Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.5 Solutions: Exploring Algebraic Identities
Rational expressions are algebraic fractions in which the numerator and denominator are algebraic expressions, and they are simplified by factorising and cancelling common factors.
Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.5 Solutions connect this idea with algebraic identities, factorisation and simplifying rational expressions from Exploring Algebraic Identities.
Exercise 4.5 is based on simplifying rational expressions Class 9 using factorisation. In Ganita Manjari Class 9 Chapter 4 Exercise 4.5, students first factor the numerator and denominator separately, then cancel only the common factors, assuming that the denominator is not zero.
These Class 9 Maths Chapter 4 Exercise 4.5 Solutions from Exploring Algebraic Identities Exercise 4.5 help students revise rational expressions Class 9, common factors in rational expressions, and factorisation of algebraic expressions Class 9. The exercise also prepares students for later algebra questions where expressions must be simplified before solving.
Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.5 - NCERT Solutions
Exercise 4.5 Question 1
Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:
(i) (3p² − 3pq − 18q²) / (p² + 3pq − 10q²)
Solution:
Factor the numerator first.
3p² − 3pq − 18q²
= 3(p² − pq − 6q²)
Now factor p² − pq − 6q².
We need two terms whose product is −6q² and whose sum is −q.
The terms are −3q and 2q.
So,
p² − pq − 6q² = (p − 3q)(p + 2q)
Therefore:
3p² − 3pq − 18q² = 3(p − 3q)(p + 2q)
Now factor the denominator.
p² + 3pq − 10q²
= (p + 5q)(p − 2q)
So,
(3p² − 3pq − 18q²) / (p² + 3pq − 10q²)
= 3(p − 3q)(p + 2q) / [(p + 5q)(p − 2q)]
There is no common factor to cancel.
Answer: 3(p − 3q)(p + 2q) / [(p + 5q)(p − 2q)]
(ii) (n³ − 3n²m + 3nm² − m³) / (5m² − 10mn + 5n²)
Solution:
The numerator matches the identity:
(a − b)³ = a³ − 3a²b + 3ab² − b³
So,
n³ − 3n²m + 3nm² − m³ = (n − m)³
Now factor the denominator.
5m² − 10mn + 5n²
= 5(m² − 2mn + n²)
= 5(m − n)²
Since (m − n)² = (n − m)²,
5(m − n)² = 5(n − m)²
Therefore:
(n³ − 3n²m + 3nm² − m³) / (5m² − 10mn + 5n²)
= (n − m)³ / [5(n − m)²]
= (n − m) / 5
Answer: (n − m) / 5
(iii) (w³ − v³ + x³ + 3wvx) / (w² + v² + x² − 2wv − 2vx + 2wx)
Solution:
The numerator can be written using the identity:
a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
Take:
a = w
b = −v
c = x
Then:
w³ − v³ + x³ + 3wvx
= (w − v + x)(w² + v² + x² + wv + vx − wx)
Now factor the denominator.
w² + v² + x² − 2wv − 2vx + 2wx
= (w − v + x)²
So,
(w³ − v³ + x³ + 3wvx) / (w² + v² + x² − 2wv − 2vx + 2wx)
= [(w − v + x)(w² + v² + x² + wv + vx − wx)] / (w − v + x)²
Cancel the common factor w − v + x.
Answer: (w² + v² + x² + wv + vx − wx) / (w − v + x)
(iv) (4y² − 20yz + 25z²) / (25z² − 4y²)²
Solution:
Factor the numerator.
4y² − 20yz + 25z²
= (2y)² − 2(2y)(5z) + (5z)²
= (2y − 5z)²
Now factor the denominator.
25z² − 4y²
= (5z)² − (2y)²
= (5z − 2y)(5z + 2y)
So,
(25z² − 4y²)² = [(5z − 2y)(5z + 2y)]²
Since:
5z − 2y = −(2y − 5z)
we get:
(5z − 2y)² = (2y − 5z)²
Therefore:
(25z² − 4y²)² = (2y − 5z)²(5z + 2y)²
Now simplify:
(4y² − 20yz + 25z²) / (25z² − 4y²)²
= (2y − 5z)² / [(2y − 5z)²(5z + 2y)²]
= 1 / (5z + 2y)²
Answer: 1 / (5z + 2y)²
(v) [(x² + x − 6)(x² − 7x + 12)] / [(x² − 6x + 8)(x² − 9)]
Solution:
Factor each expression separately.
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)
Now substitute the factors.
[(x² + x − 6)(x² − 7x + 12)] / [(x² − 6x + 8)(x² − 9)]
= [(x + 3)(x − 2)(x − 3)(x − 4)] / [(x − 2)(x − 4)(x − 3)(x + 3)]
All factors cancel.
Answer: 1
(vi) (p⁴ − 16) / (p² − 4p + 4)
Solution:
Factor the numerator using difference of squares.
p⁴ − 16 = (p²)² − 4²
= (p² − 4)(p² + 4)
Now factor p² − 4.
p² − 4 = (p − 2)(p + 2)
So,
p⁴ − 16 = (p − 2)(p + 2)(p² + 4)
Now factor the denominator.
p² − 4p + 4 = (p − 2)²
Therefore:
(p⁴ − 16) / (p² − 4p + 4)
= [(p − 2)(p + 2)(p² + 4)] / (p − 2)²
Cancel one common factor p − 2.
Answer: [(p + 2)(p² + 4)] / (p − 2)
Final Answers for Exercise 4.5
| Question | Final Answer |
| 1(i) | 3(p − 3q)(p + 2q) / [(p + 5q)(p − 2q)] |
| 1(ii) | (n − m) / 5 |
| 1(iii) | (w² + v² + x² + wv + vx − wx) / (w − v + x) |
| 1(iv) | 1 / (5z + 2y)² |
| 1(v) | 1 |
| 1(vi) | [(p + 2)(p² + 4)] / (p − 2) |
Concept Used in Exploring Algebraic Identities Exercise 4.5
Exploring Algebraic Identities Exercise 4.5 is based on simplifying rational expressions Class 9. A rational expression is simplified by factorising the numerator and denominator and then cancelling the common factors.
For example:
(x² + x − 6)(x² − 7x + 12) / [(x² − 6x + 8)(x² − 9)]
First factor each part:
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)
After substitution, all common factors cancel and the answer becomes 1.
The key point is that a factor can be cancelled only when it is multiplied with the whole numerator and denominator. Individual terms cannot be cancelled separately.
NCERT Solutions for Class 9 Maths Chapter 4 - Related Links