Expansion Class 9 ICSE — Arun Deep Chapter 4
Expansion class 9 ICSE — Chapter 4 of Arun Deep Mathematics — shows how to multiply algebraic expressions without long multiplication. The chapter opens with a second method proof using continued proportion. It then introduces the sum-and-product rule for binomial products. This page explains each section in the book’s order, with worked patterns you can follow. You will also see the exercise workload at a glance and the pattern types each block trains.
Download Expansions Class 9 ICSE Mathematics Chapter 4 PDF
Download the Expansions Class 9 ICSE Mathematics chapter PDF from the Arun Deep textbook to revise offline or print for practice. It is the full chapter as printed in the book — 15 pages, 1 sections, 50 questions.
What is inside Chapter 4
15 pages · 1 sections · 50 exercise questions
| Section | Title |
|---|---|
| 1 | Second Method |
- EXERCISE 4 (A): 22 questions
- EXERCISE 4 (B): 14 questions
- EXERCISE 4 (C): 14 questions
1 Second Method
This section proves an identity using continued proportion. Three numbers a, b, c are in continued proportion when the ratios are equal: \frac{a}{b} = \frac{b}{c}. Cross-multiplying gives b^2 = ac. This single fact drives the whole proof.
The key move is treating (a + c) as one unit. The product becomes a difference of squares. This works because the two middle terms, +b and -b, are identical apart from sign.
(a + b + c)(a - b + c) = [(a + c) + b][(a + c) - b] = (a + c)^2 - b^2
Since b^2 = ac, substitute and expand. The 2ac term appears from squaring (a + c). Subtracting ac leaves exactly one ac.
(a + c)^2 - ac = a^2 + 2ac + c^2 - ac = a^2 + ac + c^2
Replace ac with b^2 to reach a^2 + b^2 + c^2. The identity holds because the middle term cancels exactly. No approximation is involved.
Try it with a = 2, b = 4, c = 8. Check the condition: \frac{2}{4} = \frac{4}{8}, so b^2 = 16 = 2 \times 8. Now compute both sides:
(2 + 4 + 8)(2 - 4 + 8) = 14 \times 6 = 84
a^2 + b^2 + c^2 = 4 + 16 + 64 = 84
Both sides match. The common mistake is expanding (a + c)^2 as a^2 + c^2. That drops the 2ac middle term. Always write all three terms.
The difference of squares identity shortens this proof considerably. That is its practical value. A direct expansion would need six product terms. The regrouping collapses them into two.
This continued-proportion pattern recurs across the Class 9 ICSE Maths syllabus. Check the official CISCE curriculum for the full list of expansion topics.
What to practise from Expansion Class 9 ICSE
Exercise 4 (C) trains the sum-and-product rule. The core pattern is:
(x + a)(x + b) = x^2 + (a + b)x + ab
The coefficient of x is the sum of the constants. The constant term is their product. When both constants are negative, both terms in the result stay positive. That is why (x - 5)(x - 7) gives +35.
Some products appear in reversed order. Factor out -1 to align them. For (2 - x)(4 - x), write -(x - 2) and -(x - 4). The negatives cancel, leaving (x - 2)(x - 4).
Quartic products use substitution. Let x^2 = y and expand in y. Then substitute back. Two-variable products like (ab + 3)(ab - 2) treat ab as the single variable.
Leading-coefficient products need term-by-term expansion. Multiply each term of the first bracket by each term of the second. Combine like terms at the end.
Use the sum-and-product rule as a mental check before writing any answer. For example, take (x + 4)(x + 7). The sum is 4 + 7 = 11. The product is 4 \times 7 = 28. So the answer is x^2 + 11x + 28. If the coefficient or constant does not match, recheck your signs.
The full chapter has three exercise blocks. See the Class 9 ICSE books list for related algebra chapters.
- Exercise 4 (A) — 22 questions
- Exercise 4 (B) — 14 questions
- Exercise 4 (C) — 14 questions
Expansion Class 9 ICSE frequently asked questions
What is the expansion formula (x + a)(x + b) in class 9 ICSE?
The formula is (x + a)(x + b) = x^2 + (a + b)x + ab. The coefficient of x is the sum a + b. The constant term is the product ab. For (x + 6)(x + 2), you get x^2 + 8x + 12.
How do you prove (a + b + c)(a – b + c) = a² + b² + c² when a/b = b/c?
Group (a + c) as one unit. The product becomes [(a + c) + b][(a + c) - b]. That is (a + c)^2 - b^2. Since b^2 = ac, expand and simplify to reach a^2 + b^2 + c^2. The condition \frac{a}{b} = \frac{b}{c} is what makes b^2 = ac true.
How many questions are in Arun Deep Class 9 maths chapter 4 exercises?
Three exercise blocks make up the chapter. Exercise 4 (A) has 22 questions. Exercise 4 (B) has 14. Exercise 4 (C) has 14. That is 50 questions in total. Exercise 4 (C) focuses on the sum-and-product rule.
Why do we rewrite (2 – x)(4 – x) as (x – 2)(x – 4) before expanding?
Each factor has a negative x term. Factor out -1 from both: -(x - 2) \times -(x - 4). The negatives cancel, giving (x - 2)(x - 4). Now the sum-and-product rule applies directly. This is the edge case where a negative sign is factored out first.
Reference: ICSE Class 9 Mathematics textbook material, chapter 4: Expansions — CISCE-aligned.