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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.

What is an Identity?

An algebraic identity is an equality that holds true for all possible values of the variables involved. Unlike equations, which are true only for specific values, identities are universally valid. For example, the expression (a + b)² = a² + 2ab + b² is an identity because it is true for any values of a and b. This section explains the concept of identities by comparing them with equations and demonstrating how identities can be verified by substituting values. It also discusses the importance of identities in simplifying algebraic expressions and solving problems efficiently. The section introduces the idea that identities can be used to expand or factorize expressions without performing multiplication or division explicitly, saving time and effort in calculations.

📊 Diagram: No diagrams; the section uses algebraic expressions and substitution examples.

🧪 Activity: Activity to verify identities by substituting different values for variables.

🔗 Connection: Prepares the learner to understand and prove specific algebraic identities in the next section.

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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