# How do you simplify (sqrt [6] + 2sqrt [2])(4sqrt[6] - 3sqrt2)?

Jun 28, 2018

$12 + 5 \sqrt{12}$

#### Explanation:

Given: $\left(\sqrt{6} + 2 \sqrt{2}\right) \left(4 \sqrt{6} - 3 \sqrt{2}\right)$.

Use the $\text{FOIL}$ theorem, which states that $\left(a + b\right) \left(c + d\right) = a c + a d + b c + b d$.

So, we get:

$= \sqrt{6} \cdot 4 \sqrt{6} - 3 \sqrt{2} \cdot \sqrt{6} + 2 \sqrt{2} \cdot 4 \sqrt{6} - 2 \sqrt{2} \cdot 3 \sqrt{2}$

$= 4 \cdot 6 - 3 \sqrt{12} + 8 \sqrt{12} - 6 \cdot 2$

$= 24 - 12 + 5 \sqrt{12}$

$= 12 + 5 \sqrt{12}$

Jun 28, 2018

color(crimson)(=> 2 (6 - 5 sqrt 3)

#### Explanation:

$\left(\sqrt{6} + 2 \sqrt{2}\right) \left(4 \sqrt{6} - 3 \sqrt{2}\right)$

$\implies \sqrt{6} \cdot 4 \sqrt{6} + 2 \sqrt{2} \cdot 4 \sqrt{6} - \sqrt{6} \cdot 3 \sqrt{2} - 2 \sqrt{2} \cdot 3 \sqrt{2}$

$\implies 4 \cdot 6 + 8 \sqrt{12} - 3 \sqrt{12} - 6 \cdot 2$

$\implies 24 - 12 + 8 \sqrt{12} - 3 \sqrt{12}$

$\implies 12 + 5 \sqrt{4 \cdot 3}$

$\implies 12 - 10 \sqrt{3}$

color(crimson)(=> 2 (6 - 5 sqrt 3)