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

1 Answer

Jun 29, 2016

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

Explanation:

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!

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