Question

-6\sqrt4{32x^5}

Explanation:

Step1: Factor the radicand

Factor \(32\) and \(x^5\) to find perfect fourth - powers. We know that \(32 = 16\times2=2^{4}\times2\) and \(x^{5}=x^{4}\times x\). So, \(\sqrt[4]{32x^{5}}=\sqrt[4]{2^{4}\times2\times x^{4}\times x}\).

Step2: Use the property of radicals \(\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\) (\(a\geq0,b\geq0\))

\(\sqrt[4]{2^{4}\times2\times x^{4}\times x}=\sqrt[4]{2^{4}}\cdot\sqrt[4]{x^{4}}\cdot\sqrt[4]{2x}\). By the property \(\sqrt[n]{a^{n}} = a\) (for \(n\) even and \(a\geq0\)), we have \(\sqrt[4]{2^{4}} = 2\) and \(\sqrt[4]{x^{4}}=x\) (assuming \(x\geq0\)). So \(\sqrt[4]{2^{4}}\cdot\sqrt[4]{x^{4}}\cdot\sqrt[4]{2x}=2x\sqrt[4]{2x}\).

Step3: Multiply by the coefficient outside the radical

The original expression is \(- 6\sqrt[4]{32x^{5}}\), substituting the simplified radical we found above, we get \(-6\times(2x\sqrt[4]{2x})\).

Step4: Calculate the product of the coefficients

\(-6\times2x=-12x\). So the simplified form of \(-6\sqrt[4]{32x^{5}}\) is \(-12x\sqrt[4]{2x}\).

Answer:

\(-12x\sqrt[4]{2x}\)