# What is the answer to this problem with explanation?

## $\left(3 \sqrt{10} + 5\right) \left(4 \sqrt{6} - 2\right)$

Jun 18, 2018

$24 \sqrt{15} - 6 \sqrt{10} + 20 \sqrt{6} - 10$

#### Explanation:

$\text{using the "color(blue)"law of radicals}$

•color(white)(x)sqrtaxxsqrtbhArrsqrt(ab)

$\text{expand the factors using FOIL}$

$= \left(3 \sqrt{10} \times 4 \sqrt{6}\right) + \left(- 2 \times 3 \sqrt{10}\right) + \left(5 \times 4 \sqrt{6}\right)$
$\textcolor{w h i t e}{=} + \left(5 \times - 2\right)$

$= 12 \sqrt{60} - 6 \sqrt{10} + 20 \sqrt{6} - 10$

$\sqrt{10} \text{ and "sqrt6" are in simplest form}$

$\sqrt{60} = \sqrt{4 \times 15} = \sqrt{4} \times \sqrt{15} = 2 \sqrt{15}$

$\sqrt{15} \text{ is in simplest form}$

$= \left(12 \times 2 \sqrt{15}\right) - 6 \sqrt{10} + 20 \sqrt{6} - 10$

$= 24 \sqrt{15} - 6 \sqrt{10} + 20 \sqrt{6} - 10$

Jun 18, 2018

$48 \sqrt{15} - 6 \sqrt{10} + 20 \sqrt{6} - 10$
See below

#### Explanation:

We have to operate in this expresion usin distributive law for real numbers. This property stablish

$\left(a + d\right) \left(b + c\right) = a b + a c + \mathrm{db} + \mathrm{dc}$

(3sqrt10+5)(4sqrt6-2)=3sqrt10·4sqrt6-3sqrt10·2+5·4sqrt6-5·2=12sqrt10sqrt6-6sqrt10+20sqrt6-10

But sqrt10sqrt6=sqrt60=sqrt(5·2^2·3)=2sqrt15

Then

$12 \sqrt{10} \sqrt{6} - 6 \sqrt{10} + 20 \sqrt{6} - 10 = 24 \sqrt{15} - 6 \sqrt{10} + 20 \sqrt{6} - 10$