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

Answer 1

Eli Assouline

#=(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!

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Answer 2

Savannah Anderson

To find the product of (2n - 7)^2, you need to square the expression:

(2n - 7)^2 = (2n - 7)(2n - 7)

Using the FOIL method (First, Outer, Inner, Last), you multiply each term in the first expression by each term in the second expression:

(2n - 7)(2n - 7) = 4n^2 - 14n - 14n + 49

Simplify by combining like terms:

4n^2 - 28n + 49

So, the product of (2n - 7)^2 is 4n^2 - 28n + 49.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.