Question

simplify.
\sqrt{32x^{5}w^{2}}
assume that all variables represent positive real numbers.

Explanation:

Step1: Factor the radicand

Factor \(32x^{5}w^{2}\) into perfect - square factors and non - perfect - square factors.
We know that \(32 = 16\times2\), \(x^{5}=x^{4}\times x\), and \(w^{2}\) is already a perfect square.
So, \(32x^{5}w^{2}=16\times2\times x^{4}\times x\times w^{2}\)

Step2: Apply the square - root property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (\(a\geq0,b\geq0\))

\(\sqrt{32x^{5}w^{2}}=\sqrt{16\times2\times x^{4}\times x\times w^{2}}=\sqrt{16}\times\sqrt{x^{4}}\times\sqrt{w^{2}}\times\sqrt{2x}\)

Step3: Simplify the square roots of perfect squares

We know that \(\sqrt{16} = 4\), \(\sqrt{x^{4}}=x^{2}\) (since \(x\) is a positive real number, \(\sqrt{x^{4}}=(x^{4})^{\frac{1}{2}}=x^{2}\)), and \(\sqrt{w^{2}} = w\) (since \(w\) is a positive real number, \(\sqrt{w^{2}}=(w^{2})^{\frac{1}{2}}=w\))

So, \(\sqrt{16}\times\sqrt{x^{4}}\times\sqrt{w^{2}}\times\sqrt{2x}=4\times x^{2}\times w\times\sqrt{2x}\)

Answer:

\(4x^{2}w\sqrt{2x}\)