Question 70

Expand: $$(4a+3b+2c)^{2}$$

Solution

$$(4a+3b+2c)^{2}$$

= $$(4a)^2 + (3b)^2 + (2c)^2 + 2(4a.3b + 3b.2c + 2c.4a)$$

($$\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2(a + b + c))$$

= $$16a^2 + 9b^2 + 4c^2 + 2(12ab + 6bc + 8ac)$$

= $$16a^2 + 9b^2 + 4c^2 + 24ab + 12bc + 16ac$$

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