 6a² − 24b² (ii) 3ps² − 15ps + 12p

Volume = length × breadth × height, so factor each expression into three factors.
(i) 6a² − 24b²

= 6(a² − 4b²)

= 6(a − 2b)(a + 2b)

Length = 6 units, Breadth = (a + 2b) units, Height = (a − 2b) units

(ii) 3ps² − 15ps + 12p

= 3p(s² − 5s + 4)

Need two numbers with sum −5 and product 4: they are −1 and −4

= 3p(s − 1)(s − 4)
Length = 3p units, Breadth = (s − 1) units, Height = (s − 4) units

Tip: Always pull out the common numerical or letter factor first — in (i) it supplies
one whole dimension, and in (ii) it supplies 3p. What is left is then a quadratic you
already know how to split.

Q7 The village playground is shaped as a square of side 40 metres. A path of width s
metres is created around the playground for people to walk. Find an expression for
the area of the path in terms of s.

The path is the region between two squares: the outer square (playground + path on both sides)
and the playground itself.

Page 50 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

40 + 2s

path, width s s

40 playground 40 × 40

The playground, 40 m square, with a path of width s all round. The outer square has side 40 + 2s.

Side of outer square = 40 + s + s = (40 + 2s) m

Area of outer square = (40 + 2s)² = 40² + 2(40)(2s) + (2s)²

= 1600 + 160s + 4s²

Area of playground = 40² = 1600

Area of path = (1600 + 160s + 4s²) − 1600

= (4s² + 160s) m² = 4s(s + 40) m²

Why the side gains 2s and not s: the path runs all the way round, so it adds s
metres on the left and s metres on the right. The same happens top and bottom.
Forgetting the second s is the commonest slip in path problems.

Page 51 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Check it yourself: take s = 1. The path should be a 1 m border round a 40 m square:
four strips of 40 × 1 plus four corner squares of 1 × 1, that is 160 + 4 = 164 m². The
formula gives 4(1) + 160(1) = 164 m². ✓ The 4s² is exactly the four corners.

Q8 If a number plus its reciprocal equals 10/3, find the number.

Let the number be x. Then its reciprocal is 1/x, and x ≠ 0.

x + 1/x = 10/3

Multiply throughout by 3x:

3x² + 3 = 10x

3x² − 10x + 3 = 0

Split the middle term: two numbers with product 3 × 3 = 9 and sum −10 are −1 and −9.

3x² − 9x − x + 3 = 0

3x(x − 3) − 1(x − 3) = 0

(x − 3)(3x − 1) = 0

x = 3 or x = 1/3

Both answers are genuine, and they are reciprocals of each other.

Check: 3 + 1/3 = 10/3 ✓ 1/3 + 3 = 10/3 ✓

Why two answers had to appear: the condition x + 1/x = 10/3 does not distinguish
between x and 1/x — swapping them leaves the left side unchanged. So the
solutions must come in reciprocal pairs, and here that pair is 3 and 1/3. Notice their
product is 1, which matches the constant term 3 divided by the leading coefficient 3.

Page 52 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Q9 A rectangular pool has area 2x² + 7x + 3 square hastas. If its width is 2x + 1 hastas,
find its length. Hasta was a unit used to measure length.

Length = area ÷ width, so factor the area and see which factor is the given width.

2x² + 7x + 3

Split 7x using two numbers with product 2 × 3 = 6 and sum 7: they are 1 and 6

= 2x² + x + 6x + 3

= x(2x + 1) + 3(2x + 1)

= (2x + 1)(x + 3)

Width = (2x + 1) hastas ⇒ Length = (x + 3) hastas

Did you know? The hasta (from the Sanskrit for ‘hand’) was a standard length in
ancient India, measured from the elbow to the tip of the middle finger — roughly 45
cm. It appears throughout the Śulba Sūtras, where altar areas are prescribed in
square hastas.

Q10 *If both x − 2 and x − 1/2 are factors of px² + 5x + r, show that p = r.

If x − 2 is a factor, the expression must be zero at x = 2; likewise it must be zero at x = 1/2.

At x = 2: p(2)² + 5(2) + r = 0 ⇒ 4p + r = −10 …(1)

At x = 1/2: p(1/2)² + 5(1/2) + r = 0 ⇒ p/4 + r = −5/2

Multiply by 4: p + 4r = −10 …(2)

Subtract (2) from (1):

Page 53 of 58

as e
Class 9 Maths Chapter 4 Exploring Algebraic Identities
a g l AglaSem · NCERT Solutions

co m
e m.
(4p + r) − (p + 4r) = −10 − (−10)
m l as
3p − 3r = 0
m .co a g
l a se
g
p=r ■
a

co m
The two equations even pin down the values: putting r = p in (1) gives 5p = −10, so p = r = −2 and

m .
the expression is −2x² + 5x − 2 = −(2x − 1)(x − 2). Both stated factors are visible.
e ag
g l as
a
Why ‘factor’ means ‘root’: if x − k is a factor, then px² + 5x + r = (x − k) × (something).
Substituting x = k makes the first bracket zero, so the whole product is zero. This is
co m
m.
the Factor Theorem, and it converts a statement about factors into two ordinary

m as e
.co l
equations you can solve.
a g
a s em
a gl Tip: Notice how the symmetry does the work. The two roots, 2 and 1/2, are
reciprocals, and for a quadratic the product of the roots is (constant)/(leading
m a s
.co agl
coefficient) = r/p. Reciprocal roots multiply to 1, so r/p = 1 — the same conclusion in
one line.
se m
g l a
a

co m
Q11 *If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ − 3abc = − 25.
m .
as e
. com a g l
s em

gl a
a
Use the three-cubes identity from Section 4.7:

se m
com g l a
. a
a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
m
ase
agl
The first bracket is given. For the second, first get a² + b² + c² from the square of the sum:

co m
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
m .
m as e
.co
5² = a² + b² + c² + 2(10)
a g l
se m 25 = a² + b² + c² + 20
g l a
a c
a² + b² + c² = 5
m .
m a s e
Now substitute both pieces:
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 54 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

a³ + b³ + c³ − 3abc = (5)(5 − 10)

= 5 × (−5)

= −25 ■

Why no individual value of a, b or c is needed: the identity expresses the answer
entirely in terms of the two symmetric quantities a + b + c and ab + bc + ca, and both
are given. Symmetric expressions in three letters can almost always be rebuilt from
these building blocks — that is what makes identities such powerful shortcuts.

Did you know? The numbers here are unusual: a² + b² + c² = 5 while ab + bc + ca = 10
forces a, b, c to be non-real. The proof does not care — an identity holds for all
values, and only the two given sums were ever used.

Q12 *By factoring the expression, check that n³ − n is always divisible by 6 for all
natural numbers n. Give reasons.

Factor first.

n³ − n = n(n² − 1)

= n(n − 1)(n + 1)

= (n − 1) · n · (n + 1)

So n³ − n is the product of three consecutive natural numbers. Now argue divisibility.

Divisible by 2. Among any two consecutive integers one is even, so among three there is
certainly an even one. Hence 2 divides the product.
Divisible by 3. Any integer leaves remainder 0, 1 or 2 on division by 3. Among three
consecutive integers, all three remainders occur exactly once, so one of them is a multiple of
3. Hence 3 divides the product.
Therefore divisible by 6. The product is divisible by 2 and by 3, and 2 and 3 have no
common factor other than 1, so their product 6 divides it.

Page 55 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

N (N − 1)N(N + 1) N³ − N ÷6

2 1×2×3 6 1

5 4×5×6 120 20

10 9 × 10 × 11 990 165

Why ‘2 divides it and 3 divides it’ gives 6: this step needs 2 and 3 to be coprime. If
a number is divisible by 4 and by 6 you may not conclude it is divisible by 24 — 12 is a
counter-example. Here gcd(2, 3) = 1, so the conclusion is safe.

Try This: The same argument shows n⁵ − n is always divisible by 30, since n⁵ − n = (n
− 1)n(n + 1)(n² + 1) and one can show 5 always divides it too.

Q13 *Find the value of (i) x³ + y³ − 12xy + 64, when x + y = − 4 (ii) x³ − 8y³ − 36xy − 216,
when x = 2y + 6

Both parts are the identity x³ + y³ + z³ − 3xyz in disguise, with the third letter chosen to make the
first bracket vanish.
(i) x³ + y³ − 12xy + 64, when x + y = −4

Notice 64 = 4³ and 12xy = 3(x)(y)(4). Take z = 4:

x³ + y³ + 4³ − 3(x)(y)(4) = x³ + y³ + 64 − 12xy ✓

= (x + y + 4)(x² + y² + 16 − xy − 4y − 4x)

Given x + y = −4, the first bracket is (−4) + 4 = 0

Value = 0

(ii) x³ − 8y³ − 36xy − 216, when x = 2y + 6

Page 56 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

Write −8y³ = (−2y)³ and −216 = (−6)³. Then with the three letters x, −2y, −6:

−3(x)(−2y)(−6) = −36xy ✓

x³ + (−2y)³ + (−6)³ − 3(x)(−2y)(−6)

= (x − 2y − 6)(x² + 4y² + 36 + 2xy − 12y + 6x)

Given x = 2y + 6, the first bracket is (2y + 6) − 2y − 6 = 0

Value = 0

Why both answers are zero: the identity says the expression equals (sum of the
three letters) × (a second bracket). The condition supplied in each part is precisely
the statement that the sum of the three letters is zero. So the whole product
collapses, whatever the second bracket happens to be — you never need to evaluate
it. Recognising the shape is the entire problem.

Tip: The general fact behind this is worth remembering: if x + y + z = 0 then x³ + y³ +
z³ = 3xyz. Both parts above are that statement read backwards.

Chapter at a glance
An equation such as x² − 1 = 24 holds only for x = 5 or x = −5; an identity such as (x + y)² = x²
+ 2xy + y² holds for every value of x and y. That is the whole difference.
A square of side (a + b) cuts into a², b² and two ab rectangles, so (a + b)² = a² + 2ab + b².
Replacing b by −b — legitimate, because an identity is true for all values — gives (a − b)² = a²
− 2ab + b². A square of side (a + b + c) cuts into nine pieces and gives the three-letter
version.
Read backwards, every identity is a factorisation rule. Recognising a² + 2ab + b² inside 9x² +
24xy + 16y² is what lets you write it as (3x + 4y)².
Algebra tiles turn factorising x² + 7x + 12 into a jigsaw: the only split of 7x that makes the
unit tiles fill a rectangle is 3x + 4x, because 3 + 4 = 7 and 3 × 4 = 12. Without tiles this is the
‘splitting the middle term’ method: find a and b with a + b = the x-coefficient and ab = the
constant.
Cubes come from the same idea one dimension up: a cube of edge (a + b) splits into two
cubes and six cuboids, giving (a + b)³ = a³ + 3a²b + 3ab² + b³.
Identities make arithmetic quick — 205² = (200 + 5)², 104 × 96 = 100² − 4², 199³ = (200 − 1)³ —
and they simplify rational expressions once numerator and denominator are factorised and

Page 57 of 58

Class 9 Maths Chapter 4 Exploring Algebraic Identities AglaSem · NCERT Solutions

a common non-zero factor is cancelled.

Quick revision

IDENTITY STATEMENT WHERE IT COMES USE IT FOR
FROM

Square of a sum (a + b)² = a² + 2ab + b² Fig. 4.2 — square of side a 64², expanding (7x + 4y)²
+ b cut into a², b² and two
ab rectangles

Square of a (a − b)² = a² − 2ab + b² Fig. 4.3, or put −b for b in 79², factoring 16y² − 24y +
difference the first identity 9

Square of a (a + b + c)² = a² + b² + c² + Fig. 4.4 — a 3 × 3 grid on a 117², factoring 9a² + 4b² +
trinomial 2ab + 2bc + 2ca square of side a + b + c c² − 12ab + 6ac − 4bc

Difference of two a² − b² = (a + b)(a − b) Fig. 4.5 — cut a strip off a 104 × 96, factoring 36s² −
squares square and slide it round 49t²

Śhrīdharāchārya’s a² = (a + b)(a − b) + b² the same figure, read the 35², 55², 105² in the head
form other way

Product of two (x + a)(x + b) = x² + (a + Fig. 4.7 — algebra tiles factoring x² + 7x + 12, x² −
binomials b)x + ab forming a rectangle 5x + 6

General binomial (px + a)(qx + b) = pq·x² + Fig. 4.8 — tiles for (2x + 3) factoring 6x² + 7x + 2, 10x²
product (pb + aq)x + ab (3x + 1) − 11x − 6

Cube of a sum (a + b)³ = a³ + 3a²b + 3ab² + Fig. 4.10 — a cube of edge 147³, factoring p³ + 6p²q +
b³ a + b in 2 cubes + 6 12pq² + 8q³
cuboids

Cube of a difference (a − b)³ = a³ − 3a²b + 3ab² − put −b for b in the cube of 199³, factoring 8n³ −
b³ a sum 60n²m + 150nm² − 125m³

Sum and difference x³ − y³ = (x − y)(x² + xy + multiply out with the factoring 27b³ − 1/(64b³),
of cubes y²) distributive property; the 64y³ + z³/125
x³ + y³ = (x + y)(x² − xy + middle terms cancel
y²)

Three cubes x³ + y³ + z³ − 3xyz = (x + y multiply out; all nine cross factoring p³ + 27q³ + r³ −
+ z)(x² + y² + z² − xy − yz − terms cancel in threes 9pqr; proving
zx) a³+b³+c³−3abc = −25

Page 58 of 58