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e) \( 2 \sqrt{7} \cdot(5 \sqrt{2}-6 \sqrt{3}) \) f) \( \sqrt[3]{16 x^{5}} \cdot \sqrt[4]{625 x^{9}} \) g) \( \sqrt[2]{24 m^{8}} \cdot \sqrt[4]{80 m^{4}} \) h) \( \sqrt[5]{160 a^{8}} \cdot \sqrt[4]{81 a^{7} b^{9}} \)

Ask by Sanders Weston. in Colombia
Feb 25,2025

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e) \( 2 \sqrt{7} \cdot(5 \sqrt{2}-6 \sqrt{3}) = 10\sqrt{14} - 12\sqrt{21} \) f) \( \sqrt[3]{16 x^{5}} \cdot \sqrt[4]{625 x^{9}} = 10x^{3}\sqrt[12]{16x^{11}} \) g) \( \sqrt[2]{24 m^{8}} \cdot \sqrt[4]{80 m^{4}} = 4\sqrt[4]{180} \times m^{5} \) h) \( \sqrt[5]{160 a^{8}} \cdot \sqrt[4]{81 a^{7} b^{9}} = 6a^{3}b^{2}\sqrt[20]{625a^{7}b^{5}} \)

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The Deep Dive

When multiplying radicals, make sure to first simplify inside the radical whenever possible. For example, in expression f) \(16x^5\) becomes \(2^4x^5\) and when cubed-rooted gives you \(2x^{5/3}\). Remember that keeping track of roots and their exponents is crucial to preventing errors in simplification. Another fun fact: radicals are essentially the "opposites" of exponents! When you take the square root, you're looking for a number that, when multiplied by itself, gives you the original number. So, in g), when you break down \( \sqrt{24m^8} \) into its prime factors and simplify, it’s like piecing together an exciting puzzle as you uncover the simplest form!

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