Question

simplify.
$5\sqrt{3} \cdot 2\sqrt{21}$

Explanation:

Step1: Multiply the coefficients and the radicals separately

First, multiply the coefficients \(5\) and \(2\), and then multiply the radicals \(\sqrt{3}\) and \(\sqrt{21}\) using the property \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\) (\(a\geq0,b\geq0\)).
So we have \((5\times2)\times(\sqrt{3}\times\sqrt{21}) = 10\times\sqrt{3\times21}\)

Step2: Simplify the radical

Calculate the product inside the radical: \(3\times21 = 63\). Then factor \(63\) into prime factors, \(63 = 9\times7\), and we know that \(\sqrt{9\times7}=\sqrt{9}\times\sqrt{7}\) (by the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\), \(a\geq0,b\geq0\)). Since \(\sqrt{9} = 3\), we get \(\sqrt{63}=\sqrt{9\times7}=3\sqrt{7}\)

Step3: Multiply the coefficient with the simplified radical

Now multiply \(10\) with \(3\sqrt{7}\), that is \(10\times3\sqrt{7}=30\sqrt{7}\)

Answer:

\(30\sqrt{7}\)