5.3 Multiplying Radical Expressions

Learning Outcomes

Simplify radical expressions by factoring and using the Product Rule for radicals.
Multiply and simplify radical expressions by using the Product Rule for radicals.

How do we multiply radical expressions together? For example, consider

$\sqrt{2}\cdot \sqrt{18}$

It is tempting to just multiply the two numbers together, but is that correct? The key is to use what we learned in the previous section. Remember that since radicals are really equivalent to rational exponents, every exponent property gives us a corresponding radical property! First recall our Power of a Product rule for exponents, which works for any rational exponent $m$ as long as both $a^m$ and $b^m$ are real numbers:

$a^m \cdot b^m = (ab)^{m}$

Now, evaluate the above expression using an exponent of $\dfrac{1}{2}$ to represent the square roots:

$\begin{align} &\quad\sqrt{2}\cdot \sqrt{18} \\ =&\quad 2^{1/2}\cdot18^{1/2} \\ =&\quad(2\cdot18)^{1/2} && \color{blue}{\textsf{Power of a Product used here}}\\ =&\quad(36)^{1/2} \\ =&\quad \sqrt{36} \\ =&\quad 6 \end{align}$

So we have shown that you can indeed multiply radicals by multiplying the radicands. It was crucial in our process that the exponents were equal to be able to use the Power of a Product rule. Thus radical multiplication only works this way if the indices are the same.

PRODUCT RULE FOR RADICALS

For any numbers $a$ and $b$ and any positive integer $n$ , if both $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers, then

$\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{ab}$

The Product Rule for radicals is important because we can use it to multiply radical expressions. We will also use it later in reverse to simplify radicals. Note that the indices of the radicals must match in order to multiply them using this rule. The condition that $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers is necessary to avoid situations where $n$ is even and one of the radicands is negative, which would mean the root is not a real number.

Simplifying Radicals

If you can identify perfect square factors within a radical, as with $\sqrt{72} = \sqrt{8\cdot 9} = \sqrt{2 \cdot (2\cdot 2)\cdot (3\cdot 3)}=\sqrt{2\cdot 2^2 \cdot 3^2}$ , you can rewrite the expression as the product of multiple radicals, some of which can be evaluated:

$= \sqrt{2^2}\cdot\sqrt{3^2}\cdot\sqrt{2}=2\cdot 3\cdot\sqrt{2} = 6\sqrt{2}$ .

The final result of $6\sqrt{2}$ is called simplified. We can read it as “six square root of two.” All perfect square factors have been removed from the radical. It is standard to place the removed factor to the left of the radical so we don’t confuse whether it is under the radical symbol or not. A number appearing next to a radical sign implies multiplication, similar to how $3x = 3\cdot x$ .

SIMPLIFYING RADICALS

We say a radical $\sqrt[n]{a}$ is simplified if $a$ does not contain any perfect $n$ th powers as factors. To simplify, look for factors of $a$ that are perfect $n$ th powers and rewrite the radical using the power rule to place the perfect $n$ th power inside of its own radical.

It may be helpful to keep a list of perfect powers handy while completing exercises.
Perfect Squares: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, ...$
Perfect Cubes: $1, 8, 27, 64, 125, 216, ...$
Perfect $4$ th Powers: $1, 16, 81, 256, 625, ...$
Perfect $5$ th Powers: $1, 32, 243, ...$

Here are some examples to illustrate the process.

Example

Simplify each expression.

$\sqrt{63}$
$10\sqrt{96}$
$\sqrt[3]{-108}$

Show Solution

$63$ is not a perfect square so we can use the Product Rule for Radicals to simplify any factors that are perfect squares.
Factor $63$ into $7$ and $9,$ then break into multiple radicals.
$\sqrt{63}=\sqrt{7\cdot 9}=\sqrt{7}\cdot \sqrt{{{3}^{2}}}=\sqrt{7}\cdot 3$
We usually rearrange so the integer factor appears before the radical. This is done so that it is clear that only the $7$ is under the radical, not the $3$ .
$=3\cdot \sqrt{7}$
$96$ is not a perfect square. We check divisibility by $4$ and find that it divides evenly as $4\cdot 24$ . The number $24$ further factors as $4\cdot 6$ so we now write
$10\sqrt{96}=10\sqrt{4\cdot 4\cdot 6}=10\sqrt{4}\cdot\sqrt{4}\cdot\sqrt{6}$
$=10\cdot 2\cdot 2\cdot \sqrt{6}=40\sqrt{6}$ .
Note that the original external factor of $10$ got multiplied by all the other factors that were brought out.
We begin checking $108$ for perfect cube factors. It does not divide $8$ evenly, but does factor as $27\cdot 4$ , so rewrite
$\sqrt[3]{-108}=\sqrt[3]{-27\cdot 4} = \sqrt[3]{-27}\cdot\sqrt[3]{4}=-3\sqrt[3]{4}$

The following video shows more examples of how to simplify square roots that do not have perfect square radicands. The video illustrates a technique that some students find helpful – completely factoring a number into its prime factorization using a factor tree to look for pairs of prime factors.

Example

Simplify $\sqrt{a^3b^5c^2}.$ Do not assume variables are nonnegative (which means you must consider use of absolute values).

Show Solution

Factor to find variables with even exponents.

$\sqrt{{{a}^{3}}{{b}^{5}}{{c}^{2}}}=\sqrt{{{a}^{2}}\cdot a\cdot {{b}^{4}}\cdot b\cdot {{c}^{2}}}$

Rewrite $b^{4}$ as $\left(b^{2}\right)^{2}$ .

$=\sqrt{{{a}^{2}}\cdot a\cdot {{({{b}^{2}})}^{2}}\cdot b\cdot {{c}^{2}}}$

Separate the squared factors into individual radicals.

$=\sqrt{{{a}^{2}}}\cdot \sqrt{{{({{b}^{2}})}^{2}}}\cdot \sqrt{{{c}^{2}}}\cdot \sqrt{a\cdot b}$

Take the square root of each radicand. Remember that $\sqrt{{{a}^{2}}}=\left| a \right|$ .

$=\left| a \right|\cdot {{b}^{2}}\cdot \left|{c}\right|\cdot \sqrt{a\cdot b}$

Simplify and multiply.

$=\left| ac \right|{{b}^{2}}\sqrt{ab}$

In the previous example, why did we not write $b^2$ as $|b^2|$ ? Because when you square a number, you will always get a nonnegative result, so we already know that $b^2$ is a nonnegative quantity (putting the absolute values on $b^2$ is not wrong, it is just unnecessary).

In the following video, you will see more examples of how to simplify radical expressions with variables.

The same process applies for cube roots. Note that in this example, it does not matter whether we specify that the variables are nonnegative, the absolute values should not be used regardless since the index is odd.

EXAMPLE

Simplify $\sqrt[3]{32x^{10}y^{13}z^9}.$

Show Solution

We can factor $32=2^5.$ Since we are looking for perfect third powers, we want to separate factors in sets of $3.$ This means rewriting each power as a multiple of $3$ plus a remainder.

$\begin{align}\sqrt[3]{32x^{10}y^{13}z^9}=&\quad\sqrt[3]{2^5x^{10}y^{13}z^9}\\ =&\quad\sqrt[3]{2^3 2^2x^9x^1y^{12}y^1z^9}\\ =&\quad\sqrt[3]{2^3x^9y^{12}z^9}\cdot\sqrt[3]{2^2xy}\\ =&\quad\sqrt[3]{(2)^3(x^3)^3(y^4)^3(z^3)^3}\cdot\sqrt[3]{2^2xy}\\ =&\quad 2x^3y^4z^3\sqrt[3]{4xy}\end{align}$

Multiplying and Simplifying Radical Expressions

When multiplying radical expressions, we can first multiply radicands and then use our simplification process.

Example

Simplify. $3\sqrt{21}\cdot 2\sqrt{12}$

Show Solution

First use the rule $\sqrt{a}\cdot \sqrt{b}=\sqrt{ab}$ to multiply the radicands. The outer factors can multiply by each other as well.

$3\sqrt{21}\cdot 2\sqrt{12}=2\cdot3\cdot\sqrt{21\cdot 12}$

It is easier to factor the two radicands first rather than multiplying them, but that is up to you.

$=6\sqrt{(3\cdot 7) \cdot (2 \cdot 2 \cdot 3)}$

Identify perfect square factors in the radicand.

$=6\sqrt{2^2 \cdot 3^2 \cdot 7}$

Rewrite as the product of radicals.

$\begin{align}&=6\sqrt{2^2}\cdot \sqrt{3^2}\cdot \sqrt{7}\\&=6\cdot 2\cdot 3\cdot \sqrt{7}\\&=36\sqrt{7}\end{align}$

You may have also noticed that $\sqrt{12}$ can be simplified as $\sqrt{4\cdot 3} = 2\sqrt{3}$ before multiplying. How would the result change if you simplified each radical first, before multiplying? You can verify that you arrive at the same product, $36\sqrt{7}$ . It does not matter whether you multiply the radicands or simplify each radical first.

You multiply radical expressions that contain variables in the same manner. As long as the indices 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, we multiply radicands right away and then the expression is simplified. Note that we specify that the variable is nonnegative, thus allowing us to avoid the need for absolute values. This will be common with most examples in this chapter.

Example

Simplify $\sqrt{12{{x}^{4}}}\cdot \sqrt{3x^2}.$ Assume all variables represent nonnegative quantities.

Show Solution

We first multiply radicands,

$\begin{align}\sqrt{12{{x}^{4}}}\cdot \sqrt{3x^2} &= \sqrt{12{{x}^{4}}\cdot 3x^2}\\&=\sqrt{12\cdot 3\cdot {{x}^{4}}\cdot x^2}\end{align}$

Recall that ${{x}^{4}}\cdot x^2={{x}^{4+2}}$ .

$\begin{align} =&\quad\sqrt{36\cdot x^{4+2}}\\ =&\quad\sqrt{36\cdot x^6}\end{align}$

Look for perfect square factors in the radicand.

$\begin{align} =&\quad\sqrt{6^2}\cdot (x^3)^{2}\\ =&\quad\sqrt{6^2}\cdot \sqrt{(x^3)^2}\\ =&\quad 6\cdot x^3\end{align}$

In our next example, we will multiply two cube roots. Remember that for odd index roots, we never need to consider use of absolute values.

Example

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

Show Solution

Notice that both radicals are cube roots, so you can use the Product Rule for Radicals to multiply the radicands. There is also no need to worry about absolute values since the index is odd.

$\begin{align} \sqrt[3]{{{x}^{5}}{{y}^{2}}}\cdot 5\sqrt[3]{8{{x}^{2}}{{y}^{4}}}=&\quad 5\sqrt[3]{x^5y^2\cdot 8x^2y^4} \\ =&\quad 5\sqrt[3]{8\cdot x^5\cdot x^2\cdot y^2\cdot y^4} \\=&\quad 5\sqrt[3]{8\cdot x^7\cdot y^6} \end{align}$

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.

$\begin{align} =&\quad 5\sqrt[3]{2^3}\cdot \sqrt[3]{(x^2)^3}\cdot \sqrt[3]{(y^2)^3}\cdot \sqrt[3]{x}\\ =&\quad 5\cdot 2\cdot x^2\cdot y^2\cdot \sqrt[3]{x}\\ =&\quad 10{{x}^{2}}{{y}^{2}}\sqrt[3]{x}\end{align}$

In the following video, we present more examples of how to multiply radical expressions.

This next example 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 $2\sqrt[4]{16{{x}^{9}}}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{81{{x}^{3}}y}.$ Assume all variables represent nonnegative quantities.

Show Solution

Since the index of all radicals is $4,$ we can use the Product Rule for Radicals. It is a good idea (but not necessary) to simplify each radical, if possible, before multiplying. Be looking for perfect $4$ th powers in each radicand.

$2\sqrt[4]{16{{x}^{9}}}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{81{{x}^{3}}y}=2\sqrt[4]{{{(2)}^{4}}\cdot {{({{x}^{2}})}^{4}}\cdot x}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{{{(3)}^{4}}\cdot {{x}^{3}}y}$

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 perfect $4$ th powers.

$\begin{align}=&\quad 2\cdot 2\cdot x^2\cdot\sqrt[4]{x}\cdot\sqrt[4]{y^3}\cdot 3\cdot\sqrt[4]{x^3y}\\ =&\quad 12x^2\sqrt[4]{x}\cdot\sqrt[4]{y^3}\cdot\sqrt[4]{x^3y}\\ =&\quad 12x^2\sqrt[4]{xy^3x^3y}\\ =&\quad 12x^2\sqrt[4]{x^4y^4}\\ \end{align}$

Now that the radicands have been multiplied, look again for perfect $4$ th powers.

$\begin{align}=&\quad 12x^2\sqrt[4]{x^4}\sqrt[4]{y^4}\\ =&\quad 12x^2 \cdot x \cdot y\\ =&\quad 12x^3y \end{align}$

In the following video, we show more examples of multiplying cube roots.

Summary

Radicals can be multiplied and simplified using the Product Rule for Radicals, $\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{ab}$ . A radical is considered simplified if all perfect power factors have been removed from the radical.

Module 5: Radical Expressions, Functions, and Equations