How do you find the product of $(2n-7)^2$ ?

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Answer 1 · 2016-06-29T19:19:18

$=(2n-7)^2=4n^2-28n+49.$

Reqd. Product $=(2n-7)^2=(2n-7)*(2n-7)=2n(2n-7)-7(2n-7)...[Distributive Law] = 4n^2-14n-14n+49=4n^2-28n+49.$

OR, we can use the formula, $(a-b)^2=a^2-2ab+b^2,$ & get,
The product $=(2n)^2-2*2n*7+7^2=4n^2-28n+49,$ as before!

How do you find the product of (2n-7)^2(2n-7)^2?