NCERT Solutions for Class 9th Maths Chapter 4 Exercise Set 4.4 — Factorisation Without Using Algebra Tiles

Book page 81–82 Updated on2026-09-08

Q1.

Fill in the blanks to complete the following identities: (i) s² − 11s + 24 = (________) (________) (ii) (________) (x + 1) = (3x² − 4x −7) (iii) 10x² − 11x − 6 = (2x − ___) (___ + 2) (iv) 6x² + 7x + 2 = (____________) (___________)

Answer

(i) s² − 11s + 24 — need a + b = −11 and ab = +24, so both are negative. The factor pairs of 24 are 1×24, 2×12, 3×8, 4×6; only 3 + 8 = 11.

a = −3, b = −8
s² − 11s + 24 = s² − 3s − 8s + 24 = s(s − 3) − 8(s − 3)
= (s − 3)(s − 8)

(ii) (________)(x + 1) = 3x² − 4x − 7 — one factor is given, so divide it out. Since (x + 1) is a factor, split −4x so that (x + 1) appears:

3x² − 4x − 7 = 3x² + 3x − 7x − 7
= 3x(x + 1) − 7(x + 1)
= (3x − 7)(x + 1)

(iii) 10x² − 11x − 6 = (2x − ___)(___ + 2) — the pattern (px + a)(qx + b) needs pq = 10, ab = −6 and pb + aq = −11. With p = 2 and b = 2, we need q = 5 and a × 2 = −6, so a = −3.

10x² − 11x − 6 = 10x² − 15x + 4x − 6 (since −15 × 4 = −60 = 10 × −6 and −15 + 4 = −11)
= 5x(2x − 3) + 2(2x − 3)
= (2x − 3)(5x + 2)
So the blanks are 3 and 5x.

(iv) 6x² + 7x + 2 — split 7x using two numbers with product 6 × 2 = 12 and sum 7: they are 3 and 4.

6x² + 7x + 2 = 6x² + 3x + 4x + 2
= 3x(2x + 1) + 2(2x + 1)
= (2x + 1)(3x + 2)

Why the product is ac and not just c when the leading coefficient is not 1: for (px + a)(qx + b), the constant is ab and the x-coefficient is pb + aq. Multiply the two split pieces: (pb)(aq) = (pq)(ab) = (leading coefficient) × (constant term). So the two numbers you look for must multiply to a·c, not to c. In (iv) that is 6 × 2 = 12, and in (iii) it is 10 × (−6) = −60.

Check it yourself: Expand each answer. (2x − 3)(5x + 2) = 10x² + 4x − 15x − 6 = 10x² − 11x − 6. ✓

Q2.

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 × 17) (iv) (135)² (v) (97)² (vi) (18 × 29) (vii) (34 × 43) (viii) (205)²

Answer

Product	Identity chosen	Working	Value
(i) 41²	(a + b)²	(40 + 1)² = 1600 + 80 + 1	1681
(ii) 27²	(a − b)²	(30 − 3)² = 900 − 180 + 9	729
(iii) 23 × 17	(a + b)(a − b)	(20 + 3)(20 − 3) = 400 − 9	391
(iv) 135²	a² = (a + b)(a − b) + b²	140 × 130 + 25 = 18200 + 25	18225
(v) 97²	(a − b)²	(100 − 3)² = 10000 − 600 + 9	9409
(vi) 18 × 29	(x + a)(x + b)	(20 − 2)(20 + 9) = 400 + 20(7) − 18	522
(vii) 34 × 43	(x + a)(x + b)	(40 − 6)(40 + 3) = 1600 + 40(−3) − 18	1462
(viii) 205²	(a + b)²	(200 + 5)² = 40000 + 2000 + 25	42025

The two unequal products are worth writing out in full, since they use the third identity of Section 4.5.

(vi) 18 × 29 with x = 20, a = −2, b = 9:
(x + a)(x + b) = x² + (a + b)x + ab
= 400 + (−2 + 9)(20) + (−2)(9)
= 400 + 140 − 18 = 522

(vii) 34 × 43 with x = 40, a = −6, b = 3:
= 1600 + (−6 + 3)(40) + (−6)(3)
= 1600 − 120 − 18 = 1462

How to pick the identity: if the two numbers are equally far from a round number, they are a ± b and the difference-of-squares identity finishes in one step (that is (iii), 23 and 17 about 20). If they are at different distances, use (x + a)(x + b) with x the round number. And for a square ending in 5, Śhrīdharāchārya’s form is fastest — 135² is 13 × 14 = 182 followed by 25.

Q3.

Factor the following: (i) 9a² + b² + 4c² − 6ab + 12ac − 4bc (ii) 16s² + 25t² − 40st (iii) r² − r − 42 (iv) 49g² + 14gh + h² (v) 64u² + 121v² + 4w² − 176uv − 32uw + 44vw

Answer

(i) 9a² + b² + 4c² − 6ab + 12ac − 4bc — three squares and three cross terms, so aim at (A + B + C)². The terms containing b are the negative ones, so give b the minus sign.

A = 3a, B = −b, C = 2c
2AB = 2(3a)(−b) = −6ab ✓ 2BC = 2(−b)(2c) = −4bc ✓ 2CA = 2(2c)(3a) = 12ac ✓
= (3a − b + 2c)²

(ii) 16s² + 25t² − 40st

= (4s)² − 2(4s)(5t) + (5t)²
= (4s − 5t)²

(iii) r² − r − 42 — need a + b = −1 and ab = −42, so the two numbers have opposite signs and differ by 1: they are −7 and 6.

r² − r − 42 = r² − 7r + 6r − 42
= r(r − 7) + 6(r − 7)
= (r − 7)(r + 6)

(iv) 49g² + 14gh + h²

= (7g)² + 2(7g)(h) + h²
= (7g + h)²

(v) 64u² + 121v² + 4w² − 176uv − 32uw + 44vw — the negative cross terms are the ones containing u, so u should be the odd one out. Take A = 8u, B = −11v, C = −2w and check:

2AB = 2(8u)(−11v) = −176uv ✓
2BC = 2(−11v)(−2w) = +44vw ✓
2CA = 2(−2w)(8u) = −32uw ✓
= (8u − 11v − 2w)²

How the sign hunt works: in (A + B + C)² a cross term is negative exactly when its two letters carry opposite signs. Here uv and uw are negative while vw is positive, so v and w agree with each other and both disagree with u — hence u positive, v and w negative (or the whole thing negated, which gives the same square). Reading the sign pattern first saves you from trial and error.

Tip: A square never has a unique-looking answer: (8u − 11v − 2w)² and (−8u + 11v + 2w)² are the same expression. Either is a correct factorisation.

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