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How do you evaluate #(1+ 2v ) ( 1- 2v )#?

Answer 1

Kennedy Adams

Is a notable identity #(1+2v)(1-2v)=1-4v^2#

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Answer 2

Julia Adams

#1 - 4v^2#

Tip: Remember this Algebraic Identity called the difference of squares: #(a+b)(a-b) = a^2 - b^2 #

sub in # a = 1, b = 2v #

#(1+2v)(1-2v) = 1^2 - (2v)^2 = 1 - 4v^2 #

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Answer 3

Savannah Anderson

#1-4v^2#

#(a+b)(a-b)# is equal to #a^2-b^2# Here, #a=1# and #b=2v# So plug that in and you get #1^2-2^2v^2# Which is equal to #1-4v^2#

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Answer 4

Kennedy Adams

To evaluate the expression (1 + 2v) * (1 - 2v), you can use the distributive property or the FOIL method. Applying either method yields the same result:

(1 + 2v) * (1 - 2v) = 1 * 1 + 1 * (-2v) + 2v * 1 + 2v * (-2v) = 1 - 2v + 2v - 4v^2 = 1 - 4v^2

So, (1 + 2v) * (1 - 2v) simplifies to 1 - 4v^2.

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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.