Factorise : 4a2 - (4b2 + 4bc + C2) - Mathematics

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Question

Factorise: 4a2 - (4b2 + 4bc + c2)

Sum

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SolutionShow Solution

4a2 - (4b2 + 4bc + c2)

= (2a)2 - [(2b)2 + (c)2 + 2 × 2b × c]

= (2a)2 - (2b + c)2

= [2a - (2b + c)][2a + (2b + c)] ...[∵ a2 - b2 = (a + b)(a - b)]

= [2a - 2b - c][2a + 2b + c]

shaalaa.comMethod of Factorisation : Difference of Two Squares Report Error Is there an error in this question or solution?Q 13Q 12Q 14Chapter 5: Factorisation - Exercise 5 (C) [Page 72]

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Selina Concise Mathematics [English] Class 9 ICSEChapter 5 FactorisationExercise 5 (C) | Q 13 | Page 72

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