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

May 20, 2018

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

May 20, 2018

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

#### Explanation:

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

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

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

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

#### Explanation:

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