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MathematicsClass 9Exploring Algebraic

Exploring Algebraic | Class 9 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 4 min read

Exploring Algebraic | Class 9 Mathematics Notes

Exploring Algebraic – this guide gives you a concise, exam-ready overview of Exploring Algebraic from Class 9 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

4.3 FACTORSATION OF ALGEBRAIC EXPRESSIONS USING IDENTITIES

This section explores how algebraic identities can be used to factorise expressions, which is the reverse process of expansion. Using the identity (a + b)² = a² + 2ab + b², expressions of the form a² + 2ab + b² can be rewritten as (a + b)², revealing their factors. Several examples illustrate this: x² + 4x + 4 is identified as (x + 2)², and 36x² + 12x + 1 as (6x + 1)². When expressions have common factors, like 50p² + 60pq + 18q², the common factor is factored out first (2 in this case), simplifying the expression inside the bracket to a perfect square, which is then factorised using the identity. The section also introduces the identity (a - b)² = a² - 2ab + b² and demonstrates its geometric visualization using squares and rectangles. This identity is similarly useful for factorisation and calculations. The section concludes by revisiting the initial pattern involving three consecutive squares and proving it algebraically using these identities.

📊 Diagram: Fig. 4.3: A square of side $a$ units

🧪 Activity: Students are encouraged to replace b by -b in the identity (a + b)² to derive (a - b)² and explore its applications.

🔗 Connection: Leads to study of more identities involving three terms and their geometric interpretations.

Frequently asked questions

1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier. (i) 117^2 (ii) 78^2 (iii) 198^2 (iv) 214^2 (v) 1104^2 (vi) 1120^2

Solutions:

(i) 117^2 Use (a + b)^2 = a^2 + 2ab + b^2 117 = 100 + 17 So, 117^2 = 100^2 + 210017 + 17^2 = 10000 + 3400 + 289 = 13689

(ii) 78^2 Use (a - b)^2 = a^2 - 2ab + b^2 78 = 80 - 2 So, 78^2 = 80^2 - 2802 + 2^2 = 6400 - 320 + 4 = 6084

(iii) 198^2 Use (a - b)^2 198 = 200 - 2 198^2 = 200^2 - 22002 + 2^2 = 40000 - 800 + 4 = 39204

(iv) 214^2 Use (a + b)^2 214 = 200 + 14 214^2 = 200^2 + 220014 + 14^2 = 40000 + 5600 + 196 = 45796

(v) 1104^2 Use (a + b)^2 1104 = 1100 + 4 1104^2 = 1100^2

2. Factor using suitable identities: (i) 16y^{2} - 24y + 9 (ii) (9/4) s^2 + 6st + 4t^2 (iii) (m^2)/9 + (mk)/3 + (k^2)/4 + 3nk + 2mn + 9n^2 (iv) (p^2)/16 - 2 + 16/(p^2) (v) 9a^{2} + 4b^{2} + c^{2} - 12ab + 6ac - 4bc

Solutions:

(i) 16y^2 - 24y + 9 = (4y)^2 - 24y3 + 3^2 = (4y - 3)^2

(ii) (9/4)s^2 + 6st + 4t^2 = (3s/2)^2 + 2(3s/2)2t + (2t)^2 = (3s/2 + 2t)^2

(iii) (m^2)/9 + (mk)/3 + (k^2)/4 + 3nk + 2mn + 9n^2 Group terms: = (m^2)/9 + (mk)/3 + (k^2)/4 + 2mn + 3nk + 9n^2 Try to write as square of (m/3 + k/2 + 3n)^2 Calculate: (m/3)^2 + (k/2)^2 + (3n)^2 + 2(m/3)(k/2) + 2(k/2)(3n) + 2(m/3)(3n) = (m^2)/9 + (k^2)/4 + 9n^2 + (mk)/3 + 3nk + 2mn Matches given expression exactly. So factorization is (m/3 + k

3. Expand the following using the identity (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca: (i) (p + 3q + 7r)^2 (ii) (3x - 2y + 4z)^2

Solutions:

(i) (p + 3q + 7r)^2 = p^2 + (3q)^2 + (7r)^2 + 2p3q + 23q7r + 27rp = p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14rp

(ii) (3x - 2y + 4z)^2 = (3x)^2 + (-2y)^2 + (4z)^2 + 23x(-2y) + 2(-2y)4z + 24z3x = 9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx

4. Is this an identity? (a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2 = 2a^2 + 2b^2 + 2c^2.

Yes, this is an identity.

Proof: Expand each square:

(a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2ac - 2bc (a - b + c)^2 = a^2 + b^2 + c^2 - 2ab + 2ac - 2bc (a - b - c)^2 = a^2 + b^2 + c^2 - 2ab - 2ac + 2bc

Add all three: = (a^2 + b^2 + c^2 + 2ab - 2ac - 2bc) + (a^2 + b^2 + c^2 - 2ab + 2ac - 2bc) + (a^2 + b^2 + c^2 - 2ab - 2ac + 2bc) = 3a^2 + 3b^2 + 3c^2 + (2ab - 2ab - 2ab) + (-2ac + 2ac - 2ac) + (-2bc - 2bc + 2bc) = 3a^2 + 3b^2 + 3c^2 - 2ab - 2ac - 2bc But carefully summing the cross terms: 2ab

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