Question

Simplify.
Multiply and remove all perfect squares from inside the square roots. Assume
\[a\] is positive.
\[\sqrt{2a}\cdot\sqrt{14a^3}\cdot\sqrt{5a}=\]

Answer 1

To simplify the expression \(\sqrt{2a} \cdot \sqrt{14a^3} \cdot \sqrt{5a}\), we can first combine the square roots:

\[
\sqrt{2a} \cdot \sqrt{14a^3} \cdot \sqrt{5a} = \sqrt{(2a) \cdot (14a^3) \cdot (5a)}
\]

Now, let's multiply the terms inside the square root:

\[
(2a) \cdot (14a^3) \cdot (5a) = 2 \cdot 14 \cdot 5 \cdot a \cdot a^3 \cdot a = 140 \cdot a^{1+3+1} = 140 \cdot a^5
\]

Thus, we have:

\[
\sqrt{2a} \cdot \sqrt{14a^3} \cdot \sqrt{5a} = \sqrt{140a^5}
\]

Next, we need to simplify \(\sqrt{140a^5}\). We first break this down:

\[
\sqrt{140a^5} = \sqrt{140} \cdot \sqrt{a^5}
\]

Calculating \(\sqrt{a^5}\):

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

Now, we focus on \(\sqrt{140}\). We can factor 140 to find its perfect squares:

\[
140 = 2 \cdot 70 = 2 \cdot 2 \cdot 35 = 2^2 \cdot 35
\]

Thus, we have:

\[
\sqrt{140} = \sqrt{2^2 \cdot 35} = \sqrt{2^2} \cdot \sqrt{35} = 2\sqrt{35}
\]

Combining everything, we have:

\[
\sqrt{140a^5} = \sqrt{140} \cdot \sqrt{a^5} = 2\sqrt{35} \cdot (a^2 \sqrt{a}) = 2a^2\sqrt{35a}
\]

Thus, the final simplified expression is:

\[
\boxed{2a^2\sqrt{35a}}
\]