# How do you simplify (sqrt12 - sqrt2)(sqrt12 + sqrt2)?

Apr 28, 2018

$\left(\sqrt{12} - \sqrt{2}\right) \left(\sqrt{12} + \sqrt{2}\right) = 10$

#### Explanation:

To simplify $\left(\sqrt{12} - \sqrt{2}\right) \left(\sqrt{12} + \sqrt{2}\right)$, we use the difference of two squares

$\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$

So

$\left(\sqrt{12} - \sqrt{2}\right) \left(\sqrt{12} + \sqrt{2}\right) = {\left(\sqrt{12}\right)}^{2} - {\left(\sqrt{2}\right)}^{2} = 12 - 2 = 10$

Apr 28, 2018

$10$

#### Explanation:

$\left(\sqrt{12} - \sqrt{2}\right) \left(\sqrt{12} + \sqrt{2}\right)$

There are two ways to solve this, and I'm going to show the longer way first.

METHOD 1:

To solve this, we distribute and expand using the FOIL method:

First, multiply the "firsts":
$\sqrt{12} \cdot \sqrt{12} = \sqrt{144} = 12$

Then the "outers":
$\sqrt{12} \cdot \sqrt{2} = \sqrt{24}$

Then the "inners":
$- \sqrt{2} \cdot \sqrt{12} = - \sqrt{24}$

Finally the "lasts":
$- \sqrt{2} \cdot \sqrt{2} = - \sqrt{4} = - 2$

When we combine these expressions we get:
$12 + \sqrt{24} - \sqrt{24} - 2$

The $\sqrt{24}$ and $- \sqrt{24}$ cancel each other out, so we're left with:
$12 - 2$

Which simplifies down to:
$10$

METHOD 2:

To solve this, we use:

This expression $\left(\sqrt{12} - \sqrt{2}\right) \left(\sqrt{12} + \sqrt{2}\right)$ is in the form $\left(a - b\right) \left(a + b\right)$, which is the same as $\left(a + b\right) \left(a - b\right)$.

As we can see, this is equivalent to ${a}^{2} - {b}^{2}$, so it becomes:
${\sqrt{12}}^{2} - {\sqrt{2}}^{2}$

Simplify:
$12 - 2$

$10$