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MathematicsClass 9Exploring Algebraic Identities

Exploring Algebraic Identities | Class 9 Mathematics Notes

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

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

Standard Algebraic Identities

This section introduces three fundamental algebraic identities that are widely used in algebraic manipulations. These identities are: (1) The square of a sum: (a + b)² = a² + 2ab + b², (2) The square of a difference: (a - b)² = a² - 2ab + b², and (3) The difference of squares: a² - b² = (a - b)(a + b). Each identity is explained in detail with step-by-step expansion and factorization. The section demonstrates how these identities can be derived by multiplying binomials and how they can be used to simplify expressions or solve problems without direct multiplication. The importance of these identities in algebra is emphasized, as they form the basis for many higher-level algebraic concepts. The section also encourages students to memorize these identities and understand their proofs to apply them confidently.

📊 Diagram: Diagrams illustrating the geometric interpretation of (a + b)² as the area of a square with side (a + b), subdivided into smaller squares and rectangles representing a², b², and 2ab. Similarly, difference of squares is shown as the difference between two squares with sides a and b.

🧪 Activity: Activity involving expanding and factorizing expressions using these identities.

🔗 Connection: Leads to the next section where more complex identities and their applications are explored.

Frequently asked questions

Find the values of the following using the identity (a – b)2 = a2 – 2ab + b2. (i) (79)2 (ii) (193)2 (iii) (299)2

Using the identity (a - b)^2 = a^2 - 2ab + b^2, we rewrite each number as (a - b) and calculate:

(i) 79^2 = (80 - 1)^2 = 80^2 - 2801 + 1^2 = 6400 - 160 + 1 = 6241

(ii) 193^2 = (200 - 7)^2 = 200^2 - 22007 + 7^2 = 40000 - 2800 + 49 = 37249

(iii) 299^2 = (300 - 1)^2 = 300^2 - 23001 + 1^2 = 90000 - 600 + 1 = 89401

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier. (i) 1172 (ii) 782 (iii) 1982 (iv) 2142 (v) 11042 (vi) 11202

Use the identity (a + b)^2 = a^2 + 2ab + b^2 or (a - b)^2 = a^2 - 2ab + b^2 as appropriate:

(i) 117^2 = (100 + 17)^2 = 10000 + 210017 + 289 = 10000 + 3400 + 289 = 13689

(ii) 78^2 = (80 - 2)^2 = 6400 - 320 + 4 = 6084

(iii) 198^2 = (200 - 2)^2 = 40000 - 800 + 4 = 39204

(iv) 214^2 = (200 + 14)^2 = 40000 + 5600 + 196 = 45796

(v) 1104^2 = (1100 + 4)^2 = 1,210,000 + 8,800 + 16 = 1,218,816

(vi) 1120^2 = (1100 + 20)^2 = 1,210,000 + 44,000 + 400 = 1,254,400

Factor using suitable identities: (i) 16y2 – 24y + 9 (ii) s2 + 6st + 4t2 (iii) m2 + mk + k2 + 3nk + 2mn + 9n2 (iv) p3 − 2p2 + 3p − 4 (v) 9a2 + 4b2 + c2 − 12ab + 6ac − 4bc

Factorization:

(i) 16y^2 - 24y + 9 = (4y - 3)^2 (perfect square trinomial)

(ii) s^2 + 6st + 4t^2 = (s + 2t)(s + 2t) or (s + 2t)^2

(iii) m^2 + mk + k^2 + 3nk + 2mn + 9n^2 Group terms: (m^2 + mk + k^2) + (3nk + 2mn + 9n^2) Try to factor as (m + an + bk)^2 or factor by grouping: Rewrite as (m + 3n + k)(m + 3n + k) = (m + 3n + k)^2

(iv) p^3 − 2p^2 + 3p − 4 Group: (p^3 - 2p^2) + (3p - 4) = p^2(p - 2) + 1(3p - 4) No common factor; try factor by grouping or synthetic division: Try p = 1: 1 - 2 + 3

Expand the following using the identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca: (i) (p + 3q + 7r)2 (ii) (3x – 2y + 4z)2

(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

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