To simplify the radical \( \sqrt{28x^4y^3} \), let's break it down step by step:
- Factor the constant and variables inside the radical:
- \( 28 = 4 \times 7 \)
- \( x^4 \) is already a perfect square \( (x^2)^2 \)
- \( y^3 = y^2 \times y \), where \( y^2 \) is also a perfect square.
So we can rewrite the expression as: \[ \sqrt{28x^4y^3} = \sqrt{4 \times 7 \times (x^2)^2 \times (y^2) \times y} \]
- Apply the square root to each factor: \[ \sqrt{4} = 2, \quad \sqrt{(x^2)^2} = x^2, \quad \sqrt{y^2} = y \]
Putting it all together gives: \[ \sqrt{28x^4y^3} = 2x^2y\sqrt{7y} \]
Thus, the simplified form of \( \sqrt{28x^4y^3} \) is: \[ 2x^2y\sqrt{7y} \]
This matches the format of the choices you provided and it can be expressed as: 2xy√7y
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