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Question
Expand:
`(a/2 + 2b - 4c)^2`
Sum
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Solution
`(a/2 + 2b - 4c)^2`
Here, let a = `a/2`, b = 2b, c = −4c
Using the identity,
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
`(a/2 + 2b - 4c)^2 = (a/2)^2 + (2b)^2 + (-4c)^2 + 2(a/2)(2b) + 2(2b)(-4c) + 2(-4c)(a/2)`
= `a^2/4 + 4b^2 + 16c^2 + ((2a)/2)(2b) + 2(-8bc) + 2((-4ac)/2)`
= `a^2/4 + 4b^2 + 16c^2 + (a)(2b) - 16bc + 2(-2ac)`
∴ `a^2/4 + 4b^2 + 16c^2 + 2ab - 16bc - 4ac`
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Chapter 3: Expansions - EXERCISE C [Page 38]
