# What is the product (in simplest radical form) of the square route of 6 and the square root of 15?

Apr 9, 2018

$\sqrt{6} \cdot \sqrt{15} = 3 \sqrt{10}$

#### Explanation:

By radical laws, $\sqrt{a} \cdot \sqrt{b} = \sqrt{a b}$.

Using this law:
$\sqrt{6} \cdot \sqrt{15}$
$= \sqrt{90}$

$\sqrt{90}$ has $9$, a perfect square, as one of its factors, so it can be simplified.
$\sqrt{90}$
$= \sqrt{9 \cdot 10}$
$= \sqrt{9} \cdot \sqrt{10}$
$= 3 \sqrt{10}$