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:
\[ \boxed{2a^2\sqrt{35a}} \]
1 answer