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Question
simplify. 9\sqrt{56x^{7}y^{12}} assume x and y are nonnegative. 18x^{3}y^{6}\sqrt{7x} 18x^{3}y^{6}\sqrt{14x} 18x^{3}y^{6}\sqrt{14} 2x^{3}y^{6}\sqrt{14x}
Step1: Factorize the number inside the square - root
Factorize 56 as \(56 = 2^3\times7\), so \(9\sqrt{56x^{7}y^{12}}=9\sqrt{2^3\times7\times x^{7}y^{12}}\).
Step2: Use the property of square - roots \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\)
\(9\sqrt{2^3\times7\times x^{7}y^{12}} = 9\sqrt{2^2\times2\times7\times x^{6}\times x\times y^{12}}\).
Step3: Take out the perfect - square factors from the square - root
We know that \(\sqrt{2^2}=2\), \(\sqrt{x^{6}} = x^{3}\), \(\sqrt{y^{12}}=y^{6}\). So \(9\sqrt{2^2\times2\times7\times x^{6}\times x\times y^{12}}=9\times2\times x^{3}\times y^{6}\sqrt{2\times7\times x}\).
Step4: Simplify the coefficient
\(9\times2 = 18\), and \(9\times2\times x^{3}\times y^{6}\sqrt{2\times7\times x}=18x^{3}y^{6}\sqrt{14x}\).
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\(18x^{3}y^{6}\sqrt{14x}\) (corresponding to the second option)