Question

question
what is the product of $4\sqrt{21}$ and $9\sqrt{6}$ in simplest radical form?

Explanation:

Step1: Multiply the coefficients and the radicals separately

To find the product of \(4\sqrt{21}\) and \(9\sqrt{6}\), we use the property of radicals \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\) and the commutative property of multiplication. First, multiply the coefficients (\(4\) and \(9\)) and then multiply the radicals (\(\sqrt{21}\) and \(\sqrt{6}\)):

\((4\sqrt{21})\times(9\sqrt{6})=(4\times9)\times(\sqrt{21}\times\sqrt{6})\)

Step2: Calculate the product of the coefficients and the product inside the radical

Calculate \(4\times9 = 36\). For the radicals, using the property \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\), we have \(\sqrt{21}\times\sqrt{6}=\sqrt{21\times6}\). Calculate \(21\times6 = 126\), so now we have \(36\sqrt{126}\).

Step3: Simplify the radical \(\sqrt{126}\)

We need to factor \(126\) into its prime factors to simplify the square root. The prime factorization of \(126\) is \(126 = 2\times3^2\times7\). So, \(\sqrt{126}=\sqrt{2\times3^2\times7}\). Using the property \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (for non - negative \(a\) and \(b\)) and \(\sqrt{a^2}=a\) (for \(a\geq0\)), we get \(\sqrt{2\times3^2\times7}=\sqrt{3^2}\times\sqrt{2\times7}=3\sqrt{14}\).

Step4: Multiply the coefficient with the simplified radical

Now, substitute \(\sqrt{126}=3\sqrt{14}\) back into \(36\sqrt{126}\):

\(36\sqrt{126}=36\times3\sqrt{14}\)

Calculate \(36\times3 = 108\). So, the product in simplest radical form is \(108\sqrt{14}\).

Answer:

\(108\sqrt{14}\)