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# Multiplying and Dividing Radical Expressions

### Learning Outcomes

• Use properties of exponents to multiply radical expressions
• Use properties of exponents to divide radical expressions
Multiply and Divide
You can do more than just simplify radical expressions. You can multiply and divide them, too. Multiplying radicals is very simple if the index on all the radicals match. The product raised to a power rule that we discussed previously will help us find products of radical expressions. Recall the rule:

### A Product Raised to a Power Rule

For any numbers a and b and any integer x: ${{(ab)}^{x}}={{a}^{x}}\cdot {{b}^{x}}$ For any numbers a and b and any positive integer x: ${{(ab)}^{\frac{1}{x}}}={{a}^{\frac{1}{x}}}\cdot {{b}^{\frac{1}{x}}}$ For any numbers a and b and any positive integer x: $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$
The Product Raised to a Power Rule is important because you can use it to multiply radical expressions. Note that the roots are the same—you can combine square roots with square roots, or cube roots with cube roots, for example. But you can’t multiply a square root and a cube root using this rule.  The indices of the radicals must match in order to multiply them. In the following example, we will work with integers, and then we will move on to expressions with variable radicands.

### Example

Simplify. $\sqrt{18}\cdot \sqrt{16}$

Answer: Use the rule $\sqrt[x]{a}\cdot \sqrt[x]{b}=\sqrt[x]{ab}$ to multiply the radicands.

$\begin{array}{r}\sqrt{18\cdot 16}\\\\\sqrt{288}\end{array}$

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

$\sqrt{144\cdot 2}$

Identify perfect squares.

$\sqrt{{{(12)}^{2}}\cdot 2}$

Rewrite as the product of two radicals.

$\sqrt{{{(12)}^{2}}}\cdot \sqrt{2}$

Simplify, using $\sqrt{{{x}^{2}}}=\left| x \right|$.

$\begin{array}{r}\left| 12 \right|\cdot \sqrt{2}\\12\cdot \sqrt{2}\end{array}$

[latex-display] \sqrt{18}\cdot \sqrt{16}=12\sqrt{2}[/latex-display]

Using the Product Raised to a Power Rule, you can take a seemingly complicated expression, $\sqrt{18}\cdot \sqrt{16}$, and turn it into something more manageable, $12\sqrt{2}$. You may have also noticed that both $\sqrt{18}$ and $\sqrt{16}$ can be written as products involving perfect square factors. How would the expression change if you simplified each radical first, before multiplying?  In the next example, we will use the same product from above to show that you can simplify before multiplying and get the same result.

### Example

Simplify $\sqrt{18}\cdot \sqrt{16}$

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

$\begin{array}{r}\sqrt{9\cdot 2}\cdot \sqrt{4\cdot 4}\\\\\sqrt{3\cdot 3\cdot 2}\cdot \sqrt{4\cdot 4}\end{array}$

Identify perfect squares.

$\sqrt{{{(3)}^{2}}\cdot 2}\cdot \sqrt{{{(4)}^{2}}}$

Rewrite as the product of radicals.

$\sqrt{{{(3)}^{2}}}\cdot \sqrt{2}\cdot \sqrt{{{(4)}^{2}}}$

Simplify, using $\sqrt{{{x}^{2}}}=\left| x \right|$.

$\begin{array}{c}\left|3\right|\cdot\sqrt{2}\cdot\left|4\right|\\3\cdot\sqrt{2}\cdot4\end{array}$

Multiply.

$12\sqrt{2}$

[latex-display] \sqrt{18}\cdot \sqrt{16}=12\sqrt{2}[/latex-display]

In both cases, you arrive at the same product, $12\sqrt{2}$. It does not matter whether you multiply the radicands or simplify each radical first.

### Try It

[ohm_question]18625[/ohm_question]
You multiply radical expressions that contain variables in the same manner. As long as the roots of the radical expressions are the same, you can use the Product Raised to a Power Rule to multiply and simplify. Look at the two examples that follow. In both problems, the Product Raised to a Power Rule is used right away and then the expression is simplified.  Note that we specify that the variable is non-negative, $x\ge 0$, thus allowing us to avoid the need for absolute value.

### Example

Simplify. $\sqrt{12{{x}^{3}}}\cdot \sqrt{3x}$, $x\ge 0$

Answer: Use the rule $\sqrt[x]{a}\cdot \sqrt[x]{b}=\sqrt[x]{ab}$ to multiply the radicands.

$\sqrt{12{{x}^{3}}\cdot 3x}\\\\\sqrt{12\cdot 3\cdot {{x}^{3}}\cdot x}$

Recall that ${{x}^{3}}\cdot x={{x}^{3+1}}$.

$\begin{array}{r}\sqrt{36\cdot {{x}^{3+1}}}\\\\\sqrt{36\cdot {{x}^{4}}}\end{array}$

Look for perfect squares in the radicand.

$\sqrt{{{(6)}^{2}}\cdot {{({{x}^{2}})}^{2}}}$

Rewrite as the product of the result.

$\begin{array}{c}\sqrt{{{(6)}^{2}}}\cdot \sqrt{{{({{x}^{2}})}^{2}}}\\6\cdot {{x}^{2}}\end{array}$

[latex-display] \sqrt{12{{x}^{3}}}\cdot \sqrt{3x}=6{{x}^{2}}[/latex-display]

Note that even though our answer contained a variable with an odd exponent that was simplified from an even indexed root, we don't need to write our answer with absolute value because we specified before we simplified that $x\ge 0$. It is important to read the problem very well when you are doing math.  Even the smallest statement like $x\ge 0$ can influence the way you write your answer. In this video example, we multiply more square roots with and without variables. https://youtu.be/PQs10_rFrSM Now, let's look at multiplying radical expressions that are cube roots.

### Example

Multiply $2\sqrt[3]{18}\cdot-7\sqrt[3]{15}$

Answer: Multiply the factors outside the radicals, and factor the radicands. [latex-display]-14\sqrt[3]{2\cdot3^2}\sqrt[3]{3\cdot5}[/latex-display] Combine the radicands into one radical, and reorganize to see if there are any cubes. [latex-display]-14\sqrt[3]{2\cdot3^2\cdot3\cdot5}=-14\sqrt[3]{2\cdot3^3\cdot5}[/latex-display] Apply the cube root to $3^3$, and simplify the radicand. [latex-display]-14\cdot3\sqrt[3]{2\cdot5}=-42\sqrt[3]{10}[/latex-display]

[latex-display]2\sqrt[3]{18}\cdot-7\sqrt[3]{15}=-42\sqrt[3]{10}[/latex-display]

We will show two more examples of multiplying cube root radicals, this time we will include variables.

### Example

Multiply $\sqrt[3]{4x^3}\cdot\sqrt[3]{2x^2}$

Answer: Factor the radicands, keeping in mind you want to find cubes. [latex-display]\sqrt[3]{4x^3}\cdot\sqrt[3]{2x^2}=\sqrt[3]{2^2\cdot{x^3}}\cdot\sqrt[3]{2x^2}[/latex-display] Combine the radicands into one radical, and reorganize into cubes where possible. [latex-display]\begin{array}{c}\sqrt[3]{2^2\cdot{x^3}}\cdot\sqrt[3]{2x^2}\\\\=\sqrt[3]{2^2\cdot{x^3}\cdot2\cdot{x^2}}\\\\=\sqrt[3]{2^3\cdot{x^3}\cdot{x^2}}\end{array}[/latex-display] Apply the cube root to $2^3$, and $x^3$ and simplify the radicand. [latex-display]\sqrt[3]{2^3\cdot{x^3}\cdot{x^2}}=2\cdot{x}\sqrt[3]{x^2}[/latex-display]

[latex-display]\sqrt[3]{4x^3}\cdot\sqrt[3]{2x^2}=2\cdot{x}\sqrt[3]{x^2}[/latex-display]

### Example

Simplify. $\sqrt[3]{{{x}^{5}}{{y}^{2}}}\cdot 5\sqrt[3]{8{{x}^{2}}{{y}^{4}}}$

Answer: Notice that both radicals are cube roots, so you can use the rule  to multiply the radicands.

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

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

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

Rewrite as the product of radicals.

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

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

In the next video, we present more examples of multiplying cube roots. https://youtu.be/cxRXofdelIM This final example of multiplying radical expressions is slightly more complicated because there are more than two radicals being multiplied. In this case, notice how the radicals are simplified before multiplication takes place. Remember that the order you choose to use is up to you—you will find that sometimes it is easier to multiply before simplifying, and other times it is easier to simplify before multiplying. With some practice, you may be able to tell which is easier before you approach the problem, but either order will work for all problems.

### Example

Simplify. $2\sqrt[4]{16{{x}^{9}}}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{81{{x}^{3}}y}$, $x\ge 0$, $y\ge 0$

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

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

Rewrite as the product of radicals.

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

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

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

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

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

Now that the radicands have been multiplied, look again for powers of $4$, 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 $x\ge 0$, $y\ge 0$.

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

The answer is $12{{x}^{3}}y,\,\,x\ge 0,\,\,y\ge 0$.

You can use the same ideas to help you figure out how to simplify and divide radical expressions. Recall that the Product Raised to a Power Rule states that $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$. Well, what if you are dealing with a quotient instead of a product? There is a rule for that, too. The Quotient Raised to a Power Rule states that ${{\left( \frac{a}{b} \right)}^{x}}=\frac{{{a}^{x}}}{{{b}^{x}}}$. Again, if you imagine that the exponent is a rational number, then you can make this rule applicable for roots as well:

${{\left( \frac{a}{b} \right)}^{\frac{1}{x}}}=\frac{{{a}^{\frac{1}{x}}}}{{{b}^{\frac{1}{x}}}}$

Therefore

$\sqrt[x]{\frac{a}{b}}=\frac{\sqrt[x]{a}}{\sqrt[x]{b}}$.

### A Quotient Raised to a Power Rule

For any real numbers a and b (b ≠ 0) and any positive integer x: ${{\left( \frac{a}{b} \right)}^{\frac{1}{x}}}=\frac{{{a}^{\frac{1}{x}}}}{{{b}^{\frac{1}{x}}}}$ For any real numbers a and b (b ≠ 0) and any positive integer x: $\sqrt[x]{\frac{a}{b}}=\frac{\sqrt[x]{a}}{\sqrt[x]{b}}$
As you did with multiplication, you will start with some examples featuring integers before moving on to radicals with variables and more complex expressions like $\frac{\sqrt[3]{24x{{y}^{4}}}}{\sqrt[3]{8y}}$.

### Example

Simplify. $\sqrt{\frac{48}{25}}$

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

$\frac{\sqrt{48}}{\sqrt{25}}$

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

$\begin{array}{c}\frac{\sqrt{16\cdot 3}}{\sqrt{25}}\\\\\text{or}\\\\\frac{\sqrt{4\cdot 4\cdot 3}}{\sqrt{5\cdot 5}}\end{array}$

Identify and pull out perfect squares.

$\begin{array}{r}\frac{\sqrt{{{(4)}^{2}}\cdot 3}}{\sqrt{{{(5)}^{2}}}}\\\\\frac{\sqrt{{{(4)}^{2}}}\cdot \sqrt{3}}{\sqrt{{{(5)}^{2}}}}\end{array}$

Simplify.

$\frac{4\cdot \sqrt{3}}{5}$

[latex-display] \sqrt{\frac{48}{25}}=\frac{4\sqrt{3}}{5}[/latex-display]

### Example

Simplify. $\sqrt[3]{\frac{640}{40}}$

Answer: Rewrite using the Quotient Raised to a Power Rule.

$\frac{\sqrt[3]{640}}{\sqrt[3]{40}}$

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

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

Identify and pull out perfect cubes.

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

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

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

Rewrite the numerator as a product of factors.

$\frac{2\cdot 2\sqrt[3]{5}\cdot \sqrt[3]{2}}{2\sqrt[3]{5}}$

Identify factors of $1$, and simplify.

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

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

That was a lot of effort, but you were able to simplify using the Quotient Raised to a Power Rule. What if you found the quotient of this expression by dividing within the radical first and then took the cube root of the quotient? Now take another look at that problem using this approach.

### Example

Simplify. $\frac{\sqrt[3]{640}}{\sqrt[3]{40}}$

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

$\sqrt[3]{\frac{640}{40}}$

Within the radical, divide $640$ by $40$.

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

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

$\sqrt[3]{8\cdot2}$

Identify perfect cubes and pull them out.

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

Simplify.

$2\cdot\sqrt[3]{2}$

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

Notice how much more straightforward the approach was. As with multiplication, the main idea here is that sometimes it makes sense to divide and then simplify, and other times it makes sense to simplify and then divide. Whichever order you choose, though, you should arrive at the same final expression. In the next video, we show more examples of simplifying a radical that contains a quotient. https://youtu.be/SxImTm9GVNo Now let’s turn to some radical expressions containing variables. Notice that the process for dividing these is the same as it is for dividing integers.

### Example

Simplify. $\frac{\sqrt{30x}}{\sqrt{10x}},x>0$

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

$\sqrt{\frac{30x}{10x}}$

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

$\begin{array}{r}\sqrt{\frac{3\cdot10x}{10x}}\\\\\sqrt{3\cdot\frac{10x}{10x}}\\\\\sqrt{3\cdot1}\end{array}$

[latex-display] \frac{\sqrt{30x}}{\sqrt{10x}}=\sqrt{3}[/latex-display]

https://youtu.be/QwUsRWCNt24

As you become more familiar with dividing and simplifying radical expressions, make sure you continue to pay attention to the roots of the radicals that you are dividing. For example, you can think of this expression:

$\frac{\sqrt{8{{y}^{2}}}}{\sqrt{225{{y}^{4}}}}$

As equivalent to:

$\sqrt{\frac{8{{y}^{2}}}{225{{y}^{4}}}}$

This is because both the numerator and the denominator are square roots.

Notice that you cannot express this expression:

$\frac{\sqrt{8{{y}^{2}}}}{\sqrt[4]{225{{y}^{4}}}}$

In this format:

$\sqrt[4]{\frac{8{{y}^{2}}}{225{{y}^{4}}}}$.

This is because the numerator is a square root and the denominator is a fourth root. In this last video, we show more examples of simplifying a quotient with radicals.

### Example

Simplify. $\frac{\sqrt[3]{24x{{y}^{4}}}}{\sqrt[3]{8y}},\,\,y\ne 0$

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

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

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

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

Identify perfect cubes and pull them out of the radical.

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

Simplify.

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

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

### Try It

[ohm_question]91537[/ohm_question]
In our last video, we show more examples of simplifying radicals that contain quotients with variables. https://youtu.be/04X-hMgb0tA

## Summary

The Product Raised to a Power Rule and the Quotient Raised to a Power Rule can be used to simplify radical expressions as long as the roots of the radicals are the same. The Product Rule states that the product of two or more numbers raised to a power is equal to the product of each number raised to the same power. The same is true of roots: $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$. When dividing radical expressions, the rules governing quotients are similar: $\sqrt[x]{\frac{a}{b}}=\frac{\sqrt[x]{a}}{\sqrt[x]{b}}$.

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