Question

Multiply. Assume b is greater than or equal to zero, and write your answer in simplest form.
square root of 6*square root of 3b5

Answer 1

To multiply \(\sqrt{6}\) and \(\sqrt{3b^5}\), you can use the property of square roots that states:

\[
\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}
\]

So:

\[
\sqrt{6} \cdot \sqrt{3b^5} = \sqrt{6 \cdot 3b^5}
\]

Now, let's multiply \(6\) and \(3\):

\[
6 \cdot 3 = 18
\]

Thus:

\[
\sqrt{6 \cdot 3b^5} = \sqrt{18b^5}
\]

Next, we can simplify \(\sqrt{18b^5}\). The number \(18\) can be factored as \(9 \cdot 2\), and the square root of \(9\) is \(3\):

\[
\sqrt{18b^5} = \sqrt{9 \cdot 2 \cdot b^5}
\]

Now we can separate the square root:

\[
\sqrt{18b^5} = \sqrt{9} \cdot \sqrt{2} \cdot \sqrt{b^5}
\]

This simplifies to:

\[
3 \cdot \sqrt{2} \cdot \sqrt{b^5}
\]

Now, to simplify \(\sqrt{b^5}\), we recognize that:

\[
b^5 = b^4 \cdot b = (b^2)^2 \cdot b
\]

So:

\[
\sqrt{b^5} = \sqrt{b^4 \cdot b} = \sqrt{b^4} \cdot \sqrt{b} = b^2 \cdot \sqrt{b}
\]

Putting it all together, we have:

\[
3 \cdot \sqrt{2} \cdot b^2 \cdot \sqrt{b} = 3b^2\sqrt{2b}
\]

Thus, the final result in simplest form is:

\[
\boxed{3b^2\sqrt{2b}}
\]