NCERT Solutions for Class 9th Maths Chapter 4 In-text Questions — Factorisation Using Algebra Tiles
Book page 80 Updated on2026-09-08
Q1.
Now consider the case where we have a rectangle of sidelengths 2x + 3 and 3x + 1, as shown in Fig. 4.8. What can you say about its area (2x + 3)(3x + 1)?
Answer
Count the tiles in Fig. 4.8. Two x-columns and three unit-columns across; three x-rows and one unit-row down.
The rectangle (2x + 3) by (3x + 1) built from 6 x²-tiles, 11 x-tiles and 3 unit tiles.
Why the middle coefficient is 11 and not 3 + 1: the two ends of the rectangle are no longer plain x-tiles — there are 2 of them one way and 3 the other. So an x-tile arises in two different ways, 3 × 3 = 9 of them from the wide side and 2 × 1 = 2 from the tall side. This is why the general rule needs (pb + aq), not (a + b).
Q2.
Fill in the blanks with the appropriate expressions to make the equation true: (px + a)(qx + b) = (_____)x² + (_____)x + _____ . Also, verify your answer using the distributive property.
Answer
(px + a)(qx + b) = (pq)x² + (pb + aq)x + ab
Verification by the distributive property:
(px + a)(qx + b) = px(qx + b) + a(qx + b)
= pq·x² + pb·x + aq·x + ab
= pq·x² + (pb + aq)x + ab
Check against Fig. 4.8, where p = 2, a = 3, q = 3, b = 1:
Why the three blanks are what they are: the product of two brackets has four terms. Only one of them, px × qx, has two x’s — that is the x² coefficient pq. Two of them have one x each — those combine into pb + aq. One has none — that is ab. Setting p = q = 1 recovers the earlier rule (x + a)(x + b) = x² + (a + b)x + ab, so this is its general form.
Tip: When you factor 6x² + 7x + 2 later, you are solving pq = 6, ab = 2, pb + aq = 7. This identity tells you exactly which three conditions to satisfy.