Whatis the simplest tom of the radical expression \( \sqrt{48 x^{4} y^{3}} \) ? Assumex andy are oositive. \( .4 x^{2} y^{2} \sqrt{3 y} \) \( \circ 4 x^{2} y \sqrt{y} \) \( \circ 4 x^{2} y \sqrt{3 y} \) \( -4 x^{2} y \sqrt{x y} \)

To simplify the radical expression \( \sqrt{48 x^{4} y^{3}} \), we start by breaking it down. We can factor out the components under the square root: \( 48 = 16 \times 3 \), \( x^{4} = (x^{2})^{2} \), and \( y^{3} = y^{2} \times y \). Hence, we have: \[ \sqrt{48 x^{4} y^{3}} = \sqrt{16 \times 3 \times (x^{2})^{2} \times y^{2} \times y} \] Taking the square root of each part, we get: \[ \sqrt{16} = 4, \quad \sqrt{(x^{2})^{2}} = x^{2}, \quad \text{and} \quad \sqrt{y^{2}} = y \] The remaining part under the square root is \( \sqrt{3y} \): Combining these results, we arrive at the simplified expression: \[ 4 x^{2} y \sqrt{3y} \] So, the final answer is \( 4 x^{2} y \sqrt{3 y} \). --- Did you know that the simplification of radicals dates back thousands of years? Ancient Babylonians were already using methods for calculating square roots around 1800 BC! They divided numbers and approximated roots using a process that laid the groundwork for modern algebra. If you're keen to learn more about radicals, check out some classic algebra textbooks or even online platforms focused on mathematics education. Websites like Khan Academy offer great tutorials on simplifying radicals, complete with practice problems! And don’t forget YouTube channels like 3Blue1Brown for visual explanations that turn complex math into something fun to understand!