Use the rule [latex] \sqrt[x]{a}\cdot \sqrt[x]{b}=\sqrt[x]{ab}[/latex] to multiply the radicands.

[latex]\begin{array}{r}\sqrt{18\cdot 16}\\\sqrt{288}\end{array}[/latex]

Look for perfect squares in the radicand, and rewrite the radicand as the product of two factors.

[latex] \sqrt{144\cdot 2}[/latex]

Identify perfect squares.

[latex] \sqrt{{{(12)}^{2}}\cdot 2}[/latex]

Rewrite as the product of two radicals.

[latex] \sqrt{{{(12)}^{2}}}\cdot \sqrt{2}[/latex]

Simplify, using [latex] \sqrt{{{x}^{2}}}=\left| x \right|[/latex].

[latex]\begin{array}{r}\left| 12 \right|\cdot \sqrt{2}\\12\cdot \sqrt{2}\end{array}[/latex]

The answer is [latex]12\sqrt{2}[/latex].

Look for perfect squares in each radicand, and rewrite as the product of two factors.

[latex] \begin{array}{r}\sqrt{9\cdot 2}\cdot \sqrt{4\cdot 4}\\\sqrt{3\cdot 3\cdot 2}\cdot \sqrt{4\cdot 4}\end{array}[/latex]

Identify perfect squares.

[latex] \sqrt{{{(3)}^{2}}\cdot 2}\cdot \sqrt{{{(4)}^{2}}}[/latex]

Rewrite as the product of radicals.

[latex] \sqrt{{{(3)}^{2}}}\cdot \sqrt{2}\cdot \sqrt{{{(4)}^{2}}}[/latex]

Simplify, using [latex] \sqrt{{{x}^{2}}}=\left| x \right|[/latex].

[latex]\begin{array}{c}\left|3\right|\cdot\sqrt{2}\cdot\left|4\right|\\3\cdot\sqrt{2}\cdot4\end{array}[/latex]

Multiply.

[latex]12\sqrt{2}[/latex]

Use the rule [latex] \sqrt[x]{a}\cdot \sqrt[x]{b}=\sqrt[x]{ab}[/latex] to multiply the radicands.

[latex] \sqrt{12{{x}^{4}}\cdot 3x^2}\\\sqrt{12\cdot 3\cdot {{x}^{4}}\cdot x^2}[/latex]

Recall that [latex] {{x}^{4}}\cdot x^2={{x}^{4+2}}[/latex].

[latex]\begin{array}{r}\sqrt{36\cdot {{x}^{4+2}}}\\\sqrt{36\cdot {{x}^{6}}}\end{array}[/latex]

Look for perfect squares in the radicand.

[latex] \sqrt{{{(6)}^{2}}\cdot {{({{x}^{3}})}^{2}}}[/latex]

Rewrite as the product of radicals.

[latex] \begin{array}{c}\sqrt{{{(6)}^{2}}}\cdot \sqrt{{{({{x}^{3}})}^{2}}}\\6\cdot {{x}^{3}}\end{array}[/latex]

The answer is [latex]6{{x}^{3}}[/latex].

Notice that both radicals are cube roots, so you can use the rule [latex] [/latex] to multiply the radicands.

[latex]\begin{array}{l}5\sqrt[3]{{{x}^{5}}{{y}^{2}}\cdot 8{{x}^{2}}{{y}^{4}}}\\5\sqrt[3]{8\cdot {{x}^{5}}\cdot {{x}^{2}}\cdot {{y}^{2}}\cdot {{y}^{4}}}\\5\sqrt[3]{8\cdot {{x}^{5+2}}\cdot {{y}^{2+4}}}\\5\sqrt[3]{8\cdot {{x}^{7}}\cdot {{y}^{6}}}\end{array}[/latex]

Look for perfect cubes in the radicand. Since [latex] {{x}^{7}}[/latex] is not a perfect cube, it has to be rewritten as [latex] {{x}^{6+1}}={{({{x}^{2}})}^{3}}\cdot x[/latex].

[latex] 5\sqrt[3]{{{(2)}^{3}}\cdot {{({{x}^{2}})}^{3}}\cdot x\cdot {{({{y}^{2}})}^{3}}}[/latex]

Rewrite as the product of radicals.

[latex] \begin{array}{r}5\sqrt[3]{{{(2)}^{3}}}\cdot \sqrt[3]{{{({{x}^{2}})}^{3}}}\cdot \sqrt[3]{{{({{y}^{2}})}^{3}}}\cdot \sqrt[3]{x}\\5\cdot 2\cdot {{x}^{2}}\cdot {{y}^{2}}\cdot \sqrt[3]{x}\end{array}[/latex]

The answer is [latex]10{{x}^{2}}{{y}^{2}}\sqrt[3]{x}[/latex].

Notice this expression is multiplying three radicals with the same (fourth) root. Simplify each radical, if possible, before multiplying. Be looking for powers of [latex]4[/latex] in each radicand.

[latex] 2\sqrt[4]{{{(2)}^{4}}\cdot {{({{x}^{2}})}^{4}}\cdot x}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{{{(3)}^{4}}\cdot {{x}^{3}}y}[/latex]

Rewrite as the product of radicals.

[latex] 2\sqrt[4]{{{(2)}^{4}}}\cdot \sqrt[4]{{{({{x}^{2}})}^{4}}}\cdot \sqrt[4]{x}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{{{(3)}^{4}}}\cdot \sqrt[4]{{{x}^{3}}y}[/latex]

Identify and pull out powers of [latex]4[/latex], using the fact that [latex] \sqrt[4]{{{x}^{4}}}=\left| x \right|[/latex].

[latex] \begin{array}{r}2\cdot \left| 2 \right|\cdot \left| {{x}^{2}} \right|\cdot \sqrt[4]{x}\cdot \sqrt[4]{{{y}^{3}}}\cdot \left| 3 \right|\cdot \sqrt[4]{{{x}^{3}}y}\\2\cdot 2\cdot {{x}^{2}}\cdot \sqrt[4]{x}\cdot \sqrt[4]{{{y}^{3}}}\cdot 3\cdot \sqrt[4]{{{x}^{3}}y}\end{array}[/latex]

Since all the radicals are fourth roots, you can use the rule [latex] \sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}[/latex] to multiply the radicands.

[latex]\begin{array}{r}2\cdot 2\cdot 3\cdot {{x}^{2}}\cdot \sqrt[4]{x\cdot {{y}^{3}}\cdot {{x}^{3}}y}\\12{{x}^{2}}\sqrt[4]{{{x}^{1+3}}\cdot {{y}^{3+1}}}\end{array}[/latex]

Now that the radicands have been multiplied, look again for powers of [latex]4[/latex], and pull them out. We can drop the absolute value signs in our final answer because at the start of the problem we were told [latex] x\ge 0[/latex], [latex] y\ge 0[/latex].

[latex] \begin{array}{l}12{{x}^{2}}\sqrt[4]{{{x}^{4}}\cdot {{y}^{4}}}\\12{{x}^{2}}\sqrt[4]{{{x}^{4}}}\cdot \sqrt[4]{{{y}^{4}}}\\12{{x}^{2}}\cdot \left| x \right|\cdot \left| y \right|\end{array}[/latex]

The answer is [latex]12{{x}^{3}}y,\,\,x\ge 0,\,\,y\ge 0[/latex].

Use the rule [latex] \sqrt[x]{\frac{a}{b}}=\frac{\sqrt[x]{a}}{\sqrt[x]{b}}[/latex] to create two radicals; one in the numerator and one in the denominator.

[latex] \frac{\sqrt{48}}{\sqrt{25}}[/latex]

Simplify each radical. Look for perfect square factors in the radicand, and rewrite the radicand as a product of factors.

[latex] \begin{array}{c}\frac{\sqrt{16\cdot 3}}{\sqrt{25}}\\\\\text{or}\\\\\frac{\sqrt{4\cdot 4\cdot 3}}{\sqrt{5\cdot 5}}\end{array}[/latex]

Identify and pull out perfect squares.

[latex] \begin{array}{r}\frac{\sqrt{{{(4)}^{2}}\cdot 3}}{\sqrt{{{(5)}^{2}}}}\\\\\frac{\sqrt{{{(4)}^{2}}}\cdot \sqrt{3}}{\sqrt{{{(5)}^{2}}}}\end{array}[/latex]

Simplify.

[latex] \frac{4\cdot \sqrt{3}}{5}[/latex]

The answer is [latex]\frac{4\sqrt{3}}{5}[/latex].

Rewrite using the Quotient Raised to a Power Rule.

[latex] \frac{\sqrt[3]{640}}{\sqrt[3]{40}}[/latex]

Simplify each radical. Look for perfect cubes in the radicand, and rewrite the radicand as a product of factors.

[latex] \frac{\sqrt[3]{64\cdot 10}}{\sqrt[3]{8\cdot 5}}[/latex]

Identify and pull out perfect cubes.

[latex] \begin{array}{r}\frac{\sqrt[3]{{{(4)}^{3}}\cdot 10}}{\sqrt[3]{{{(2)}^{3}}\cdot 5}}\\\\\frac{\sqrt[3]{{{(4)}^{3}}}\cdot \sqrt[3]{10}}{\sqrt[3]{{{(2)}^{3}}}\cdot \sqrt[3]{5}}\\\\\frac{4\cdot \sqrt[3]{10}}{2\cdot \sqrt[3]{5}}\end{array}[/latex]

You can simplify this expression even further by looking for common factors in the numerator and denominator.

[latex] \frac{4\sqrt[3]{10}}{2\sqrt[3]{5}}[/latex]

Rewrite the numerator as a product of factors.

[latex] \frac{2\cdot 2\sqrt[3]{5}\cdot \sqrt[3]{2}}{2\sqrt[3]{5}}[/latex]

Identify factors of [latex]1[/latex], and simplify.

[latex] \begin{array}{r}2\cdot \frac{2\sqrt[3]{5}}{2\sqrt[3]{5}}\cdot \sqrt[3]{2}\\\\2\cdot 1\cdot \sqrt[3]{2}\end{array}[/latex]

The answer is [latex]2\sqrt[3]{2}[/latex].

Since both radicals are cube roots, you can use the rule [latex] \frac{\sqrt[x]{a}}{\sqrt[x]{b}}=\sqrt[x]{\frac{a}{b}}[/latex] to create a single rational expression underneath the radical.

[latex] \sqrt[3]{\frac{640}{40}}[/latex]

Within the radical, divide [latex]640[/latex] by [latex]40[/latex].

[latex] \begin{array}{r}640\div 40=16\\\sqrt[3]{16}\end{array}[/latex]

Look for perfect cubes in the radicand, and rewrite the radicand as a product of factors.

[latex]\sqrt[3]{8\cdot2}[/latex]

Identify perfect cubes and pull them out.

[latex] \begin{array}{r}\sqrt[3]{{{(2)}^{3}}\cdot 2}\\\sqrt[3]{{(2)}^{3}}\cdot\sqrt[3]{2}\end{array}[/latex]

Simplify.

[latex]2\cdot\sqrt[3]{2}[/latex]

The answer is [latex]2\sqrt[3]{2}[/latex].

Use the Quotient Raised to a Power Rule to rewrite this expression.

[latex]\sqrt{\frac{30x}{10x}}[/latex]

Simplify [latex] \sqrt{\frac{30x}{10x}}[/latex] by identifying similar factors in the numerator and denominator and then identifying factors of [latex]1[/latex].

[latex]\begin{array}{r}\sqrt{\frac{3\cdot10x}{10x}}\\\\\sqrt{3\cdot\frac{10x}{10x}}\\\\\sqrt{3\cdot1}\end{array}[/latex]

The answer is [latex]\sqrt{3}[/latex].

Use the Quotient Raised to a Power Rule to rewrite this expression.

[latex] \sqrt[3]{\frac{24x{{y}^{4}}}{8y}}[/latex]

Simplify [latex] \sqrt[3]{\frac{24x{{y}^{4}}}{8y}}[/latex] by identifying similar factors in the numerator and denominator and then identifying factors of [latex]1[/latex].

[latex]\begin{array}{l}\sqrt[3]{\frac{8\cdot 3\cdot x\cdot {{y}^{3}}\cdot y}{8\cdot y}}\\\\\sqrt[3]{\frac{3\cdot x\cdot {{y}^{3}}}{1}\cdot \frac{8y}{8y}}\\\\\sqrt[3]{\frac{3\cdot x\cdot {{y}^{3}}}{1}\cdot 1}\end{array}[/latex]

Identify perfect cubes and pull them out of the radical.

[latex] \sqrt[3]{3x{{y}^{3}}}\\\sqrt[3]{{{(y)}^{3}}\cdot \,3x}[/latex]

Simplify.

[latex] \sqrt[3]{{{(y)}^{3}}}\cdot \,\sqrt[3]{3x}[/latex]

The answer is [latex]y\,\sqrt[3]{3x}[/latex].