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simplify. $6\\sqrt{7} \\cdot 2\\sqrt{8}$

Question

simplify.
$6\sqrt{7} \cdot 2\sqrt{8}$

Explanation:

Step1: Multiply the coefficients and the radicals separately

First, multiply the coefficients \(6\) and \(2\), and multiply the radicals \(\sqrt{7}\) and \(\sqrt{8}\) using the property \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\).
So we have \((6\times2)\times(\sqrt{7}\cdot\sqrt{8}) = 12\times\sqrt{7\times8}\)

Step2: Simplify the radical

Simplify \(\sqrt{7\times8}=\sqrt{56}\). Then, we can further simplify \(\sqrt{56}\) by factoring out the perfect square factor. Since \(56 = 4\times14\), we have \(\sqrt{56}=\sqrt{4\times14}=\sqrt{4}\times\sqrt{14}=2\sqrt{14}\)

Step3: Multiply the result with the coefficient from Step1

Now, multiply \(12\) with \(2\sqrt{14}\) (from the simplified radical). \(12\times2\sqrt{14}=24\sqrt{14}\)

Answer:

\(24\sqrt{14}\)