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Question
Find the G.C.D of the following:
2a2 + a, 4a2 – 1
Sum
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Solution
2a2 + a = a(2a + 1)
4a2 – 1 = (2a)2 – 12 .....[Using a2 – b2 = (a + b)(a – b)]
= (2a + 1)(2a – 1)
G.C.D = 2a + 1
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Chapter 3: Algebra - Exercise 3.9 [Page 115]
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