---
title: "NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.5"
url: https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-5
dateModified: 2026-10-07T15:43:39+00:00
---

# NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.5

Chapter 4: Exploring Algebraic Identities. Every question from Exercise 4.5, with full working and the final answer.

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## 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.

## Related pages

- [All Chapter 4 NCERT solutions](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions)
- [Exercise 4.1](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-1)
- [Exercise 4.2](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-2)
- [Exercise 4.3](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-3)
- [Exercise 4.4](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-4)

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