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

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

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Answer 1 · 2018-05-20T18:54:13

Is a notable identity $(1 + 2 v) (1 - 2 v) = 1 - 4 v^{2}$ $(1+2v)(1-2v)=1-4v^2$

Answer 2 · 2018-05-20T18:56:57

$1 - 4 v^{2}$ $1 - 4v^2$

Explanation:

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

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

$(1 + 2 v) (1 - 2 v) = 1^{2} - {(2 v)}^{2} = 1 - 4 v^{2}$ $(1+2v)(1-2v) = 1^2 - (2v)^2 = 1 - 4v^2$

Answer 3 · 2018-05-20T20:17:00

$1 - 4 v^{2}$ $1-4v^2$

Explanation:

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

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

3 Answers

Explanation:

Explanation:

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

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