Question

question
what is the product of $8\sqrt{7}$ and $6\sqrt{14}$ in simplest radical form?

Explanation:

Step1: Multiply the coefficients and the radicals separately

To find the product of \(8\sqrt{7}\) and \(6\sqrt{14}\), we use the property of radicals \((a\sqrt{b})(c\sqrt{d}) = ac\sqrt{bd}\). So first, multiply the coefficients \(8\) and \(6\), and then multiply the radicals \(\sqrt{7}\) and \(\sqrt{14}\).
The product of the coefficients: \(8\times6 = 48\)
The product of the radicals: \(\sqrt{7}\times\sqrt{14}=\sqrt{7\times14}\)

Step2: Simplify the radical

Simplify \(\sqrt{7\times14}\). First, calculate \(7\times14 = 98\). Then, factor \(98\) into its prime factors: \(98 = 49\times2\), and \(49 = 7^2\). So, \(\sqrt{98}=\sqrt{49\times2}=\sqrt{49}\times\sqrt{2}=7\sqrt{2}\) (since \(\sqrt{49} = 7\)).

Step3: Multiply the result from Step1 and Step2

Now, multiply the coefficient \(48\) with the simplified radical \(7\sqrt{2}\). So, \(48\times7\sqrt{2}\). Calculate \(48\times7 = 336\). So the product is \(336\sqrt{2}\).

Answer:

\(336\sqrt{2}\)