Question

22 what is the simplest product? \\(\sqrt{50x^7y^7} \cdot \sqrt{6xy^4}\\) \\(\boldsymbol{\text{a}}\\) \\(5x^4y^6\sqrt{12}\\) \\(\boldsymbol{\text{b}}\\) \\(2x^4y^6\sqrt{75y}\\) \\(\boldsymbol{\text{c}}\\) \\(30x^4y^5\sqrt{y}\\) \\(\boldsymbol{\text{d}}\\) \\(10x^4y^5\sqrt{3y}\\)

Explanation:

Step1: Multiply the radicands

Use the property \(\sqrt{a} \cdot \sqrt{b}=\sqrt{ab}\). So, \(\sqrt{50x^{7}y^{7}} \cdot \sqrt{6xy^{4}}=\sqrt{(50x^{7}y^{7})(6xy^{4})}\). Calculate the product inside the radical: \(50\times6 = 300\), \(x^{7}\times x=x^{7 + 1}=x^{8}\), \(y^{7}\times y^{4}=y^{7+4}=y^{11}\). So we get \(\sqrt{300x^{8}y^{11}}\).

Step2: Simplify the radical

Factor the radicand into perfect squares and remaining factors. \(300 = 100\times3\), \(x^{8}=(x^{4})^{2}\), \(y^{11}=y^{10}\times y=(y^{5})^{2}\times y\). So \(\sqrt{300x^{8}y^{11}}=\sqrt{100\times3\times(x^{4})^{2}\times(y^{5})^{2}\times y}\). Use the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (for non - negative \(a,b\)): \(\sqrt{100}\times\sqrt{3}\times\sqrt{(x^{4})^{2}}\times\sqrt{(y^{5})^{2}}\times\sqrt{y}\). We know that \(\sqrt{100} = 10\), \(\sqrt{(x^{4})^{2}}=x^{4}\), \(\sqrt{(y^{5})^{2}}=y^{5}\). So this simplifies to \(10x^{4}y^{5}\sqrt{3y}\).

Answer:

D. \(10x^{4}y^{5}\sqrt{3y}\)