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Identity: (a + b)2 = a2 + b2 + 2ab
How is this identity obtained? Let's see how. Taking LHS of the identity: (a + b)2 This can also be written as: = (a + b)(a + b) Multiply as we do multiplication of two binomials or use F.O.I.L. method and we get: = a(a + b) + b(a + b) = a2 + ab + ab + b2 On adding like terms we get: = a2 + 2ab + b2 On rearranging the terms we get: = a2 + b2 + 2ab Hence, we obtain the identity i.e. (a + b)2 = a2 + b2 + 2ab Following are a few applications to this identity. Example 1: Solve (3x + 2y)2 Solution: This proceeds as: The given polynomial (3x + 2y)2 represents the identity (a + b)2 Where a = 3x and b = 2y On applying values of a and b on the identity i.e. (a + b)2 = a2 + b2 + 2ab we get: (3x + 2y)2 = (3x)2 + (2y)2 + 2(3x)(2y) Expand the exponential forms and we get: = 9x2 + 4y2 + 2(3x)(2y) On solving multiplication process we get: = 9x2 + 4y2 + 12xy Hence, (3x + 2y)2 = 9x2 + 4y2 + 12xy Example 2: Solve (6m + 9n)2 Solution: This proceeds as: The given polynomial (6m + 9n)2 represents the identity (a + b)2 Where a = 6m and b = 9n On applying values of a and b on the identity i.e. (a + b)2 = a2 + b2 + 2ab we get: (6m + 9n)2 = (6m)2 + (9n)2 + 2(6m)(9n) Expand the exponential forms and we get: = 36m2 + 81n2 + 2(6m)(9n) On solving multiplication process we get: = 36m2 + 81n2 + 108mn Hence, (6m + 9n)2 = 36m2 + 81n2 + 108mn |