Question

Simplify the radical 128x4y9−−−−−−−√
.(1 point)
Responses

14x3y62xy3−−−−√
14 x cubed y to the 6th power square root of 2 x y cubed end root

8x3y62xy3−−−−√
8 x cubed y to the 6th power square root of 2 x y cubed end root

14x2y46y−−√
14 x squared y to the 4th power square root of 6 y

8x2y42y−−√

Answer 1

To simplify the radical \(\sqrt{128x^4y^9}\), we can break it down as follows:

1. **Factor the number and use properties of square roots**:
- \(128\) can be factored into \(64 \cdot 2\) where \(64\) is a perfect square (\(8^2\)).
- The \(x^4\) can be written as \((x^2)^2\), which is also a perfect square.
- The \(y^9\) can be written as \((y^4)^2 \cdot y\), since \(y^8\) is a perfect square.

2. **Apply the square root property**:
\[
\sqrt{128x^4y^9} = \sqrt{64 \cdot 2} \cdot \sqrt{x^4} \cdot \sqrt{y^9}
\]

3. **Calculate the square roots**:
- \(\sqrt{64} = 8\)
- \(\sqrt{x^4} = x^2\)
- \(\sqrt{y^9} = y^4 \cdot \sqrt{y}\) (because \(\sqrt{y^8} = y^4\))

4. **Combine the results**:
\[
\sqrt{128x^4y^9} = 8 x^2 y^4 \sqrt{2y}
\]

Thus, the simplified form is:
\[
8x^2y^4\sqrt{2y}
\]

So, the correct response is:
**8 x squared y to the 4th power square root of 2 y.**