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

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Answer 1 · 2018-05-20T18:54:13

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

Answer 2 · 2018-05-20T18:56:57

#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 #

Answer 3 · 2018-05-20T20:17:00

#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#