NCERT Solutions for Class 9th Maths Chapter 4 Exercise Set 4.2 — Factorisation of Algebraic Expressions Using Identities (2026-27)

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?)

Answer

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

49x² = (7x)², 4y² = (2y)², 2(7x)(2y) = 28xy ✓
= (7x + 2y)²

(iv) 64p² + (32/3)pq + (4/9)q²

64p² = (8p)², (4/9)q² = (2q/3)², 2(8p)(2q/3) = 32pq/3 ✓
= (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 out a common factor first. Take out 1/3:

3a² + 4ab + (4/3)b² = (1/3)(9a² + 12ab + 4b²)
9a² = (3a)², 4b² = (2b)², 2(3a)(2b) = 12ab ✓
= (1/3)(3a + 2b)²

*(vi) (9/5)s² + 6sv + 5v² — take out 1/5:

(9/5)s² + 6sv + 5v² = (1/5)(9s² + 30sv + 25v²)
9s² = (3s)², 25v² = (5v)², 2(3s)(5v) = 30sv ✓
= (1/5)(3s + 5v)²

Why the common factor has to come out first: in Example 7 the book met 50p², whose square root is √50 p — a surd, which the identity a² + 2ab + b² cannot use tidily. Taking out 2 turned 50p² into 25p², a genuine square. Parts (v) and (vi) are the 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 is exactly the one that clears the denominators and leaves square coefficients.

Check it yourself: Put a = b = 1 in (v). The original gives 3 + 4 + 4/3 = 25/3; the answer gives (1/3)(3 + 2)² = 25/3. ✓

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

Answer

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.