Question

1. for ( x geq 0 ), the expression ( sqrt{32x^5} ) is equivalent to

(1) ( 4x^2sqrt{2x} )
(3) ( 8xsqrt{3x^2} )
(2) ( 4x^3sqrt{2} )
(4) ( 2xsqrt{8x^2} )

Explanation:

Step1: Factor the radicand

Factor \(32x^5\) into perfect squares and remaining factors: \(32 = 16\times2\) and \(x^5 = x^4\times x\). So, \(\sqrt{32x^5}=\sqrt{16\times2\times x^4\times x}\).

Step2: Apply square - root property

Using the property \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (\(a\geq0,b\geq0\)), we have \(\sqrt{16\times2\times x^4\times x}=\sqrt{16}\times\sqrt{x^4}\times\sqrt{2x}\).
Since \(\sqrt{16} = 4\) and \(\sqrt{x^4}=x^2\) (because \(x\geq0\)), then \(\sqrt{16}\times\sqrt{x^4}\times\sqrt{2x}=4x^2\sqrt{2x}\).

Answer:

(1) \(4x^{2}\sqrt{2x}\)