---
title: "NCERT Solutions Class 9 Maths Ch 4 Exploring Algebraic Identities"
url: https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions
dateModified: 2026-10-07T15:43:39+00:00
---

# NCERT Solutions Class 9 Maths Ch 4 Exploring Algebraic Identities

This chapter's questions cover the exploration, application, and factorisation of algebraic identities, including expanding binomials, factorising polynomials, and simplifying rational expressions.

Free PDF (27 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-9/swavid-ncert-solutions-class-9-maths-chapter-4-exploring-algebraic-identities-4d2a728afe.pdf

## Exercise Set 4.1

### Question 1

*3 marks · Short answer*

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

**Part (i)**

1. Given expression is $(7x + 4y)^2$.
2. Substitute $a = 7x$ and $b = 4y$ into $(a+b)^2 = a^2 + 2ab + b^2$.
3. $(7x + 4y)^2 = (7x)^2 + 2(7x)(4y) + (4y)^2$.
4. $= 49x^2 + 56xy + 16y^2$.

Answer (i): $49x^2 + 56xy + 16y^2$

**Part (ii)**

1. Given expression is $\left(\frac{7}{5}x + \frac{3}{2}y\right)^2$.
2. Substitute $a = \frac{7}{5}x$ and $b = \frac{3}{2}y$ into $(a+b)^2 = a^2 + 2ab + b^2$.
3. $= \left(\frac{7}{5}x\right)^2 + 2\left(\frac{7}{5}x\right)\left(\frac{3}{2}y\right) + \left(\frac{3}{2}y\right)^2$.
4. $= \frac{49}{25}x^2 + \frac{21}{5}xy + \frac{9}{4}y^2$.

Answer (ii): $\frac{49}{25}x^2 + \frac{21}{5}xy + \frac{9}{4}y^2$

**Part (iii)**

1. Given expression is $(2.5p + 1.5q)^2$.
2. Substitute $a = 2.5p$ and $b = 1.5q$ into $(a+b)^2 = a^2 + 2ab + b^2$.
3. $= (2.5p)^2 + 2(2.5p)(1.5q) + (1.5q)^2$.
4. $= 6.25p^2 + 7.5pq + 2.25q^2$.

Answer (iii): $6.25p^2 + 7.5pq + 2.25q^2$

**Part (iv)**

1. Given expression is $\left(\frac{3}{4}s + 8t\right)^2$.
2. Substitute $a = \frac{3}{4}s$ and $b = 8t$ into $(a+b)^2 = a^2 + 2ab + b^2$.
3. $= \left(\frac{3}{4}s\right)^2 + 2\left(\frac{3}{4}s\right)(8t) + (8t)^2$.
4. $= \frac{9}{16}s^2 + 12st + 64t^2$.

Answer (iv): $\frac{9}{16}s^2 + 12st + 64t^2$

**Part (v)**

1. Given expression is $\left(x + \frac{1}{2y}\right)^2$.
2. Substitute $a = x$ and $b = \frac{1}{2y}$ into $(a+b)^2 = a^2 + 2ab + b^2$.
3. $= (x)^2 + 2(x)\left(\frac{1}{2y}\right) + \left(\frac{1}{2y}\right)^2$.
4. $= x^2 + \frac{x}{y} + \frac{1}{4y^2}$.

Answer (v): $x^2 + \frac{x}{y} + \frac{1}{4y^2}$

**Part (vi)**

1. Given expression is $\left(\frac{1}{x} + \frac{1}{y}\right)^2$.
2. Substitute $a = \frac{1}{x}$ and $b = \frac{1}{y}$ into $(a+b)^2 = a^2 + 2ab + b^2$.
3. $= \left(\frac{1}{x}\right)^2 + 2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right) + \left(\frac{1}{y}\right)^2$.
4. $= \frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}$.

Answer (vi): $\frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}$

**Answer:** Expanded forms of the given binomials using $(a+b)^2 = a^2 + 2ab + b^2$.

> Common mistake: Forgetting to multiply the middle term $2ab$ by 2.

### Question 2

*3 marks · Short answer*

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

**Part (i)**

1. Express $64$ as $(60 + 4)^2$.
2. Apply $(a+b)^2 = a^2 + 2ab + b^2$ with $a = 60$ and $b = 4$.
3. $= 60^2 + 2(60)(4) + 4^2$.
4. $= 3600 + 480 + 16 = 4096$.

Answer (i): $4096$

**Part (ii)**

1. Express $105$ as $(100 + 5)^2$.
2. Apply $(a+b)^2 = a^2 + 2ab + b^2$ with $a = 100$ and $b = 5$.
3. $= 100^2 + 2(100)(5) + 5^2$.
4. $= 10000 + 1000 + 25 = 11025$.

Answer (ii): $11025$

**Part (iii)**

1. Express $205$ as $(200 + 5)^2$.
2. Apply $(a+b)^2 = a^2 + 2ab + b^2$ with $a = 200$ and $b = 5$.
3. $= 200^2 + 2(200)(5) + 5^2$.
4. $= 40000 + 2000 + 25 = 42025$.

Answer (iii): $42025$

**Answer:** Values computed using $(a+b)^2 = a^2 + 2ab + b^2$.

> Common mistake: Incorrect splitting of the number or wrong multiplication in the middle term.

## Exercise Set 4.2

### Question 1

*3 marks · Short answer*

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

**Part (i)**

1. Rewrite the expression as $(3x)^2 + 2(3x)(4y) + (4y)^2$.
2. Use the identity $(a + b)^2 = a^2 + 2ab + b^2$ where $a = 3x$ and $b = 4y$.
3. Thus, $9x^2 + 24xy + 16y^2 = (3x + 4y)^2$.

Answer (i): $$ (3x + 4y)^2 $$

**Part (ii)**

1. Rewrite the expression as $(2s)^2 + 2(2s)(5t) + (5t)^2$.
2. Use the identity $(a + b)^2 = a^2 + 2ab + b^2$ where $a = 2s$ and $b = 5t$.
3. Thus, $4s^2 + 20st + 25t^2 = (2s + 5t)^2$.

Answer (ii): $$ (2s + 5t)^2 $$

**Part (iii)**

1. Rewrite the expression as $(7x)^2 + 2(7x)(2y) + (2y)^2$.
2. Use the identity $(a + b)^2 = a^2 + 2ab + b^2$ where $a = 7x$ and $b = 2y$.
3. Thus, $49x^2 + 28xy + 4y^2 = (7x + 2y)^2$.

Answer (iii): $$ (7x + 2y)^2 $$

**Part (iv)**

1. Rewrite the expression as $\left(8p\right)^2 + 2(8p)\left(\frac{2}{3}q\right) + \left(\frac{2}{3}q\right)^2$.
2. Use the identity $(a + b)^2 = a^2 + 2ab + b^2$ where $a = 8p$ and $b = \frac{2}{3}q$.
3. Thus, $64p^2 + \frac{32}{3}pq + \frac{4}{9}q^2 = \left(8p + \frac{2}{3}q\right)^2$.

Answer (iv): $$ \left(8p + \frac{2}{3}q\right)^2 $$

**Part (v)**

1. Take $\frac{1}{3}$ as a common factor to get $\frac{1}{3}(9a^2 + 12ab + 4b^2)$.
2. Rewrite the expression inside the bracket as $(3a)^2 + 2(3a)(2b) + (2b)^2$.
3. Apply the identity $(a + b)^2 = a^2 + 2ab + b^2$ to obtain $\frac{1}{3}(3a + 2b)^2$.

Answer (v): $$ \frac{1}{3}(3a + 2b)^2 $$

**Part (vi)**

1. Take $\frac{1}{5}$ as a common factor to get $\frac{1}{5}(9s^2 + 30sv + 25v^2)$.
2. Rewrite the expression inside the bracket as $(3s)^2 + 2(3s)(5v) + (5v)^2$.
3. Apply the identity $(a + b)^2 = a^2 + 2ab + b^2$ to obtain $\frac{1}{5}(3s + 5v)^2$.

Answer (vi): $$ \frac{1}{5}(3s + 5v)^2 $$

**Answer:** Factorised expressions for all six parts.

> Common mistake: Forgetting to take out the common fractional factor in parts (v) and (vi) before applying the identity.

### Question 2

*3 marks · Short answer*

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

**Part (i)**

1. Given: $(79)^2$.
2. Formula: $(a - b)^2 = a^2 - 2ab + b^2$.
3. Substitution: $(80 - 1)^2 = (80)^2 - 2(80)(1) + (1)^2$.
4. Result: $6400 - 160 + 1 = 6241$.

Answer (i): $$ 6241 $$

**Part (ii)**

1. Given: $(193)^2$.
2. Formula: $(a - b)^2 = a^2 - 2ab + b^2$.
3. Substitution: $(200 - 7)^2 = (200)^2 - 2(200)(7) + (7)^2$.
4. Result: $40000 - 2800 + 49 = 37249$.

Answer (ii): $$ 37249 $$

**Part (iii)**

1. Given: $(299)^2$.
2. Formula: $(a - b)^2 = a^2 - 2ab + b^2$.
3. Substitution: $(300 - 1)^2 = (300)^2 - 2(300)(1) + (1)^2$.
4. Result: $90000 - 600 + 1 = 89401$.

Answer (iii): $$ 89401 $$

**Answer:** Calculated values for all three parts.

> Common mistake: Errors in arithmetic expansion, especially missing the subtraction of the middle term.

## Exercise Set 4.3

### Question 1

*3 marks · Short answer*

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.
(i) $117^2$
(ii) $78^2$
(iii) $198^2$
(iv) $214^2$
(v) $1104^2$
(vi) $1120^2$

**Part (i)**

1. Write $117^2$ as $(100 + 17)^2$ or $(120 - 3)^2$. Using $(a + b)^2 = a^2 + 2ab + b^2$ with $a = 100$ and $b = 17$.
2. $117^2 = (100 + 17)^2 = 100^2 + 2(100)(17) + 17^2$
3. $117^2 = 10000 + 3400 + 289 = 13689$

Answer (i): 13689

**Part (ii)**

1. Write $78^2$ as $(80 - 2)^2$. Using $(a - b)^2 = a^2 - 2ab + b^2$ with $a = 80$ and $b = 2$.
2. $78^2 = (80 - 2)^2 = 80^2 - 2(80)(2) + 2^2$
3. $78^2 = 6400 - 320 + 4 = 6084$

Answer (ii): 6084

**Part (iii)**

1. Write $198^2$ as $(200 - 2)^2$. Using $(a - b)^2 = a^2 - 2ab + b^2$ with $a = 200$ and $b = 2$.
2. $198^2 = (200 - 2)^2 = 200^2 - 2(200)(2) + 2^2$
3. $198^2 = 40000 - 800 + 4 = 39204$

Answer (iii): 39204

**Part (iv)**

1. Write $214^2$ as $(200 + 14)^2$. Using $(a + b)^2 = a^2 + 2ab + b^2$ with $a = 200$ and $b = 14$.
2. $214^2 = (200 + 14)^2 = 200^2 + 2(200)(14) + 14^2$
3. $214^2 = 40000 + 5600 + 196 = 45796$

Answer (iv): 45796

**Part (v)**

1. Write $1104^2$ as $(1100 + 4)^2$. Using $(a + b)^2 = a^2 + 2ab + b^2$ with $a = 1100$ and $b = 4$.
2. $1104^2 = (1100 + 4)^2 = 1100^2 + 2(1100)(4) + 4^2$
3. $1104^2 = 1210000 + 8800 + 16 = 1218816$

Answer (v): 1218816

**Part (vi)**

1. Write $1120^2$ as $(1100 + 20)^2$. Using $(a + b)^2 = a^2 + 2ab + b^2$ with $a = 1100$ and $b = 20$.
2. $1104^2 = (1100 + 20)^2 = 1100^2 + 2(1100)(20) + 20^2$
3. $1120^2 = 1210000 + 44000 + 400 = 1254400$

Answer (vi): 1254400

**Answer:** (i) 13689, (ii) 6084, (iii) 39204, (iv) 45796, (v) 1218816, (vi) 1254400

> Common mistake: Choosing numbers that lead to difficult squares of large two-digit numbers.

### Question 2

*3 marks · Short answer*

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

**Part (i)**

1. $16y^2 - 24y + 9 = (4y)^2 - 2(4y)(3) + 3^2$
2. Using the identity $a^2 - 2ab + b^2 = (a - b)^2$, we get $(4y - 3)^2$.

Answer (i): $(4y - 3)^2$

**Part (ii)**

1. $\frac{9}{4}s^2 + 6st + 4t^2 = \left(\frac{3}{2}s\right)^2 + 2\left(\frac{3}{2}s\right)(2t) + (2t)^2$
2. Using the identity $a^2 + 2ab + b^2 = (a + b)^2$, we get $\left(\frac{3}{2}s + 2t\right)^2$.

Answer (ii): $\left(\frac{3}{2}s + 2t\right)^2$

**Part (iii)**

1. Rewrite the expression as $\left(\frac{m}{3}\right)^2 + \left(\frac{k}{2}\right)^2 + (3n)^2 + 2\left(\frac{m}{3}\right)\left(\frac{k}{2}\right) + 2\left(\frac{k}{2}\right)(3n) + 2(3n)\left(\frac{m}{3}\right)$
2. Using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$, we get $\left(\frac{m}{3} + \frac{k}{2} + 3n\right)^2$.

Answer (iii): $\left(\frac{m}{3} + \frac{k}{2} + 3n\right)^2$

**Part (iv)**

1. $\frac{p^2}{16} - 2 + \frac{16}{p^2} = \left(\frac{p}{4}\right)^2 - 2\left(\frac{p}{4}\right)\left(\frac{4}{p}\right) + \left(\frac{4}{p}\right)^2$
2. Using $a^2 - 2ab + b^2 = (a - b)^2$, we get $\left(\frac{p}{4} - \frac{4}{p}\right)^2$.

Answer (iv): $\left(\frac{p}{4} - \frac{4}{p}\right)^2$

**Part (v)**

1. $9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc = (3a)^2 + (-2b)^2 + c^2 + 2(3a)(-2b) + 2(-2b)(c) + 2(c)(3a)$
2. Using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$, we get $(3a - 2b + c)^2$.

Answer (v): $(3a - 2b + c)^2$

**Answer:** (i) $(4y - 3)^2$, (ii) $(\frac{3}{2}s + 2t)^2$, (iii) $(\frac{m}{3} + \frac{k}{2} + 3n)^2$, (iv) $(\frac{p}{4} - \frac{4}{p})^2$, (v) $(3a - 2b + c)^2$

> Common mistake: Incorrectly assigning signs to terms with negative products in three-term expansions.

### Question 3

*3 marks · Short answer*

Expand the following using the identity $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$:
(i) $(p + 3q + 7r)^2$
(ii) $(3x - 2y + 4z)^2$

**Part (i)**

1. Using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$ with $a = p$, $b = 3q$, $c = 7r$.
2. $(p + 3q + 7r)^2 = p^2 + (3q)^2 + (7r)^2 + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p) = p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr$

Answer (i): $p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr$

**Part (ii)**

1. Using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$ with $a = 3x$, $b = -2y$, $c = 4z$.
2. $(3x - 2y + 4z)^2 = (3x)^2 + (-2y)^2 + (4z)^2 + 2(3x)(-2y) + 2(-2y)(4z) + 2(4z)(3x) = 9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24xz$

Answer (ii): $9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24xz$

**Answer:** (i) $p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr$, (ii) $9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24xz$

> Common mistake: Forgetting to multiply the cross terms by 2.

### Question 4

*3 marks · Proof*

Is this an identity?
$(a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2 = 2a^2 + 2b^2 + 2c^2$.

**Solution**

1. Consider the L.H.S.: $(a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2$.
2. Expand each term using $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$.
3. $(a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2bc - 2ca$
4. $(a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ca$
5. $(a - b - c)^2 = a^2 + b^2 + c^2 - 2ab + 2bc - 2ca$
6. Add the three expansions: $(a^2 + b^2 + c^2 + 2ab - 2bc - 2ca) + (a^2 + b^2 + c^2 - 2ab - 2bc + 2ca) + (a^2 + b^2 + c^2 - 2ab + 2bc - 2ca)$.
7. Combine like terms: $(1+1+1)a^2 + (1+1+1)b^2 + (1+1+1)c^2 + (2-2-2)ab + (-2-2+2)bc + (-2+2-2)ca$.
8. Simplify to get $3a^2 + 3b^2 + 3c^2 - 2bc - 2ca - 2ab$. Since this does not identically equal $2a^2 + 2b^2 + 2c^2$ for all values, it is not an identity in that form (note: the text prints the question as an exploration of whether it equals $2a^2 + 2b^2 + 2c^2$, and the result shows it is False).

**Answer:** No, it is not an identity.

> Common mistake: Mistakes in handling negative signs while squaring trinomials.

## Exercise Set 4.4

### Question 1

*1 mark · Fill in the blank*

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

**Part (i)**

1. We need two numbers whose sum is $-11$ and product is $24$.
2. The numbers are $-3$ and $-8$.
3. $s^2 - 11s + 24 = (s - 3)(s - 8)$.

Answer (i): $(s - 3)(s - 8)$

**Part (ii)**

1. We factorise $3x^2 - 4x - 7$ by splitting the middle term to get $(3x - 7)(x + 1)$.
2. Comparing with $(_____)(x + 1) = (3x^2 - 4x - 7)$, the missing expression is $3x - 7$.

Answer (ii): $3x - 7$

**Part (iii)**

1. We factorise $10x^2 - 11x - 6$ as $(2x - 3)(5x + 2)$.
2. Comparing with $(2x - _____)(_____ + 2)$, the blanks are $3$ and $5x$.

Answer (iii): $3$ and $5x$

**Part (iv)**

1. We factorise $6x^2 + 7x + 2$ by splitting the middle term as $4x + 3x$.
2. $6x^2 + 7x + 2 = (2x + 1)(3x + 2)$.

Answer (iv): $(2x + 1)(3x + 2)$

**Answer:** Completed identities for parts (i) to (iv).

> Common mistake: Incorrect signs while splitting the middle term.

### Question 2

*3 marks · Short answer*

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) $(18 \times 29)$
(vii) $(34 \times 43)$
(viii) $(205)^2$

**Part (i)**

1. $(41)^2 = (40 + 1)^2 = 40^2 + 2(40)(1) + 1^2$
2. $= 1600 + 80 + 1 = 1681$

Answer (i): $1681$

**Part (ii)**

1. $(27)^2 = (30 - 3)^2 = 30^2 - 2(30)(3) + 3^2$
2. $= 900 - 180 + 9 = 729$

Answer (ii): $729$

**Part (iii)**

1. $23 \times 17 = (20 + 3)(20 - 3) = 20^2 - 3^2$
2. $= 400 - 9 = 391$

Answer (iii): $391$

**Part (iv)**

1. $(135)^2 = (130 + 5)^2 = 130^2 + 2(130)(5) + 5^2$
2. $= 16900 + 1300 + 25 = 18225$

Answer (iv): $18225$

**Part (v)**

1. $(97)^2 = (100 - 3)^2 = 100^2 - 2(100)(3) + 3^2$
2. $= 10000 - 600 + 9 = 9409$

Answer (v): $9409$

**Part (vi)**

1. $18 \times 29$ cannot be directly written as an identity of identical binomials, but using $(x+a)(x+b)$ or simple expansion: $18 \times 29 = (20 - 2)(30 - 1)$, or more directly as $(20 - 2) \times 29 = 580 - 58$
2. $= 522$

Answer (vi): $522$

**Part (vii)**

1. $34 \times 43$: Rewrite as $(40 - 6)(40 + 3)$ or use $(ax+b)(cx+d)$, or evaluate as $(30 + 4)(40 + 3) = 1200 + 90 + 160 + 12$
2. $= 1462$

Answer (vii): $1462$

**Part (viii)**

1. $(205)^2 = (200 + 5)^2 = 200^2 + 2(200)(5) + 5^2$
2. $= 40000 + 2000 + 25 = 42025$

Answer (viii): $42025$

**Answer:** Calculated values for parts (i) to (viii).

> Common mistake: Choosing the wrong identity or incorrect arithmetic expansion.

### Question 3

*3 marks · Short answer*

Factor the following:
(i) $9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc$
(ii) $16s^2 + 25t^2 - 40st$
(iii) $r^2 - r - 42$
(iv) $49g^2 + 14gh + h^2$
(v) $64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw$

**Part (i)**

1. Given expression: $9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc$
2. Rewrite as $(3a)^2 + (-b)^2 + (2c)^2 + 2(3a)(-b) + 2(-b)(2c) + 2(2c)(3a)$
3. Using $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$, we get $(3a - b + 2c)^2$

Answer (i): $(3a - b + 2c)^2$

**Part (ii)**

1. Given expression: $16s^2 + 25t^2 - 40st$
2. Rewrite as $(4s)^2 - 2(4s)(5t) + (5t)^2$
3. Using $(a - b)^2 = a^2 - 2ab + b^2$, we get $(4s - 5t)^2$

Answer (ii): $(4s - 5t)^2$

**Part (iii)**

1. Given expression: $r^2 - r - 42$
2. Split the middle term $-r$ using $-7r$ and $6r$, since $(-7) \times 6 = -42$ and $-7 + 6 = -1$.
3. $r^2 - 7r + 6r - 42 = r(r - 7) + 6(r - 7) = (r - 7)(r + 6)$

Answer (iii): $(r - 7)(r + 6)$

**Part (iv)**

1. Given expression: $49g^2 + 14gh + h^2$
2. Rewrite as $(7g)^2 + 2(7g)(h) + (h)^2$
3. Using $(a + b)^2 = a^2 + 2ab + b^2$, we get $(7g + h)^2$

Answer (iv): $(7g + h)^2$

**Part (v)**

1. Given expression: $64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw$
2. Notice the negative signs for terms containing $u$: $(8u)^2 + (11v)^2 + (2w)^2 + 2(8u)(-11v) + 2(-11v)(2w) + 2(2w)(8u)$
3. Using $(x + y + z)^2$, we get $(8u - 11v + 2w)^2$

Answer (v): $(8u - 11v + 2w)^2$

**Answer:** Factorised expressions for parts (i) to (v).

> Common mistake: Getting signs wrong when applying the three-variable square identity.

## Exercise Set 4.5

### Question 1

*3 marks · Short answer*

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

**Part (i)**

1. Factor the numerator by taking 3 common and splitting the middle term: $3(p^2 - pq - 6q^2) = 3(p - 3q)(p + 2q)$.
2. Factor the denominator by splitting the middle term: $p^2 + 3pq - 10q^2 = (p + 5q)(p - 2q)$.
3. Cancel common factors or state the expression as $\frac{3(p - 3q)(p + 2q)}{(p + 5q)(p - 2q)}$.

Answer (i): $$\frac{3(p - 3q)(p + 2q)}{(p + 5q)(p - 2q)}$$

**Part (ii)**

1. Recognize the numerator as the cube of a binomial: $n^3 - 3n^2m + 3nm^2 - m^3 = (n - m)^3$.
2. Factor the denominator by taking 5 common: $5(m^2 - 2mn + n^2) = 5(m - n)^2 = 5(n - m)^2$.
3. Cancel the common factor $(n - m)^2$ from numerator and denominator.

Answer (ii): $$\frac{n - m}{5}$$

**Part (iii)**

1. Rearrange the numerator using the identity for three variables: $w^3 + x^3 - v^3 + 3wvx$ or apply standard grouping.
2. Factor the denominator using the trinomial square identity with signs: $(w - v + x)^2$.
3. Simplify the rational expression by cancelling common factors.

Answer (iii): $$\frac{w^2 - wv + v^2 + wx + vx + x^2}{w - v + x}$$

**Part (iv)**

1. Factor the numerator using the identity $(a - b)^2 = a^2 - 2ab + b^2$: $(2y - 5z)^2$.
2. Factor the denominator using $a^2 - b^2 = (a + b)(a - b)$: $(5z + 2y)(5z - 2y)$ or $- (2y + 5z)(2y - 5z)$.
3. Cancel the common term $(2y - 5z)$ taking care of negative signs.

Answer (iv): $$\frac{2y - 5z}{- (5z + 2y)}$$

**Part (v)**

1. Factor each quadratic expression: $x^2 + x - 6 = (x + 3)(x - 2)$, $x^2 - 7x + 12 = (x - 3)(x - 4)$.
2. Factor the denominator terms: $x^2 - 6x + 8 = (x - 4)(x - 2)$ and $x^2 - 9 = (x + 3)(x - 3)$.
3. Substitute the factors and cancel common terms $(x + 3)$, $(x - 2)$, and $(x - 4)$.

Answer (v): $$1$$

**Part (vi)**

1. Factor the numerator using $a^2 - b^2$: $p^4 - 16 = (p^2 - 4)(p^2 + 4) = (p - 2)(p + 2)(p^2 + 4)$.
2. Factor the denominator as a square of a binomial: $p^2 - 4p + 4 = (p - 2)^2$.
3. Cancel the common factor $(p - 2)$ to get the simplified expression.

Answer (vi): $$\frac{(p + 2)(p^2 + 4)}{p - 2}$$

**Answer:** Simplified forms of the given rational expressions.

> Common mistake: Forgetting to check the signs when factoring differences of squares or cubes.

## End-of-Chapter Exercises

### Question 1

*3 marks · Short answer*

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

**Part (i)**

1. Using the identity $(x-y)^2 = x^2 - 2xy + y^2$, let $x = 3x$ and $y = 4$.
2. Substitute the values: $(-3x+4)^2 = (-3x)^2 - 2(-3x)(4) + (4)^2$.
3. Simplify the terms to get $9x^2 + 24x + 16$.

Answer (i): $9x^2 + 24x + 16$

**Part (ii)**

1. Using the identity $(x+y)(x-y) = x^2 - y^2$, let $x = 2s$ and $y = 7$.
2. Substitute the values: $(2s+7)(2s-7) = (2s)^2 - (7)^2$.
3. Simplify to get $4s^2 - 49$.

Answer (ii): $4s^2 - 49$

**Part (iii)**

1. Using the identity $(x+y)(x-y) = x^2 - y^2$, let $x = p^2$ and $y = \frac{1}{2}$.
2. Substitute the values: $\left(p^2 + \frac{1}{2}\right)\left(p^2 - \frac{1}{2}\right) = (p^2)^2 - \left(\frac{1}{2}\right)^2$.
3. Simplify to get $p^4 - \frac{1}{4}$.

Answer (iii): $p^4 - \frac{1}{4}$

**Part (iv)**

1. Using the identity $(x+y)(x-y) = x^2 - y^2$, let $x = 2n$ and $y = 7$.
2. Substitute the values: $(2n+7)(2n-7) = (2n)^2 - (7)^2$.
3. Simplify to get $4n^2 - 49$.

Answer (iv): $4n^2 - 49$

**Part (v)**

1. Recognise the identity $(x-y)(x^2 + xy + y^2) = x^3 - y^3$ with $x = s$ and $y = 2t$.
2. Substitute the values: $(s-2t)(s^2 + s(2t) + (2t)^2) = s^3 - (2t)^3$.
3. Simplify to get $s^3 - 8t^3$.

Answer (v): $s^3 - 8t^3$

**Part (vi)**

1. Using the identity $(x-y)^2 = x^2 - 2xy + y^2$, let $x = \frac{1}{2r}$ and $y = 4r$.
2. Substitute the values: $\left(\frac{1}{2r} - 4r\right)^2 = \left(\frac{1}{2r}\right)^2 - 2\left(\frac{1}{2r}\right)(4r) + (4r)^2$.
3. Simplify to get $\frac{1}{4r^2} - 4 + 16r^2$.

Answer (vi): \frac{1}{4r^2} - 4 + 16r^2

**Part (vii)**

1. Using the identity $(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$, let $x = -3m$, $y = 4k$, and $z = -l$.
2. Substitute the values and expand each term.
3. Simplify to get $9m^2 + 16k^2 + l^2 - 24mk - 8kl + 6lm$.

Answer (vii): $9m^2 + 16k^2 + l^2 - 24mk - 8kl + 6lm$

**Part (viii)**

1. Using the identity $(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$, let $x = x$ and $y = \frac{1}{3}$.
2. Substitute the values: $\left(x - \frac{1}{3}\right)^3 = x^3 - 3(x)^2\left(\frac{1}{3}\right) + 3(x)\left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^3$.
3. Simplify to get $x^3 - x^2 + \frac{1}{3}x - \frac{1}{27}$.

Answer (viii): $x^3 - x^2 + \frac{1}{3}x - \frac{1}{27}$

**Part (ix)**

1. Using the identity $(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$, let $x = \frac{7}{2}k$ and $y = \frac{2}{3}m$.
2. Substitute the values and expand each term.
3. Simplify to get $\frac{343}{8}k^3 - \frac{49}{2}k^2m + \frac{14}{3}km^2 - \frac{8}{27}m^3$.

Answer (ix): \frac{343}{8}k^3 - \frac{49}{2}k^2m + \frac{14}{3}km^2 - \frac{8}{27}m^3

**Answer:** Expanded forms of the given expressions using suitable identities.

> Common mistake: Errors in applying signs while expanding binomials and trinomials.

### Question 2

*3 marks · Short answer*

Find the values using suitable identities:
(i) $17 \times 21$
(ii) $104 \times 96$
(iii) $24 \times 16$
(iv) $147^3$
(v) $199^3$
(vi) $127^3$
(vii) $(-107)^3$
(viii) $(-299)^3$

**Part (i)**

1. Write as $(20-3)(20+1)$ or use $(20-3)(20+1)$ via distributive property, or better: $17 \times 21 = (19-2)(19+2) = 19^2 - 2^2$.
2. Calculate $361 - 4 = 357$.

Answer (i): 357

**Part (ii)**

1. Write as $(100+4)(100-4)$.
2. Apply the identity $(a+b)(a-b) = a^2 - b^2$: $100^2 - 4^2 = 10000 - 16$.
3. Calculate $9984$.

Answer (ii): 9984

**Part (iii)**

1. Write as $(20+4)(20-4)$.
2. Apply the identity $(a+b)(a-b) = a^2 - b^2$: $20^2 - 4^2 = 400 - 16$.
3. Calculate $384$.

Answer (iii): 384

**Part (iv)**

1. Write as $(150-3)^3$.
2. Apply $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ with $a=150$ and $b=3$.
3. Calculate $3375000 - 202500 + 4050 - 27 = 3176523$.

Answer (iv): 3176523

**Part (v)**

1. Write as $(200-1)^3$.
2. Apply $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ with $a=200$ and $b=1$.
3. Calculate $8000000 - 120000 + 600 - 1 = 7880599$.

Answer (v): 7880599

**Part (vi)**

1. Write as $(120+7)^3$ or $(130-3)^3$. Let us use $(127) = 127^3$. Alternatively, use $(100+27)^3$ or standard expansion. Let us use $(130-3)^3 = 130^3 - 3(130)^2(3) + 3(130)(3)^2 - 3^3$.
2. Calculate $2197000 - 152100 + 3510 - 27 = 2048383$.

Answer (vi): 2048383

**Part (vii)**

1. Write as $(-107)^3 = -(107)^3 = -(100+7)^3$.
2. Expand using $(a+b)^3$: $-(100^3 + 3(100)^2(7) + 3(100)(7^2) + 7^3)$.
3. Calculate $-(1000000 + 210000 + 14700 + 343) = -1225043$.

Answer (vii): -1225043

**Part (viii)**

1. Write as $(-299)^3 = - (300-1)^3$.
2. Expand using $(a-b)^3$: $-(300^3 - 3(300)^2(1) + 3(300)(1^2) - 1^3)$.
3. Calculate $-(27000000 - 270000 + 900 - 1) = -26730899$.

Answer (viii): -26730899

**Answer:** Calculated values using algebraic identities.

> Common mistake: Choosing incorrect base numbers for splitting in identities.

### Question 3

*3 marks · Short answer*

Factor the following algebraic expressions:
(i) $4y^2 + 1 + \frac{1}{16y^2}$
(ii) $9m^2 - \frac{1}{25n^2}$
(iii) $27b^3 - \frac{1}{64b^3}$
(iv) $x^2 + \frac{5x}{6} + \frac{1}{6}$
(v) $27u^3 - \frac{1}{125} - \frac{27u^2}{5} + \frac{9u}{25}$
(vi) $64y^3 + \frac{1}{125}z^3$
(vii) $p^3 + 27q^3 + r^3 - 9pqr$
(viii) $9m^2 - 12m + 4$
(ix) $9x^3 - \frac{8}{3}y^3 + \frac{z^3}{3} + 6xyz$
(x) $49x^2 + 36y^2 + 36z^2 + 12xz + 24xy + 36yz$
(xi) $27u^3 - \frac{1}{216} - \frac{9u^2}{2} + \frac{u}{4}$

**Part (i)**

1. Rewrite the expression as $(2y)^2 + 2(2y)\left(\frac{1}{4y}\right) + \left(\frac{1}{4y}\right)^2$.
2. Use the identity $(a + b)^2 = a^2 + 2ab + b^2$ with $a = 2y$ and $b = \frac{1}{4y}$.
3. Obtain $\left(2y + \frac{1}{4y}\right)^2$.

Answer (i): $-\left(2y + \frac{1}{4y}\right)^2$

**Part (ii)**

1. Rewrite the expression as $(3m)^2 - \left(\frac{1}{5n}\right)^2$.
2. Use the identity $a^2 - b^2 = (a + b)(a - b)$ with $a = 3m$ and $b = \frac{1}{5n}$.
3. Obtain $\left(3m + \frac{1}{5n}\right)\left(3m - \frac{1}{5n}\right)$.

Answer (ii): $-\left(3m + \frac{1}{5n}\right)\left(3m - \frac{1}{5n}\right)$

**Part (iii)**

1. Rewrite the expression as $(3b)^3 - \left(\frac{1}{4b}\right)^3$.
2. Use the identity $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$ with $x = 3b$ and $y = \frac{1}{4b}$.
3. Obtain $\left(3b - \frac{1}{4b}\right)\left(9b^2 + \frac{3}{4} + \frac{1}{16b^2}\right)$.

Answer (iii): $-\left(3b - \frac{1}{4b}\right)\left(9b^2 + \frac{3}{4} + \frac{1}{16b^2}\right)$

**Part (iv)**

1. Rewrite the quadratic expression as $x^2 + \left(\frac{1}{2} + \frac{1}{3}\right)x + \left(\frac{1}{2}\right)\left(\frac{1}{3}\right)$ by splitting the middle term.
2. Use the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$.
3. Obtain $\left(x + \frac{1}{2}\right)\left(x + \frac{1}{3}\right)$.

Answer (iv): $-\left(x + \frac{1}{2}\right)\left(x + \frac{1}{3}\right)$

**Part (v)**

1. Rearrange the expression as $(3u)^3 - 3(3u)^2\left(\frac{1}{5}\right) + 3(3u)\left(\frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^3$.
2. Use the identity $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ with $a = 3u$ and $b = \frac{1}{5}$.
3. Obtain $\left(3u - \frac{1}{5}\right)^3$.

Answer (v): $-\left(3u - \frac{1}{5}\right)^3$

**Part (vi)**

1. Rewrite the expression as $(4y)^3 + \left(\frac{z}{5}\right)^3$.
2. Use the identity $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$ with $x = 4y$ and $y = \frac{z}{5}$.
3. Obtain $\left(4y + \frac{z}{5}\right)\left(16y^2 - \frac{4}{5}yz + \frac{z^2}{25}\right)$.

Answer (vi): $-\left(4y + \frac{z}{5}\right)\left(16y^2 - \frac{4}{5}yz + \frac{z^2}{25}\right)$

**Part (vii)**

1. Rewrite the expression as $(p)^3 + (3q)^3 + (r)^3 - 3(p)(3q)(r)$.
2. Use the identity $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$.
3. Obtain $(p + 3q + r)(p^2 + 9q^2 + r^2 - 3pq - 3qr - pr)$.

Answer (vii): $-(p + 3q + r)(p^2 + 9q^2 + r^2 - 3pq - 3qr - pr)$

**Part (viii)**

1. Rewrite the expression as $(3m)^2 - 2(3m)(2) + (2)^2$.
2. Use the identity $(a - b)^2 = a^2 - 2ab + b^2$ with $a = 3m$ and $b = 2$.
3. Obtain $(3m - 2)^2$.

Answer (viii): $-(3m - 2)^2$

**Part (ix)**

1. Rewrite the expression as $\left(\sqrt[3]{9}x\right)^3$ using standard grouping or notice it comes from the expansion of $(a+b+c)^3$ or grouped algebraic forms.
2. Group terms appropriately or apply identity for sums of cubes.
3. Obtain $\left(3x - \frac{2}{3}y + \frac{z}{\sqrt[3]{3}}\right(...)$ following textbook identity conventions.

Answer (ix): $-\left(3x - \frac{2}{3}y + \frac{z}{\sqrt[3]{3}}\right)^3$

**Part (x)**

1. Rewrite as $(7x)^2 + (6y)^2 + (6z)^2 + 2(7x)(6y) + 2(6y)(6z) + 2(6z)(7x)$.
2. Use the identity $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$ with $a = 7x$, $b = 6y$, and $c = 6z$.
3. Obtain $(7x + 6y + 6z)^2$.

Answer (x): $-(7x + 6y + 6z)^2$

**Part (xi)**

1. Rewrite as $(3u)^3 - 3(3u)^2\left(\frac{1}{6}\right) + 3(3u)\left(\frac{1}{6}\right)^2 - \left(\frac{1}{6}\right)^3$.
2. Use the identity $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ with $a = 3u$ and $b = \frac{1}{6}$.
3. Obtain $\left(3u - \frac{1}{6}\right)^3$.

Answer (xi): $-\left(3u - \frac{1}{6}\right)^3$

**Answer:** Factorised expressions for all eleven given polynomials.

> Common mistake: Incorrectly identifying the middle term signs when applying binomial cube and square identities.

### Question 4

*3 marks · Short answer*

Simplify the following:
(i) $\frac{4x^2 + 4x + 1}{4x^2 - 1}$
(ii) $\frac{9(3a^3 - 24b^3)}{9a^2 - 36b^2}$
(iii) $\frac{s^3 + 125t^3}{s^2 - 2st - 35t^2}$
Note: Assume that the denominators are not equal to 0.

**Part (i)**

1. Factorise the numerator: $4x^2 + 4x + 1 = (2x + 1)^2$.
2. Factorise the denominator: $4x^2 - 1 = (2x + 1)(2x - 1)$.
3. Cancel the common factor $(2x + 1)$ to get $\frac{2x + 1}{2x - 1}$.

Answer (i): $\frac{2x + 1}{2x - 1}$

**Part (ii)**

1. Factorise the numerator: take out 3 to get $9 \cdot 3(a^3 - 8b^3) = 27(a - b)(a^2 + 2ab + 4b^2)$ using $a^3 - (2b)^3$.
2. Factorise the denominator: $9(a^2 - 4b^2) = 9(a - 2b)(a + 2b)$.
3. Simplify and cancel common terms to get $\frac{3(a^2 + 2ab + 4b^2)}{a + 2b}$.

Answer (ii): $\frac{3(a^2 + 2ab + 4b^2)}{a + 2b}$

**Part (iii)**

1. Factorise the numerator: $s^3 + 125t^3 = (s + 5t)(s^2 - 5st + 25t^2)$.
2. Factorise the denominator: $s^2 - 2st - 35t^2 = (s - 7t)(s + 5t)$ by splitting the middle term.
3. Cancel the common factor $(s + 5t)$ to get $\frac{s^2 - 5st + 25t^2}{s - 7t}$.

Answer (iii): $\frac{s^2 - 5st + 25t^2}{s - 7t}$

**Answer:** Simplified forms of the given rational expressions.

> Common mistake: Failing to completely factorise numerators and denominators before cancellation.

### Question 5

*3 marks · Short answer*

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) $25a^2 - 30ab + 9b^2$
(ii) $36s^2 - 49t^2$

**Part (i)**

1. Given area = $25a^2 - 30ab + 9b^2$.
2. Recognise the expression as a square: $(5a)^2 - 2(5a)(3b) + (3b)^2 = (5a - 3b)^2$.
3. Since area = length $\times$ breadth, the length and breadth are both $(5a - 3b)$ units.

Answer (i): Length = $5a - 3b$ units, Breadth = $5a - 3b$ units

**Part (ii)**

1. Given area = $36s^2 - 49t^2$.
2. Use the identity $x^2 - y^2 = (x+y)(x-y)$: $(6s)^2 - (7t)^2 = (6s + 7t)(6s - 7t)$.
3. Therefore, the possible length and breadth are $(6s + 7t)$ units and $(6s - 7t)$ units.

Answer (ii): Length = $6s + 7t$ units, Breadth = $6s - 7t$ units

**Answer:** Possible expressions for length and breadth.

> Common mistake: Confusing area factors with perimeter or adding terms instead of multiplying.

### Question 6

*3 marks · Short answer*

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) $6a^2 - 24b^2$
(ii) $3ps^2 - 15ps + 12p$

**Part (i)**

1. Given volume = $6a^2 - 24b^2$.
2. Take out the common factor 6: $6(a^2 - 4b^2)$.
3. Factorise further using $a^2 - b^2$: $6(a - 2b)(a + 2b)$. Thus, the dimensions are $6$, $(a - 2b)$, and $(a + 2b)$.

Answer (i): Length = $6$ units, Breadth = $a - 2b$ units, Height = $a + 2b$ units

**Part (ii)**

1. Given volume = $3ps^2 - 15ps + 12p$.
2. Take out the common factor $3p$: $3p(s^2 - 5s + 4)$.
3. Factorise the quadratic expression: $s^2 - 5s + 4 = (s - 1)(s - 4)$. Thus, the dimensions are $3p$, $(s - 1)$, and $(s - 4)$.

Answer (ii): Length = $3p$ units, Breadth = $s - 1$ units, Height = $s - 4$ units

**Answer:** Possible expressions for length, breadth, and height.

> Common mistake: Forgetting to take out common numerical and variable factors first.

### Question 7

*3 marks · Short answer*

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 $s$.

**Solution**

1. Side of the square playground = $40\text{ m}$.
2. Width of the path = $s\text{ m}$.
3. The outer square including the path has side length equal to $(40 + 2s)\text{ m}$.
4. Area of the path = (Area of the outer square) - (Area of the inner playground).
5. Area of the path = $(40 + 2s)^2 - 40^2$.
6. Using the identity $a^2 - b^2 = (a+b)(a-b)$, we get $[(40 + 2s) + 40][(40 + 2s) - 40] = (80 + 2s)(2s) = 160s + 4s^2\text{ sq. metres}$. Alternatively, expanding $(40+2s)^2 = 1600 + 160s + 4s^2$, subtracting $1600$ gives $4s^2 + 160s\text{ sq. metres}$.

**Answer:** $4s^2 + 160s\text{ sq. metres}$

> Common mistake: Adding only $s$ instead of $2s$ to the total side length of the outer square.

### Question 8

*3 marks · Short answer*

If a number plus its reciprocal equals $\frac{10}{3}$, find the number.

**Solution**

1. Let the number be $x$. According to the problem, $x + \frac{1}{x} = \frac{10}{3}$.
2. Taking LCM on the left side, we get $\frac{x^2 + 1}{x} = \frac{10}{3}$, which gives $3(x^2 + 1) = 10x$.
3. Rearranging into standard quadratic form: $3x^2 - 10x + 3 = 0$.
4. Splitting the middle term: $3x^2 - 9x - x + 3 = 0$, leading to $3x(x - 3) - 1(x - 3) = 0$, so $(3x - 1)(x - 3) = 0$.
5. Thus, $x = \frac{1}{3}$ or $x = 3$.

**Answer:** $3$ or $\frac{1}{3}$

> Common mistake: Failing to convert the fractional equation into a standard quadratic equation.

### Question 9

*3 marks · Short answer*

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

**Solution**

1. Area of the rectangular pool = $2x^2 + 7x + 3\text{ sq. hastas}$ and width = $(2x + 1)\text{ hastas}$.
2. Length = $\frac{\text{Area}}{\text{Width}} = \frac{2x^2 + 7x + 3}{2x + 1}$.
3. Factorising the numerator by splitting the middle term: $2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)$.
4. Cancelling the common factor $(2x + 1)$, we get the length as $(x + 3)\text{ hastas}$.

**Answer:** $(x + 3)\text{ hastas}$

> Common mistake: Incorrectly splitting the middle term for factorisation.

### Question 10

*4 marks · Proof*

If both $x - 2$ and $x - \frac{1}{2}$ are factors of $px^2 + 5x + r$, show that $p = r$.

**Solution**

1. Given that $x - 2$ and $x - \frac{1}{2}$ are factors of the polynomial $f(x) = px^2 + 5x + r$.
2. By the factor theorem, $f(2) = 0$ and $f\left(\frac{1}{2}\right) = 0$.
3. Substituting $x = 2$, we get $p(2)^2 + 5(2) + r = 0$, which gives $4p + r + 10 = 0$, or $r = -4p - 10$.
4. Substituting $x = \frac{1}{2}$, we get $p\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) + r = 0$, which gives $\frac{p}{4} + \frac{5}{2} + r = 0$, or $p + 10 + 4r = 0$.
5. Substituting $r = -4p - 10$ into the second equation: $p + 10 + 4(-4p - 10) = 0$, leading to $p + 10 - 16p - 40 = 0$, which simplifies to $-15p = 30$, so $p = -2$.
6. Then $r = -4(-2) - 10 = 8 - 10 = -2$.
7. Since $p = -2$ and $r = -2$, we have $p = r$. Hence proved.

**Answer:** Hence proved.

> Common mistake: Errors in algebraic substitution and sign conventions.

### Question 11

*4 marks · Proof*

If $a + b + c = 5$ and $ab + bc + ca = 10$, then prove that $a^3 + b^3 + c^3 - 3abc = -25$.

**Solution**

1. We know the identity: $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.
2. We also know that $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$.
3. Substitute the given values $a + b + c = 5$ and $ab + bc + ca = 10$ into the square identity: $5^2 = a^2 + b^2 + c^2 + 2(10)$.
4. This gives $25 = a^2 + b^2 + c^2 + 20$, which implies $a^2 + b^2 + c^2 = 5$.
5. Now substitute the values into the cube identity: $a^3 + b^3 + c^3 - 3abc = (5)[(a^2 + b^2 + c^2) - (ab + bc + ca)]$.
6. Thus, $a^3 + b^3 + c^3 - 3abc = (5)[5 - 10] = 5(-5) = -25$. Hence proved.

**Answer:** Hence proved.

> Common mistake: Forgetting to compute $a^2 + b^2 + c^2$ first before substituting into the cubic identity.

### Question 12

*4 marks · Proof*

By factoring the expression, check that $n^3 - n$ is always divisible by 6 for all natural numbers $n$. Give reasons.

**Solution**

1. Consider the expression $n^3 - n$.
2. Taking out $n$ as a common factor, we get $n(n^2 - 1)$.
3. Using the identity $x^2 - y^2 = (x + y)(x - y)$, we can rewrite the expression as $n(n - 1)(n + 1)$, or $(n - 1)n(n + 1)$.
4. Observe that $(n - 1)n(n + 1)$ represents the product of three consecutive natural numbers.
5. The product of any three consecutive natural numbers is always divisible by 2 (since at least one is even) and by 3 (since one of them must be a multiple of 3).
6. Since the number is divisible by both 2 and 3, and $\text{HCF}(2, 3) = 1$, it is always divisible by $2 \times 3 = 6$ for all natural numbers $n$. Hence proved.

**Answer:** Hence proved.

> Common mistake: Stating divisibility by 6 without showing that it is a product of three consecutive integers.

### Question 13

*3 marks · Short answer*

Find the value of
(i) $x^3 + y^3 - 12xy + 64$, when $x + y = -4$
(ii) $x^3 - 8y^3 - 36xy - 216$, when $x = 2y + 6$

**Part (i)**

1. We are given $x + y = -4$.
2. Cube both sides to get $(x + y)^3 = (-4)^3$.
3. Expand using the identity: $x^3 + y^3 + 3xy(x + y) = -64$.
4. Substitute $x + y = -4$: $x^3 + y^3 + 3xy(-4) = -64$, which gives $x^3 + y^3 - 12xy = -64$.
5. Add $64$ to both sides: $x^3 + y^3 - 12xy + 64 = -64 + 64 = 0$.

Answer (i): $0$

**Part (ii)**

1. We are given $x = 2y + 6$, which can be rewritten as $x - 2y = 6$.
2. Cube both sides to get $(x - 2y)^3 = 6^3$.
3. Expand using the identity: $x^3 - (2y)^3 - 3(x)(2y)(x - 2y) = 216$.
4. Substitute $x - 2y = 6$: $x^3 - 8y^3 - 6xy(6) = 216$, which simplifies to $x^3 - 8y^3 - 36xy = 216$.
5. Subtract $216$ from both sides: $x^3 - 8y^3 - 36xy - 216 = 216 - 216 = 0$, or rearranging gives $x^3 - 8y^3 - 36xy - 216 = 0$.

Answer (ii): $216$

**Answer:** (i) $0$, (ii) $216$

> Common mistake: Forgetting to substitute the value of $(x+y)$ or $(x-2y)$ back into the expanded product term.

## Frequently asked questions

### How many exercises and questions are in the new NCERT Class 9 Maths Chapter 4 for the 2026-27 session?

This chapter from the new NCERT book based on the NCF 2023 contains 5 main exercise sets along with End-of-Chapter Exercises. Specifically, Exercise Set 4.1 to 4.5 have 2, 2, 4, 3, and 1 questions respectively, while the end-of-chapter section has 13 questions. You can find complete step-by-step solutions and the free PDF for all these questions right here on this SwaVid page.

### What topics and question types are covered in the exercise sets of Class 9 Maths Chapter 4?

The chapter covers various concepts like evaluating squares and products using algebraic identities, expanding binomial and trinomial squares, and factorisation using identities or splitting the middle term. It also includes the factor theorem, polynomial evaluation, finding cuboid dimensions from volume, and simplifying rational algebraic expressions. The question types across these exercises include short answer (sa), fill in the blanks (fib), and proofs.

### What are the hardest question types in this chapter and how should I approach them?

The most challenging questions usually involve proofs, factorisation of cubic and quadratic expressions, and finding dimensions of cuboids from given volume expressions in the End-of-Chapter Exercises. To approach them, you should first identify the correct algebraic identity or factor theorem application needed for the expression. Breaking down the problem systematically as shown in SwaVid's free PDF solutions on this page will help you master these concepts.

### How can I write answers for Class 9 Maths Chapter 4 to score full marks in exams?

To score full marks, always state the specific algebraic identity being used before substituting any values in your steps. Clearly show each intermediate calculation, especially when dealing with trinomial expansions or factorisation by splitting the middle term. Referring to the detailed step-by-step solutions provided on this SwaVid page will guide you on the exact presentation expected by examiners.

### Is the free PDF for NCERT Solutions of Class 9 Maths Chapter 4 available here?

Yes, the complete free PDF and detailed step-by-step solutions for this chapter are available on this SwaVid page for the 2026-27 session. You can easily access them to practice all exercise sets and end-of-chapter questions based on the new NCERT syllabus. These resources are designed to help you revise algebraic identities and improve your problem-solving skills effectively.

## Related pages

- [Exercise 4.1 solutions](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-1)
- [Exercise 4.2 solutions](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-2)
- [Exercise 4.3 solutions](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-3)
- [Exercise 4.4 solutions](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-4)
- [Exercise 4.5 solutions](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-5)
- [Class 9 Maths chapters](https://www.swavid.com/maths/class/9)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student