Precalculus (6th Edition)

$\color{blue}{2x^2z^4\sqrt{2x}}$
Factor the radicand so that at least one factor is a perfect square to obtain: $=\sqrt{4(2)(x^4)(x)(z^8)} \\=\sqrt{(4x^4z^8)(2x)} \\=\sqrt{2^2(x^2)^2(z^4)^2(2x)}$ Bring out the cube root of the perfect cube factors to obtain: $=2\cdot x^2\cdot z^4\sqrt{2x} \\=\color{blue}{2x^2z^4\sqrt{2x}}$