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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.

Introduction

Algebra is a branch of mathematics that uses symbols and letters to represent numbers and quantities in formulas and equations. In this chapter, we explore algebraic identities, which are special algebraic expressions that are equal for all values of the variables involved. These identities help simplify algebraic expressions and solve problems efficiently. The chapter begins by revisiting the concept of polynomials and then introduces the idea of algebraic identities through simple examples. It emphasizes the importance of recognizing patterns in algebraic expressions to factorize or expand them quickly without performing lengthy calculations. The introduction sets the stage for understanding how these identities are derived and applied in various mathematical problems, making algebra more manageable and insightful for students.

📊 Diagram: No specific diagrams in the introduction section; the focus is on conceptual understanding.

🧪 Activity: No activity in this introductory section.

🔗 Connection: Leads to the next section which introduces specific algebraic identities and their proofs.

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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