Chapter 4 Exploring Algebraic Identities

NCERT Solutions for Class 9 Maths Chapter 4 Exploring Algebraic Identities

NCERT Solutions for Class 9 Maths Chapter 4 help students strengthen their understanding of Algebraic Identities concepts and build the confidence needed to score well in exams. These solutions align with the latest NCERT syllabus and the CBSE curriculum, ensuring students study from content that stays fully exam-relevant.

Access Detailed answers for every Important question in the chapter 4, and they download can the complete solution set as a free PDF for offline study. Subject Experts at ALLEN designs each solution to help students follow the correct method, avoid common mistakes, and revise efficiently before exams.

1.0Download NCERT Class 9 Maths Chapter 4 Exploring Algebraic Identities

Class 9 Maths Chapter 4: Exploring Algebraic Identities introduces students to algebraic identities and their applications in simplifying algebraic expressions and solving mathematical problems. Understanding these identities helps build a strong foundation for higher algebra. Download NCERT Solutions for Class 9 Maths Chapter 4 PDF to access step-by-step answers based on the latest NCERT syllabus 2026-27 and prepare confidently for your exams

NCERT Solutions Class 9 Maths Chapter 4 Algebraic Identities

2.0Learning Outcomes - NCERT Class 9 Maths Chapter 4 Exploring Algebraic Identities

Understand the concept of algebraic identities and distinguish them from algebraic equations.
Visualize algebraic identities using geometric models and algebra tiles.
Apply standard algebraic identities to expand algebraic expressions.
Factorize algebraic expressions using suitable identities.
Discover and use new algebraic identities to solve mathematical problems.
Simplify rational algebraic expressions using factorization techniques.
Solve application-based and NCERT textbook questions confidently using algebraic identities.

3.0Detailed Class 9 Maths Chapter 4 Exploring Algebraic Identities Solutions

EXERCISE : 4.1

1.Using the identity $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ , expand the following: (i) $(7 x + 4 y)^{2}$ (ii) $(\frac{7}{5} x + \frac{3}{2} y)^{2}$ (iii) $(2.5 p + 1.5 q)^{2}$ (iv) $(\frac{3}{4} s + 8 t)^{2}$ (v) $(x + \frac{1}{2 y})^{2}$ (vi) $(\frac{1}{x} + \frac{1}{y})^{2}$

Sol. (i) Using $(a + b)^{2} = a^{2} + 2 ab + b^{2}$

$Here, a = 7 x and b = 4 y = (7 x)^{2} + 2 (7 x) (4 y) + (4 y)^{2} = 49 x^{2} + 56 x y + 16 y^{2}$

(ii) Here, $a = (\frac{7}{5}) x$ and $b = (\frac{3}{2}) y$

$= [(\frac{7}{5}) x]^{2} + 2 (\frac{7}{5}) x (\frac{3}{2}) y + [(\frac{3}{2}) y]^{2} = (\frac{49}{25}) x^{2} + 2 \times (\frac{21}{10}) xy + (\frac{9}{4}) y^{2} = (\frac{49}{25}) x^{2} + (\frac{42}{10}) xy + (\frac{9}{4}) y^{2}$

Simplify:

$= (\frac{49}{25}) x^{2} + (\frac{21}{5}) x y + (\frac{9}{4}) y^{2}$

(iii) Here, $a = 2.5 p$ and $b = 1.5 q$

$= (2.5 p)^{2} + 2 (2.5 p) (1.5 q) + (1.5 q)^{2} = 6.25 p^{2} + 7.5 pq + 2.25 q^{2}$

(iv) Here, $a = (\frac{3}{4}) s$ and $b = 8 t$

$= [(\frac{3}{4}) s]^{2} + 2 \times (\frac{3}{4}) s \times 8 t + (8 t)^{2} = (\frac{9}{16}) s^{2} + 2 \times (\frac{24}{4}) s t + 64 t^{2} = (\frac{9}{16}) s^{2} + 12 s t + 64 t^{2}$

(v) Here, $a = x$ and $b = \frac{1}{( 2 y )}$

$= x^{2} + 2 \times x \times [\frac{1}{( 2 y )}] + [\frac{1}{( 2 y )}]^{2} = x^{2} + \frac{( 2 x )}{( 2 y )} + \frac{1}{( 4 y ^{2} )} = x^{2} + \frac{x}{y} + \frac{1}{( 4 y ^{2} )}$

(vi) Here, $a = \frac{1}{x}$ and $b = \frac{1}{y}$

$= (\frac{1}{x})^{2} + 2 (\frac{1}{x}) (\frac{1}{y}) + (\frac{1}{y})^{2} = \frac{1}{x ^{2}} + \frac{2}{( x y )} + \frac{1}{y ^{2}}$

2.Using the same identity, find the values of the following: (i) $(64)^{2}$ (ii) $(105)^{2}$ (iii) $(205)^{2}$

Sol. (i) (64) $^{2}$

$64 = 60 + 4 (60 + 4)^{2} = 6 0^{2} + 2 \times 60 \times 4 + 4^{2} = 3600 + 480 + 16 = 4096$

(ii) $(105)^{2}$

$105 = 100 + 5 Now (105)^{2} = (100 + 5)^{2} = 10 0^{2} + 2 \times 100 \times 5 + 5^{2} = 10000 + 1000 + 25 = 11025$

(iii) $(205)^{2}$

$205 = 200 + 5 So, (205)^{2} = (200 + 5)^{2} = 20 0^{2} + 2 \times 200 \times 5 + 5^{2} = 40000 + 2000 + 25 = 42025$

EXERCISE : 4.2

1.Factor completely: (i) $9 x^{2} + 24 x y + 16 y^{2}$ (ii) $4 s^{2} + 20 st + 25 t^{2}$ (iii) $49 x^{2} + 28 xy + 4 y^{2}$ (iv) $64 p^{2} + (\frac{32}{3}) pq + (\frac{4}{9}) q^{2}$ (v) $3 a^{2} + 4 ab + (\frac{4}{3}) b^{2}$ (vi) $(\frac{9}{5}) s^{2} + 6 sv + 5 v^{2}$

Sol. (i) $9 x^{2} + 24 xy + 16 y^{2}$ We know that:

$9 x^{2} = (3 x)^{2} 16 y^{2} = (4 y)^{2} 24 x y = 2 \times 3 x \times 4 y So, 9 x^{2} + 24 x y + 16 y^{2} = (3 x)^{2} + 2 \times 3 x \times 4 y + (4 y)^{2} = (3 x + 4 y)^{2}$

(ii) $4 s^{2} + 20 st + 25 t^{2}$

$4 s^{2} = (2 s)^{2} 25 t^{2} = (5 t)^{2} 20 s t = 2 \times 2 s \times 5 t So, 4 s^{2} + 20 s t + 25 t^{2} = (2 s)^{2} + 2 \times 2 s \times 5 t + (5 t)^{2} = (2 s + 5 t)^{2}$

(iii) $49 x^{2} + 28 xy + 4 y^{2}$

$49 x^{2} = (7 x)^{2} 4 y^{2} = (2 y)^{2} 28 x y = 2 \times 7 x \times 2 y So, 49 x^{2} + 28 x y + 4 y^{2} = (7 x)^{2} + 2 \times 7 x \times 2 y + (2 y)^{2} = (7 x + 2 y)^{2}$

(iv) $64 p^{2} + (\frac{32}{3}) pq + (\frac{4}{9}) q^{2}$

$64 p^{2} = (8 p)^{2} (\frac{4}{9}) q^{2} = (\frac{2}{3} q)^{2}$

Middle term:

$2 \times 8 p \times (\frac{2}{3} q) = \frac{32}{3} pq So, 64 p^{2} + (\frac{32}{3}) pq + (\frac{4}{9}) q^{2} = (8 p)^{2} + 2 \times 8 p \times (\frac{2}{3} q) + (\frac{2}{3} q)^{2} = (8 p + \frac{2}{3} q)^{2}$

(v) $3 a^{2} + 4 ab + (\frac{4}{3}) b^{2}$ Take common factor 3 :

$= 3 [a^{2} + (\frac{4}{3}) ab + (\frac{4}{9}) b^{2}]$

Now, $a^{2} = (a)^{2}$

$(\frac{4}{9}) b^{2} = (\frac{2}{3} b)^{2} (\frac{4}{3}) ab = 2 \times a \times (\frac{2}{3} b) So, 3 a^{2} + 4 ab + (\frac{4}{3}) b^{2} = 3 [a^{2} + (\frac{4}{3}) ab + (\frac{4}{9}) b^{2}] = 3 [a^{2} + 2 \times a \times (\frac{2}{3} b) + (\frac{2}{3} b)^{2}] = 3 (a + \frac{2}{3} b)^{2}$

(vi) $(\frac{9}{5}) s^{2} + 6 sv + 5 v^{2}$ Take common factor $\frac{1}{5}$

$= (\frac{1}{5}) [9 s^{2} + 30 sv + 25 v^{2}]$

Now, $9 s^{2} = (3 s)^{2}$

$25 v^{2} = (5 v)^{2} 30 s v = 2 \times 3 s \times 5 v So, (\frac{9}{5}) s^{2} + 6 s v + 5 v^{2} = (\frac{1}{5}) [9 s^{2} + 30 s v + 25 v^{2}] = (\frac{1}{5}) [(3 s)^{2} + 2 \times 3 s \times 5 v + (5 v)^{2}] = (\frac{1}{5}) (3 s + 5 v)^{2}$

2.Find the value of the following using the identity: $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ . (i) $(79)^{2}$ (ii) $(193)^{2}$ (iii) $(299)^{2}$

Sol. (i) (79) $^{2}$

$79 = 80 - 1 = (80 - 1)^{2} = 8 0^{2} - 2 \times 80 \times 1 + 1^{2} = 6400 - 160 + 1 = 6241$

(ii) $(193)^{2}$

$193 = 200 - 7 = (200 - 7)^{2} = 20 0^{2} - 2 \times 200 \times 7 + 7^{2} = 40000 - 2800 + 49 = 37249$

(iii) $(299)^{2}$

$299 = 300 - 1 = (300 - 1)^{2} = 30 0^{2} - 2 \times 300 \times 1 + 1^{2} = 90000 - 600 + 1 = 89401$

EXERCISE : 4.3

1.Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier. (i) $117^{2}$ (ii) $7 8^{2}$ (iii) $19 8^{2}$ (iv) $214^{2}$ $(v) 110 4^{2}$ (vi) $1120^{2}$

Sol. (i) $117^{2}$

$117 = 110 + 7 So, 11 7^{2} = (110 + 7)^{2} = 11 0^{2} + 2 (110) (7) + 7^{2} = 12100 + 1540 + 49 = 13689$

(ii) $7 8^{2}$

$78 = 80 - 2 So, 7 8^{2} = (80 - 2)^{2} = 8 0^{2} - 2 (80) (2) + 2^{2} = 6400 - 320 + 4 = 6084$

(iii) $19 8^{2}$

$198 = 200 - 2 So, 19 8^{2} = (200 - 2)^{2} = 20 0^{2} - 2 (200) (2) + 2^{2} = 40000 - 800 + 4 = 39204$

(iv) $214^{2}$

$214 = 200 + 14 So, 21 4^{2} = (200 + 14)^{2} = 20 0^{2} + 2 (200) (14) + 1 4^{2} = 40000 + 5600 + 196 = 45796$

(v) $110 4^{2}$

$1104 = 1100 + 4 So, 110 4^{2} = (1100 + 4)^{2} = 110 0^{2} + 2 (1100) (4) + 4^{2} = 1210000 + 8800 + 16 = 1218816$

(vi) $1120^{2}$

$1120 = 1100 + 20 So, 112 0^{2} = (1100 + 20)^{2} = 110 0^{2} + 2 (1100) (20) + 2 0^{2} = 1210000 + 44000 + 400 = 1254400$

2.Factor using suitable identities: (i) $16 y^{2} - 24 y + 9$ (ii) $\frac{9}{4} s^{2} + 6 st + 4 t^{2}$ (iii) $\frac{m ^{2}}{9} + \frac{mk}{3} + \frac{k ^{2}}{4} + 3 nk + 2 mn + 9 n^{2}$ (iv) $\frac{p ^{2}}{16} - 2 + \frac{16}{p ^{2}}$ (v) $9 a^{2} + 4 b^{2} + c^{2} - 12 ab + 6 ac - 4 bc$

Sol. (i) $16 y^{2} - 24 y + 9$

$16 y^{2} = (4 y)^{2} 9 = 3^{2} - 24 y = - 2 (4 y) (3)$

Therefore, $16 y^{2} - 24 y + 9$

$= (4 y)^{2} - 2 (4 y) (3) + 3^{2} = (4 y - 3)^{2}$

(ii) $(\frac{9}{4}) s^{2} + 6 st + 4 t^{2}$

$(\frac{9}{4}) s^{2} = [\frac{( 3 s )}{2}]^{2} 4 t^{2} = (2 t)^{2} 6 s t = 2 \times (\frac{3 s}{2}) \times (2 t)$

Therefore, $(\frac{9}{4}) s^{2} + 6 st + 4 t^{2}$

$= [(\frac{3 s}{2})]^{2} + 2 \times (\frac{3 s}{2}) \times (2 t) + (2 t)^{2} = [(\frac{3 s}{2} + 2 t)]^{2}$

(iii) $\frac{m ^{2}}{9} + \frac{mk}{3} + \frac{k ^{2}}{4} + 3 nk + 2 mn + 9 n^{2}$ Grouping the terms as:

$\frac{m ^{2}}{9} + \frac{mk}{3} + \frac{k ^{2}}{4} + 2 mn + 3 nk + 9 n^{2}$

We have:

$\frac{m ^{2}}{9} = (\frac{m}{3})^{2} \frac{k ^{2}}{4} = (\frac{k}{2})^{2} 9 n^{2} = (3 n)^{2}$

Now checking the cross terms:

$2 \times (\frac{m}{3}) \times (\frac{k}{2}) = \frac{mk}{3} 2 \times (\frac{m}{3}) \times (3 n) = 2 mn 2 \times (\frac{k}{2}) \times (3 n) = 3 nk So, \frac{m ^{2}}{9} + \frac{mk}{3} + \frac{k ^{2}}{4} + 3 nk + 2 mn + 9 n^{2} = \frac{m ^{2}}{9} + \frac{k ^{2}}{4} + 9 n^{2} + \frac{mk}{3} + 3 nk + 2 mn$

[Rearranging the terms]

$= (\frac{m}{3})^{2} + (\frac{k}{2})^{2} + (3 n)^{2} + 2 \times (\frac{m}{3}) \times (\frac{k}{2}) + 2 \times (\frac{m}{3}) \times (3 n) + 2 \times (\frac{k}{2}) \times (3 n) = (\frac{m}{3} + \frac{k}{2} + 3 n)^{2}$

(iv) $\frac{p ^{2}}{16} - 2 + \frac{16}{p ^{2}}$

$- 2 = - 2 \times (\frac{p}{4}) \times (\frac{4}{p})$

Now, $\frac{p ^{2}}{16} = (\frac{p}{4})^{2}$

$\frac{16}{p ^{2}} = (\frac{4}{p})^{2}$

Writing -2 as:

$So, \frac{p ^{2}}{16} - 2 + \frac{16}{p ^{2}} = (\frac{p}{4})^{2} - 2 (\frac{p}{4}) (\frac{4}{p}) + (\frac{4}{p})^{2} = (\frac{p}{4} - \frac{4}{p})^{2}$

(v) $9 a^{2} + 4 b^{2} + c^{2} - 12 ab + 6 ac - 4 bc$ We have:

$9 a^{2} = (3 a)^{2} 4 b^{2} = (- 2 b)^{2}$

[Here $4 b^{2} = (- 2 b)^{2}$ as in two negative terms -12ab and -4bc, b is common]

$c^{2} = c^{2}$

Now, $9 a^{2} + 4 b^{2} + c^{2} - 12 ab + 6 ac - 4 bc$

$= (3 a)^{2} + (- 2 b)^{2} + c^{2} + 2 (3 a) (- 2 b) + 2 (3 a) (c) + 2 (- 2 b) (c) = (3 a - 2 b + c)^{2}$

3.Expand the following using the identity

$(a + b + c) = a^{2} + b^{2} + c^{2} + 2 ab + 2 b c + 2 c a$

(i) $(p + 3 q + 7 r)^{2}$ (ii) $(3 x - 2 y + 4 z)^{2}$

Sol. (i) $(p + 3 q + 7 r)^{2}$ Here, $a = p, b = 3 q, c = 7 r$ So, $(p + 3 q + 7 r)^{2}$

$= p^{2} + (3 q)^{2} + (7 r)^{2} + 2 (p) (3 q) + 2 (3 q) (7 r) + 2 (p) (7 r) = p^{2} + 9 q^{2} + 49 r^{2} + 6 pq + 42 q r + 14 p r$

(ii) $(3 x - 2 y + 4 z)^{2}$ Here, $a = 3 x, b = - 2 y, c = 4 z$ So, $(3 x - 2 y + 4 z)^{2}$

$= (3 x)^{2} + (- 2 y)^{2} + (4 z)^{2} + 2 (3 x) (- 2 y) + 2 (- 2 y) (4 z) + 2 (3 x) (4 z) = 9 x^{2} + 4 y^{2} + 16 z^{2} - 12 x y - 16 yz + 24 x z$

4.It this an identity?

$(a + b - c)^{2} + (a - b + c)^{2} + (a - b - c)^{2} = 2 a^{2} + 2 b^{2} + 2 c^{2} .$

Sol. To check whether this is an identity, expand the left-hand side. First Part: $(a + b - c)^{2}$

$= a^{2} + b^{2} + c^{2} + 2 ab - 2 a c - 2 b c$

Second Part: $(a - b + c)^{2}$

$= a^{2} + b^{2} + c^{2} - 2 ab + 2 a c - 2 b c$

Third Part: $(a - b - c)^{2}$

$= a^{2} + b^{2} + c^{2} - 2 ab - 2 a c + 2 b c$

Now adding all three:

$LHS = (a^{2} + b^{2} + c^{2} + 2 ab - 2 ac - 2 bc)$

$+ (a^{2} + b^{2} + c^{2} - 2 ab + 2 ac - 2 bc)$ $+ (a^{2} + b^{2} + c^{2} - 2 ab - 2 ac + 2 bc)$ $= 3 a^{2} + 3 b^{2} + 3 c^{2} - 2 ab - 2 ac - 2 bc$ This is NOT equal to $2 a^{2} + 2 b^{2} + 2 c^{2}$ Hence, the given statement is not true for all values of a, b and c. So, it is not an identity.

EXERCISE : 4.4

1.Fill in the blanks to complete the following identities: (i) $s^{2} - 11 s + 24 = ($ $____$ ) ( $____$ ) (ii) ( $____$ $) (x + 1) = (3 x^{2} - 4 x - 7)$ (iii) $10 x^{2} - 11 x - 6 = (2 x -$ $____$ ) ( $____$ + 2) (iv) $6 x^{2} + 7 x + 2 = ($ $____$ )( $____$ )

Sol. (i) $s^{2} - 11 s + 24 = ($ $____$ ) ( $____$ ) We need two numbers whose: Sum = -11 and Product = 24 These are -3 and -8. So, $s^{2} - 11 s + 24$ $= s^{2} - 8 s - 3 s + 24$ $= s (s - 8) - 3 (s - 8)$ $= (s - 3) (s - 8)$ So, $s^{2} - 11 s + 24 = (s - 3) (s - 8)$ ( (ii) ( $____$ $____$ $) (x + 1) = (3 x^{2} - 4 x - 7)$ RHS: $3 x^{2} - 4 x - 7$ $= 3 x^{2} - 7 x + 3 x - 7$ $= x (3 x - 7) + 1 (3 x - 7)$ $= (3x - 7) (x + 1)$ So, $(3 x - 7) (x + 1) = (3 x^{2} - 4 x - 7)$

(iii)

$10 x^{2} - 11 x - 6 = (2 x - - L H S = 10 x^{2} - 11 x - 6 = 10 x^{2} - 15 x + 4 x - 6 = 5 x (2 x - 3) + 2 (2 x - 3) = (5 x + 2) (2 x - 3) = (2 x - 3) (5 x + 2)$

$____$ ) ( $____$ + 2) So, $10 x^{2} - 11 x - 6 = (2 x - 3) (5 x + 2)$

(iv)

$6 x^{2} + 7 x + 2 = (= 6 x^{2} + 3 x + 4 x + 2 = 3 x (2 x + 1) + 2 (2 x + 1) = (3x + 2) (2x + 1)$

) So, $6 x^{2} + 7 x + 2 = (3 x + 2) (2 x + 1)$

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

Sol.

(i) $(41)^{2}$

$41 = 40 + 1 = (40 + 1)^{2} = 4 0^{2} + 2 \times 40 \times 1 + 1^{2} = 1600 + 80 + 1 = 1681$

(ii)

$(27)^{2} 27 = 30 - 3 So, (27)^{2} = (30 - 3)^{2} = (30)^{2} - 2 \times 30 \times 3 + (3)^{2} = 900 - 180 + 9 = 729$

(iii)

$(23 \times 17) 23 \times 17 = (20 + 3) (20 - 3) = 2 0^{2} - 3^{2} = 400 - 9 = 391$

(iv)

$(135)^{2} 135 = 100 + 35 So, (135)^{2} = (100 + 35)^{2} = (100)^{2} + 2 \times 100 \times 35 + (35)^{2} = 10000 + 7000 + 1225 = 18225$

(v)

$(97)^{2} 97 = 100 - 3 So, (97)^{2} = (100 - 3)^{2} = (100)^{2} - 2 \times 100 \times 3 + (3)^{2} = 10000 - 600 + 9 = 9409$

(vi) (18 × 29)

$= (20 - 2) (20 + 9) = 2 0^{2} + (- 2 + 9) \times 20 - 2 \times 9 = 400 + 140 - 18 = 522$

(vii) (34 × 43)

$= (38 - 4) (38 + 5) = 3 8^{2} + (- 4 + 5) \times 38 - 4 \times 5 = 3 8^{2} + 38 - 20 = 1444 + 38 - 20 = 1462$

(viii) $(205)^{2}$

$205 = 200 + 5 = (200 + 5)^{2} = (200)^{2} + 2 \times 200 \times 5 + (5)^{2} = 40000 + 2000 + 25 = 42025$

3.Factor the following: (i) $9 a^{2} + b^{2} + 4 c^{2} - 6 ab + 12 ac - 4 bc$ (ii) $16 s^{2} + 25 t^{2} - 40 st$ (iii) $r^{2} - r - 42$ (iv) $49 g^{2} + 14 gh + h^{2}$ (v) $64 u^{2} + 121 v^{2} + 4 w^{2} - 176 uv - 2 uw + 44 vw$

Sol. (i)

$9 a^{2} + b^{2} + 4 c^{2} - 6 ab + 12 a c - 4 b c = (3 a)^{2} + (- b)^{2} + (2 c)^{2} + 2 (3 a) (- b) + 2 (3 a) (2 c) + 2 (- b) (2 c) = (3 a - b + 2 c)^{2}$

(ii)

$16 s^{2} + 25 t^{2} - 40 st Now, 16 s^{2} + 25 t^{2} - 40 st = 16 s^{2} - 40 st + 25 t^{2}$

[Rearranging the terms]

$= (4 s)^{2} - 2 (4 s) (5 t) + (5 t)^{2} = (4 s - 5 t)^{2}$

(iii) $r^{2} - r - 42$ We need two numbers whose Product = -42 and Sum = -1. These numbers are -7 and 6. So, $r^{2} - r - 42$ $= r^{2} - 7 r + 6 r - 42$ $= r (r - 7) + 6 (r - 7)$ $= (r - 7) (r + 6)$

(iv)

$49 g^{2} + 14 g h + h^{2} 49 g^{2} + 14 g h + h^{2} = (7 g)^{2} + 2 (7 g) (h) + (h)^{2} = (7 g + h)^{2}$

(v) $64 u^{2} + 121 v^{2} + 4 w^{2} - 176 uv - 32 uw +$ 44vw $= 64 u^{2} + 121 v^{2} + 4 w^{2} - 176 uv + 44 vw$

32uw [Rearranging the terms] $= (8 u)^{2} + (- 11 v)^{2} + (- 2 w)^{2} +$ 2(8u)(-11v) + 2(-11v)(-2w) + 2(8u)(-2w) $= (8 u - 11 v - 2 w)^{2}$

EXERCISE : 4.5

1.Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero: (i) $\frac{( 3 p ^{2} - 3 pq - 18 q ^{2} )}{( p ^{2} + 3 pq - 10 q ^{2} )}$ (ii) $\frac{( n ^{3} - 3 n ^{2} m + 3 nm ^{2} - m ^{3} )}{( 5 m ^{2} - 10 mn + 5 n ^{2} )}$ (iii) $\frac{w ^{3} - v ^{3} + x ^{3} + 3 wvx}{w ^{2} + v ^{2} + x ^{2} - 2 wv - 2 vx + 2 wx}$ (iv) $\frac{( 4 y ^{2} - 20 yz + 25 z ^{2} )}{( 25 z ^{2} - 4 y ^{2} )}$ (v) $\frac{( x ^{2} + x - 6 ) ( x ^{2} - 7 x + 12 )}{( x ^{2} - 6 x + 8 ) ( x ^{2} - 9 )}$ (vi) $\frac{p ^{2} - 16}{p ^{2} - 4 p + 4}$

Sol. (i)

$\frac{( 3 p ^{2} - 3 pq - 18 q ^{2} )}{( p ^{2} + 3 pq - 10 q ^{2} )}$

First factorising the numerator:

$3 p^{2} - 3 pq - 18 q^{2} = 3 (p^{2} - pq - 6 q^{2}) = 3 [p^{2} - 3 pq + 2 pq - 6 q^{2}] = 3 [p (p - 3 q) + 2 q (p - 3 q)] = 3 (p - 3 q) (p + 2 q)$

Now factorising the denominator:

$p^{2} + 3 pq - 10 q^{2} = p^{2} + 5 pq - 2 pq - 10 q^{2} = p (p + 5 q) - 2 q (p + 5 q) = (p + 5 q) (p - 2 q)$

So, $\frac{( 3 p ^{2} - 3 pq - 18 q ^{2} )}{( p ^{2} + 3 pq - 10 q ^{2} )}$

$= \frac{3 ( p - q ) ( p + 2 q )}{[( p + 5 q ) ( p - 2 q )]}$

(ii) $\frac{( n ^{3} - 3 n ^{2} m + 3 nm ^{2} - m ^{3} )}{( 5 m ^{2} - 10 mn + 5 n ^{2} )}$ Factorising the numerator using identity:

$n^{3} - 3 n^{2} m + 3 nm^{2} - m^{3} = (n - m)^{3}$

Factorising the denominator:

$5 m^{2} - 10 mn + 5 n^{2} = 5 (m^{2} - 2 mn + n^{2}) = 5 (m - n)^{2}$

Now, $\frac{( n - m ) ^{3}}{[ 5 ( m - n ) ^{2} ]}$

$= \frac{- ( m - n ) ^{3}}{[ 5 ( m - n ) ^{2} ]}$

[Since, $(n - m) = - (m - n)$ , so, $(n - m)^{3}$

$= - (m - n)^{3}] = \frac{- ( m - n )}{5}$

(iii) $\frac{w ^{3} - v ^{3} + x ^{3} + 3 w vx}{w ^{2} + v ^{2} + x ^{2} - 2 w v - 2 vx + 2 w x}$ . Factorising the numerator:

$w^{3} - v^{3} + x^{3} + 3 w vx = w^{3} + (- v)^{3} + x^{3} - 3 w (- v) x = (w - v + x) (w^{2} + v^{2} + x^{2} + w v + vx - w x)$

Now denominator:

$w^{2} + v^{2} + x^{2} - 2 w v - 2 vx + 2 w x = w^{2} + (- v)^{2} + x^{2} + 2 w (- v) + 2 (- v) x + 2 x w = (w - v + x)^{2}$

Therefore,

$\frac{w ^{3} - v ^{3} + x ^{3} + 3 w vx}{w ^{2} + v ^{2} + x ^{2} - 2 w v - 2 vx + 2 w x} = \frac{[ w - v + x ] ( w ^{2} + v ^{2} + x ^{2} + w v + vx - w x )}{( w - v + x ) ^{2}} = \frac{( w ^{2} + x ^{2} + v ^{2} + w x + vx - w v )}{( w + x - v )}$

(iv) $\frac{( 4 y ^{2} - 20 yz + 25 z ^{2} )}{( 25 z ^{2} - 4 y ^{2} )}$ Factor numerator:

$4 y^{2} - 20 yz + 25 z^{2} = (2 y)^{2} - 2 (2 y) (5 z) + (5 z)^{2} = (2 y - 5 z)^{2}$

Factorising the denominator:

$25 z^{2} - 4 y^{2} = (5 z - 2 y) (5 z + 2 y)$

So, $\frac{( 4 y ^{2} - 20 yz + 25 z ^{2} )}{( 25 z ^{2} - 4 y ^{2} )}$

$\frac{( 5 z - 2 y ) ^{2}}{[( 5 z - 2 y ) ( 5 z + 2 y )]}$

$[$ Since $(2 y - 5 z)^{2} = (- 1)^{2} (5 z - 2 y)^{2}]$

$= (5 z - 2 y)^{2}] = \frac{( 5 z - 2 y )}{( 5 z + 2 y )}$

(v) $\frac{( x ^{2} + x - 6 ) ( x ^{2} - 7 x + 12 )}{( x ^{2} - 6 x + 8 ) ( x ^{2} - 9 )}$ Factorising each polynomial:

$x^{2} + x - 6 = x^{2} + 3 x - 2 x - 6 = x (x + 3) - 2 (x + 3) = (x + 3) (x - 2) x^{2} - 7 x + 12 = x^{2} - 3 x - 4 x + 12 = x (x - 3) - 4 (x - 3) = (x - 3) (x - 4) x^{2} - 6 x + 8 = x^{2} - 4 x - 2 x + 8 = x (x - 4) - 2 (x - 4) = (x - 2) (x - 4) x^{2} - 9 = x^{2} - 3^{2} = (x - 3) (x + 3)$

Substituting each polynomial, we get:

$\frac{( x + 3 ) ( x - 2 ) ( x - 3 ) ( x - 4 )}{( x - 2 ) ( x - 4 ) ( x - 3 ) ( x + 3 )} = 1$

(vi) $p^{2} - 16 = (p + 4) (p - 4)$

$p^{2} - 4 p + 4 = (p - 2)^{2}$

Now $\frac{( p + 4 ) ( p - 4 )}{( p - 2 ) ^{2}}$

END OF CHAPTER EXERCISE

1.Using suitable identities to find the following products: (i) $(- 3 x + 4)^{2}$ (ii) $(2 s + 7) (2 s - 7)$ (iii) $(p^{2} + \frac{1}{2}) (p^{2} - \frac{1}{2})$ (iv) $(2 n + 7) (2 n - 7)$ (v) $(s - 2 t) (s^{2} + 2 st + 4 t^{2})$ (vi) $[\frac{1}{( 2 r )} - 4 r]^{2}$ (vii) $(- 3 m + 4 k - l)^{2}$ (viii) $(x - \frac{1}{3} y)^{3}$ (ix) $(\frac{7}{2} k - \frac{2}{3} m)^{3}$

Sol. (i)

$(- 3 x + 4)^{2} = (- 3 x)^{2} + 2 (- 3 x) (4) + 4^{2} = 9 x^{2} - 24 x + 16$

(ii)

$(2 s + 7) (2 s - 7) = (2 s)^{2} - 7^{2} = 4 s^{2} - 49$

(iii)

$(p^{2} + \frac{1}{2}) (p^{2} - \frac{1}{2}) = (p^{2})^{2} - (\frac{1}{2})^{2} = p^{4} - \frac{1}{4}$

(iv)

$(2 n + 7) (2 n - 7) = (2 n)^{2} - 7^{2} = 4 n^{2} - 49$

(v)

$(s - 2 t) (s^{2} + 2 s t + 4 t^{2}) = s^{3} - (2 t)^{3} = s^{3} - 8 t^{3}$

(vi)

$[\frac{1}{( 2 r )} - 4 r]^{2} = (\frac{1}{( 2 r )})^{2} - 2 (\frac{1}{2 r}) (4 r) + (4 r)^{2} = \frac{1}{( 4 r ^{2} )} - 4 + 16 r^{2}$

(vii) $(- 3 m + 4 k - l)^{2}$

$So, (- 3 m + 4 k - l)^{2} = (- 3 m)^{2} + (4 k)^{2} + (- l)^{2} + 2 (- 3 m) (4 k) + 2 (4 k) (- l) + 2 (- 3 m) (- l) = 9 m^{2} + 16 k^{2} + l^{2} - 24 mk - 8 k l + 6 m l$

(viii)

$(x - \frac{1}{3} y)^{3}$

$= x^{3} - 3 x^{2} (\frac{y}{3}) + 3 x (\frac{y}{3})^{2} - (\frac{y}{3})^{3} = x^{3} - x^{2} y + 3 x (\frac{y ^{2}}{9}) - \frac{y ^{3}}{27} = x^{3} - x^{2} y + (\frac{1}{3}) x y^{2} - \frac{y ^{3}}{27}$

(ix) $(\frac{7}{2} k - \frac{2}{3} m)^{3}$

$Here, a = \frac{7 k}{2}, b = \frac{2 m}{3} (\frac{7 k}{2})^{3} - (\frac{2 m}{3})^{3} - 3 (\frac{7 k}{2})^{2} (\frac{2 m}{3}) + 3 (\frac{7 k}{2}) (\frac{2 m}{3})^{2} = \frac{343}{8} k^{3} - \frac{49}{2} k^{2} m + \frac{14}{3} k m^{2} - \frac{8}{27} m^{3}$

2.Find the value using suitable identities: (i) $17 \times 21$ (ii) $104 \times 96$ (iii) 24 × 16 (iv) $14 7^{3}$ (v) $19 9^{3}$ (vi) $127^{3}$ (vii) $(- 107)^{3}$ (viii) $(- 299)^{3}$

Sol. (i) $17 \times 21$

$17 \times 21 = (19 - 2) (19 + 2) = 1 9^{2} - 2^{2} = 361 - 4 = 357$

(ii) $104 \times 96$

$104 \times 96 = (100 + 4) (100 - 4) = 10 0^{2} - 4^{2} = 10000 - 16 = 9984$

(iii) 24 × 16

$24 \times 16 = (20 + 4) (20 - 4) = 2 0^{2} - 4^{2} = 400 - 16 = 384$

(iv) $14 7^{3}$

$147 = 150 - 3 So, 14 7^{3} = (150 - 3)^{3} = 15 0^{3} - 3 \times 15 0^{2} \times 3 + 3 \times 150 \times 3^{2} - 3^{3} = 3375000 - 202500 + 4050 - 27 = 3176523$

(v) $19 9^{3}$

$199 = 200 - 1 So, 19 9^{3} = (200 - 1)^{3} = 20 0^{3} - 3 \times 20 0^{2} \times 1 + 3 \times 200 \times 1^{2} - 1^{3} = 8000000 - 120000 + 600 - 1 = 7880599$

(vi) $127^{3}$

$Here, 127 = 130 - 3 So, 12 7^{3} = (130 - 3)^{3} = 13 0^{3} - 3 \times 13 0^{2} \times 3 + 3 \times 130 \times 3^{2} - 3^{3} = 2197000 - 152100 + 3510 - 27 = 2048383$

(vii) $(- 107)^{3}$

$(- 107)^{3} = - (10 7^{3})$

Now, $107 = 100 + 7$

$10 7^{3} = (100 + 7)^{3} = 10 0^{3} + 3 \times 10 0^{2} \times 7 + 3 \times 100 \times 7^{2} + 7^{3} = 1000000 + 210000 + 14700 + 343 = 1225043 So, (- 107)^{3} = - 1225043$

(viii) $(- 299)^{3}$

$(- 299)^{3} = - (29 9^{3})$

Here, $299 = 300 - 1$

$So, 29 9^{3} = (300 - 1)^{3} = 30 0^{3} - 3 \times 30 0^{2} \times 1 + 3 \times 300 \times 1^{2} - 1^{3} = 27000000 - 270000 + 900 - 1 = 26730900 - 1 = 26730899 So, (- 299)^{3} = - 26730899$

3.Factor the following algebraic expressions: (i) $4 y^{2} + 1 + \frac{1}{16 y ^{2}}$ (ii) $9 m^{2} - \frac{1}{25 n ^{2}}$

(iii) $27 b^{3} - \frac{1}{64 b ^{3}}$ (iv) $x^{2} + \frac{5 x}{6} + \frac{1}{6}$ (v) $27 u^{3} - \frac{1}{125} - \frac{27 u ^{2}}{5} + \frac{9 u}{25}$ (vi) $64 y^{3} + \frac{1}{125} z^{3}$ (vii) $p^{3} + 27 q^{3} + r^{3} - 9 pqr$ (viii) $9 m^{2} - 12 m + 4$ (ix) $9 x^{3} - \frac{8}{3} y^{3} + \frac{z ^{3}}{3} + 6 xyz$ (x) $4 x^{2} + 9 y^{2} + 36 z^{2} + 12 x z + 36 yz + 24 x y$ (xi) $27 u^{3} - \frac{1}{216} - \frac{9 u ^{2}}{2} + \frac{u}{4}$

Sol. (i) $4 y^{2} + 1 + \frac{1}{16 y ^{2}}$ Here, we have $4 y^{2} = (2 y)^{2}$

$\frac{1}{16 y ^{2}} = (\frac{1}{4 y})^{2}$

and $2 \times 2 y \times \frac{1}{4 y} = 1$ Therefore, $4 y^{2} + 1 + \frac{1}{16 y ^{2}}$

$= (2 y + \frac{1}{4 y})^{2}$

(ii) $9 m^{2} - \frac{1}{25 n ^{2}}$

$= (3 m)^{2} - (\frac{1}{5 n})^{2} = (3 m + \frac{1}{5 n}) (3 m - \frac{1}{5 n})$

(iii) $27 b^{3} - \frac{1}{64 b ^{3}}$

$= (3 b)^{3} - (\frac{1}{4 b})^{3}$

(iv) $x^{2} + \frac{5 x}{6} + \frac{1}{6}$ We need two numbers whose sum is $\frac{5}{6}$ and product is $\frac{1}{6}$ . These numbers are $\frac{1}{2}$ and $\frac{1}{3}$ .

$So, x^{2} + \frac{5 x}{6} + \frac{1}{6} = x^{2} + \frac{x}{2} + \frac{x}{3} + \frac{1}{6}$

[Splitting the middle term]

$= x (x + \frac{1}{2}) + \frac{1}{3} (x + \frac{1}{2}) = (x + \frac{1}{3}) (x + \frac{1}{2})$

(v) $27 u^{3} - \frac{1}{125} - \frac{27 u ^{2}}{5} + \frac{9 u}{25}$ Given expression:

$27 u^{3} - \frac{1}{125} - \frac{27 u ^{2}}{5} + \frac{9 u}{25} = 27 u^{3} - \frac{27 u ^{2}}{5} + \frac{9 u}{25} - \frac{1}{125}$

[Rearranging the terms]

$= (3 u)^{3} - 3 (3 u)^{2} (\frac{1}{5}) + 3 (3 u) (\frac{1}{5})^{2} - (\frac{1}{5})^{3} = (3 u - \frac{1}{5})^{3}$

(vi)

$64 y^{3} + \frac{1}{125} z^{3} = (4 y)^{3} + (\frac{z}{5})^{3} = (4 y + \frac{z}{5}) [(4 y)^{2} - (4 y) (\frac{z}{5}) + (\frac{z}{5})^{2}] = (4 y + \frac{z}{5}) (16 y^{2} - \frac{4 yz}{5} + \frac{z ^{2}}{25})$

(vii)

$p^{3} + 27 q^{3} + r^{3} - 9 pq r = p^{3} + (3 q)^{3} + r^{3} - 3 (p) (3 q) (r) = (p + 3 q + r) [p^{2} + (3 q)^{2} + r^{2} - p (3 q) - (3 q) r - p r)] = (p + 3 q + r) (p^{2} + 9 q^{2} + r^{2} - 3 pq - 3 q r - p r)$

(viii)

$= = 9 m^{2} - 12 m + 4 (3 m)^{2} - 2 (3 m) (2) + (2)^{2} (3 m - 2)^{2}$

(ix)

$9 x^{3} - \frac{8}{3} y^{3} + \frac{z ^{3}}{3} + 6 x yz = (\frac{1}{3}) [27 x^{3} - 8 y^{3} + z^{3} + 18 x yz] = (\frac{1}{3}) [(3 x)^{3} + (- 2 y)^{3} + z^{3} - 3 (3 x) (- 2 y) z] = (\frac{1}{3}) (3 x - 2 y + z) [(3 x)^{2} + (- 2 y)^{2} + z^{2} - (3 x) (- 2 y) - (- 2 y) (z) - (z) (3 x)] = (\frac{1}{3}) (3 x - 2 y + z) [9 x^{2} + 4 y^{2} + z^{2} + 6 x y + 2 yz - 3 z x]$

(x)

$4 x^{2} + 9 y^{2} + 36 z^{2} + 12 x y + 36 yz + 24 x z = (2 x)^{2} + (3 y)^{2} + (6 z)^{2} + 2 (2 x) (3 y) + 2 (3 y) (6 z) + 2 (2 x) (6 z) = (2 x + 3 y + 6 z)^{2}$

(xi) $27 u^{3} - \frac{1}{216} - \frac{9 u ^{2}}{2} + \frac{u}{4}$ .

$= 27 u^{3} - \frac{9 u ^{2}}{2} + \frac{u}{4} - \frac{1}{216}$

[Rearranging the terms]

$= (3 u)^{3} - 3 (3 u)^{2} (\frac{1}{6})$

Simplifying the following: (i) $\frac{( 4 x ^{2} + 4 x + 1 )}{( 4 x ^{2} - 1 )}$ (ii) $\frac{9 ( 3 a ^{3} - 24 b ^{3} )}{( 9 a ^{2} - 36 b ^{2} )}$ (iii) $\frac{( s ^{3} + 125 t ^{3} )}{( s ^{2} - 2 st - 35 t ^{2} )}$

Sol. (i) Factorising numerator:

$4 x^{2} + 4 x + 1 = (2 x)^{2} + 2 (2 x) (1) + 1^{2} = (2 x + 1)^{2}$

Factorising denominator:

$4 x^{2} - 1 = (2 x + 1) (2 x - 1)$

Now the expression:

$= \frac{( 4 x ^{2} + 4 x + 1 )}{( 4 x ^{2} - 1 )}$

$= \frac{( 2 x + 1 ) ^{2}}{[( 2 x + 1 ) ( 2 x - 1 )]} = \frac{( 2 x + 1 )}{( 2 x - 1 )}$

(ii) $\frac{9 ( 3 a ^{3} - 24 b ^{3} )}{( 9 a ^{2} - 36 b ^{2} )}$ First simplify:

$9 (3 a^{3} - 24 b^{3}) = 27 (a^{3} - 8 b^{3})$

and $9 a^{2} - 36 b^{2} = 9 (a^{2} - 4 b^{2})$

$So, \frac{9 ( 3 a ^{2} - 24 b ^{3} )}{( 9 a ^{2} - 36 b ^{2} )}$

$= \frac{27 ( a ^{3} - 8 b ^{3} )}{9 ( a ^{2} - 4 b ^{2} )}$

$= \frac{3 ( a - 2 b ) ( a ^{2} + 2 ab + 4 b ^{2} )}{( a ^{2} - 4 b ^{2} )}$

$= \frac{3 ( a - 2 b ) ( a ^{2} + 2 ab + 4 b ^{2} )}{[( a - 2 b ) ( a + 2 b )]}$

$= \frac{3 ( a ^{2} + 2 ab + 4 b ^{2} )}{( a + 2 b )}$

(iii) $\frac{( s ^{3} + 125 t ^{3} )}{( s ^{2} - 2 st - 35 t ^{2} )}$ Factorising numerator:

$s^{3} + 125 t^{3} = s^{3} + (5 t)^{3} = (s + 5 t) (s^{2} - 5 s t + 25 t^{2})$

Factorising denominator:

$s^{2} - 2 s t - 35 t^{2} = s^{2} - 7 s t + 5 s t - 35 t^{2} = s (s - 7 t) + 5 t (s - 7 t) = (s + 5 t) (s - 7 t)$

Now the given expression:

$= \frac{( s ^{3} + 125 t ^{3} )}{( s ^{2} - 2 s t - 35 t ^{2} )} = \frac{( s + 5 t ) ( s ^{2} - 5 s t + 25 t ^{2} )}{[( s + 5 t ) ( 5 - 7 t )]} = \frac{( s ^{2} - 5 s t + 25 t ^{2} )}{( 5 - 7 t )}$

[Cancelling common factor $(s + 5 t)$ ]. 5. Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units. (i) $25 a^{2} - 30 ab + 9 b^{2}$ (ii) $36 s^{2} - 49 t^{2}$

Sol. Geometric Interpretations of Factorization (i) $25 a^{2} - 30 ab + 9 b^{2}$ This is a perfect square:

$25 a^{2} - 30 ab + 9 b^{2} = (5 a)^{2} - 2 (5 a) (3 b) + (3 b)^{2} = (5 a - 3 b)^{2}$

So possible length and breadth are $(5 a - 3 b)$ and $(5 a - 3 b)$ . (ii) $36 s^{2} - 49 t^{2}$

$= (6 s)^{2} - (7 t)^{2} = (6 s + 7 t) (6 s - 7 t)$

So possible length and breadth are $(6 s + 7 t)$ and $(6 s - 7 t)$ . 6. 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) $6 a^{2} - 24 b^{2}$ (ii) $3 p s^{2} - 15 p s + 12 p$

Sol. (i) $6 a^{2} - 24 b^{2}$ Take common factor:

$6 a^{2} - 24 b^{2} = 6 (a^{2} - 4 b^{2}) = 6 [a^{2} - (2 b)^{2}] = 6 (a + 2 b) (a - 2 b)$

So possible dimensions are 6, (a + 2b) and (a - 2b).

(ii) $3 ps^{2} - 15 ps + 12 p$

$= 3 p (s^{2} - 5 s + 4)$

[Taking common factor]

$= 3 p (s^{2} - 4 s - 1 s + 4) = 3 p [s (s - 4) - 1 (s - 4)] = 3 p (s - 1) (s - 4)$

So, the possible dimensions are 3p, (s -1) and (s - 4).

7. 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.

Sol. Side of playground $= 40 m$ Since a path of width s metres is made all around the outside, the side of the outer square becomes: 40 + 2s Area of outer square $= (40 + 2 s)^{2}$ Area of playground $= 4 0^{2} = 1600$ Therefore, area of the path:

$= (40 + 2 s)^{2} - 1600 = (1600 + 160 s + 4 s^{2}) - 1600 = 4 s^{2} + 160 s$

Hence, the required expression: Area of path $= 4 s^{2} + 160 s$ square metres. 8. If a number plus its reciprocal equals $\frac{10}{3}$ ,find the number.

Sol. Let the number be x. Then, according to question:

$x + \frac{1}{x} = \frac{10}{3}$

Multiply both sides by 3 x :

$3 x^{2} + 3 = 10 x$

$\Rightarrow 3 x^{2} - 10 x + 3 = 0$

$\Rightarrow 3 x^{2} - 9 x - x + 3 = 0$ $\Rightarrow 3 x (x - 3) - 1 (x - 3) = 0$ $\Rightarrow (3 x - 1) (x - 3) = 0$ So, $3 x - 1 = 0$ or $x - 3 = 0$ Hence, $x = \frac{1}{3}$ or $x = 3$ Therefore, the number is 3 or $\frac{1}{3}$ . 9. A rectangular pool has area $2 x^{2} + 7 x + 3$ square hastas. If its width is $2 x + 1$ hastas, find its length. Hasta was a unit used to measure length.

Sol. Area of pool $= 2 x^{2} + 7 x + 3$ Width $= 2 x + 1$

$Length = \frac{Area}{Width}$

$= \frac{( 2 x ^{2} + 7 x + 3 )}{( 2 x + 1 )}$

$= \frac{( 2 x ^{2} + 6 x + x + 3 )}{( 2 x + 1 )}$

$= \frac{[ 2 x ( x + 3 ) + 1 ( x + 3 )]}{( 2 x + 1 )}$

$= \frac{( 2 x + 1 ) ( x + 3 )}{( 2 x + 1 )}$

$= x + 3$

Therefore, the length of the pool is $x + 3$ hastas.

If both $x - 2$ and $x - \frac{1}{2}$ are factors of $px^{2} +$ $5 x + r$ , show that $p = r$ . Sol. Since $x - 2$ and $x - \frac{1}{2}$ are factors, the quadratic polynomial can be written as:

$p x^{2} + 5 x + r = k (x - 2) (x - \frac{1}{2}) \Rightarrow (x - 2) (x - \frac{1}{2}) = k [x^{2} - (\frac{5}{2}) x + 1] \Rightarrow x^{2} - (\frac{1}{2}) x - 2 x + 1 = k [x^{2} - (\frac{5}{2}) x + 1] \Rightarrow x^{2} - (\frac{5}{2}) x + 1 = k [x^{2} - (\frac{5}{2}) x + 1]$

Comparing the coefficients on both sides, we get: Coefficient of $x^{2}$ gives: $k = p$ Constant term gives: $k = r$ Therefore, $p = r$ Hence proved. 11. If $a + b + c = 5$ and $ab + bc + ca = 10$ , then prove that $a^{3} + b^{3} + c^{3} - 3 abc = - 25$ .

Sol. Given:

$a + b + c = 5 ab + b c + c a = 10$

First finding $a^{2} + b^{2} + c^{2}$ using:

$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (ab + b c + c a)$

Now, $a^{3} + b^{3} + c^{3} - 3 abc = (a + b + c)$

$(a^{2} + b^{2} + c^{2} - ab - b c - c a) = (5) (5 - 10) = (5) (- 5) = - 25$

Hence proved $a^{3} + b^{3} + c^{3} - 3 abc = - 25$ . 12. By factoring the expression, check that $n^{3} - n$ is always divisible by 6 for all natural numbers n. Give reasons.

Sol. We have: $n^{3} - n$

$= n (n^{2} - 1) = n (n - 1) (n + 1)$

So, $n^{3} - n = n (n - 1) (n + 1)$ Thus, $n^{3} - n$ is the product of three consecutive natural numbers: ( $n - 1$ ), n, ( $n + 1$ ). Among any three consecutive natural numbers: one is always divisible by 3 at least one is always even, so divisible by 2 Therefore, their product is always divisible by: $2 \times 3 = 6$ Hence, $n^{3} - n$ is always divisible by 6 for all natural numbers n. 13. Find : (i) $x^{3} + y^{3} - 12 xy + 64$ , when $x + y = - 4$ (ii) $x^{3} - 8 y^{3} - 36 xy - 216$ , when $x = 2 y + 6$ Sol. (i) Since $x + y = - 4$ , we have:

$(x + y)^{3} = (- 4)^{3} = - 64 So, - 64 = x^{3} + y^{3} + 3 x y (- 4)$

$\Rightarrow - 64 = x^{3} + y^{3} - 12 x y$ Therefore, $x^{3} + y^{3} - 12 xy = - 64$ Now, $x^{3} + y^{3} - 12 xy + 64$

$= -64 + 64 = 0$

Hence, the value is 0 . (ii) $x^{3} - 8 y^{3} - 36 xy - 216$ , when $x - 2 y - 6 = 0$ Given: $x - 2 y - 6 = 0$ So, $x - 2 y = 6$ Using identity

$a^{3} + b^{3} + c^{3} - 3 abc = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - bc - ca)$

If $a + b + c = 0$ then $a^{3} + b^{3} + c^{3} = 3 abc$ Here $a = x, b = - 2 y, c = - 6$ $x^{3} - 8 y^{3} + (- 6)^{3} = 36 xy$ Now put this value in

$x^{3} - 8 y^{3} - 36 x y - 216 = 0 36 x y - 36 x y = 0$

Therefore value $x^{3} - 8 y^{3} - 36 xy - 216$ is 0.

4.0Important Key Concepts Class 9 Maths Exploring Algebraic Identities - Chapter 4

Topic	What Students Learn
Visualising (a + b)² and (a − b)²	A geometric derivation of these identities using squares and rectangles, building an intuitive picture of why the formulas hold before they're ever applied numerically
Expanding Algebraic Expressions	Step-by-step application of standard identities to expand expressions containing variables, fractions, and decimal coefficients accurately
Fast Numerical Calculations	Techniques to compute squares of large numbers, such as 117² or 78², by breaking them into convenient sums or differences instead of relying on long multiplication
Factorisation Using Algebra Tiles	A hands-on, visual method of factorising quadratic expressions by physically arranging tiles that represent each term
Factorisation Without Algebra Tiles	The splitting-the-middle-term method, which factorises quadratic expressions purely through algebraic manipulation
More Identities: (a + b + c)²	Expansion of the three-term square identity and its use in simplifying more complex expressions involving three variables
Sum and Difference of Cubes	The identities for a³ + b³ and a³ − b³, along with the special three-variable identity x³ + y³ + z³ − 3xyz, and how each applies to factorisation problems
Finding New Identities	Derivation of further identities such as (a + b)³, (a − b)³, and x³ − y³ = (x − y)(x² + xy + y²) by logically extending known identities rather than memorising them
Simplifying Rational Expressions	Combining factorisation with identity-based techniques to cancel common factors and reduce algebraic fractions to their simplest form

5.0Exercise-wise NCERT Class 9 Maths Chapter 4 Exploring Algebraic Identities

Chapter 4 of the newly updated NCERT syllabus is divided into five exercises and an End-of-Chapter set, moving students from basic identities to advanced factorisation. Here's what each exercise covers.

Exercise	Topics Covered
Exercise 4.1	Introduction to algebraic identities and their geometric interpretation using algebra tiles and area models.
Exercise 4.2	Verification and application of standard algebraic identities to expand algebraic expressions.
Exercise 4.3	Factorisation of algebraic expressions using algebraic identities and simplification of expressions.
Exercise 4.4	Discovering additional algebraic identities and applying them to solve mathematical problems.
End-of-Chapter Exercises	Mixed questions covering algebraic identities, factorisation, simplification, and application-based problems from the chapter

6.0Quick Revision on Class 9 Maths Chapter 4 - Exploring Algebraic Identities

7.0Related Study Materials Class 9 Maths

Continue your learning with ALLEN’s Class 9 Maths study material as per latest NCERT syllabus. Along with NCERT Solutions, NCERT textbooks, revision notes, sample papers and previous years question papers to strengthen your concepts, revise effectively and prepare with confidence for school and CBSE examinations.

CBSE Class 9 Maths Syllabus

Class 9 Maths Revision Notes

NCERT Textbook for Class 9 Maths

CBSE Sample Papers for Class 9 Maths

8.0Advantages of Chapter 4 Maths Class 9 NCERT Solutions

Develops understanding of algebraic expressions Clearly defines variables, constants, terms and coefficients.
Simplifies Algebraic Operations: It helps students in adding, subtracting, multiplying and dividing algebraic expressions correctly.
Develops Factorisation Skills Enhances the skill of factorising algebraic expressions through different methods.
Improves Expression Simplification: Teaches systematic ways to simplify complex algebraic expressions.
Increases Conceptual Clarity : Helps the students to identify different types of algebraic expressions and their components.
Provides a Solid Base in Algebra: Prepares students for solving algebraic problems and understanding difficult algebraic concepts in later grades.

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NCERT Solutions for Class 9 Maths Chapter 4 Exploring Algebraic Identities

1.0Download NCERT Class 9 Maths Chapter 4 Exploring Algebraic Identities

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