7.3: Multiplying and Dividing Radical Expressions

section 7.3 Learning Objectives

7.3: Multiplying and Dividing Roots

Find the product of two radical terms
Multiply a radical and a sum or difference of radicals
Multiply binomials containing radicals
Simplify the square of a sum or difference of radicals
Divide radical expressions

Multiply and Divide

You can do more than just simplify radical expressions. You can also multiply and divide them. Multiplying radicals is very simple if the index on all the radicals match. The product rule of radicals can be generalized as follows

Product Rule for Radicals:

For any real numbers $\sqrt[n]{a}$ and $\sqrt[n]{b}$ ,

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

Find the product of two radical terms

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 with a cube root using this rule.

In the following example, we multiply two square roots.

Example 1

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

Show Solution

Use the rule

\sqrt[x]{a} \cdot \sqrt[x]{b} = \sqrt[x]{a b}

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

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

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

$= \sqrt{144\cdot 2}$

Identify perfect squares.

$= \sqrt{{{12}^{2}}\cdot 2}$

$=12\cdot \sqrt{2}$

Answer

$\sqrt{18}\cdot \sqrt{16}=12\sqrt{2}$

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?

Example 2

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

Show Solution

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

$\begin{array}{r}\sqrt{18}\cdot \sqrt{16}=\sqrt{9\cdot 2}\cdot \sqrt{4\cdot 4}\\\\=\sqrt{3\cdot 3\cdot 2}\cdot \sqrt{4\cdot 4}\end{array}$

Identify perfect squares.

$= \sqrt{{{3}^{2}}\cdot 2}\cdot \sqrt{{{4}^{2}}}$

$=3\cdot\sqrt{2}\cdot4$

Multiply.

$= 12\cdot\sqrt{2}$

Answer

$\sqrt{18}\cdot \sqrt{16}=12\sqrt{2}$

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.

In this video example, we multiply more square roots. Note, if you are curious how this would apply to radicals containing variables (not covered in this course), check out the last example in the video.

Example 3

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

Show Solution

Multiply the factors outside the radicals, and factor the radicands.

$2\sqrt[3]{18}\cdot-7\sqrt[3]{15}=-14\sqrt[3]{2\cdot3^2}\cdot\sqrt[3]{3\cdot5}$

Combine the radicands into one radical, and reorganize to see if there are any cubes.

$=-14\sqrt[3]{2\cdot3^2\cdot3\cdot5}=-14\sqrt[3]{2\cdot3^3\cdot5}$

Apply the cube root to $3^3$ , and simplify the radicand.

$=-14\cdot3\sqrt[3]{2\cdot5}=-42\sqrt[3]{10}$

Answer

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

In the next video, we present more examples of multiplying cube roots.

Multiply a radical and a sum or difference of radicals

When multiplying a radical by a sum or difference of radicals, we will use the Distributive Property.

Example 4

Multiply $\sqrt{3}(\sqrt{5}+\sqrt{3})$

Show Answer

Apply the Distributive Property

$\sqrt{3}(\sqrt{5}+\sqrt{3})=\sqrt{3}\cdot \sqrt{5}+\sqrt{3}\cdot \sqrt{3}$

Simplify each radical, if possible

$=\sqrt{15}+\sqrt{9}$

$=\sqrt{15}+3$

Solution:

$\sqrt{3}(\sqrt{5}+\sqrt{3})=\sqrt{15}+\sqrt{9}$

Example 5

Multiply $3\sqrt{5}(\sqrt{8}-\sqrt{7})$

Show Answer

Apply the Distributive Property

$3\sqrt{5}(\sqrt{8}-\sqrt{7})=3\sqrt{5}\cdot \sqrt{8}-3\sqrt{5}\cdot \sqrt{7}$

Prime Factor each radicand

$=3\sqrt{40}-3\sqrt{35}$

Identify pairs of identical factors

$=3\sqrt{2\cdot 2\cdot 2\cdot 5}-3\sqrt{5\cdot 7}$

$=3\sqrt{2^2\cdot 2\cdot 5}-3\sqrt{5\cdot 7}$

$=3\cdot 2\sqrt{2\cdot 5}-3\sqrt{5\cdot 7}$

$=6\sqrt{10}-3\sqrt{35}$

Solution:

$3\sqrt{5}(\sqrt{8}-\sqrt{7})=6\sqrt{10}-3\sqrt{35}$

Example 6

Multiply $\sqrt[3]{2}(\sqrt[3]{4}-2\sqrt[3]{28})$

Show Answer

Apply the Distributive Property

$\sqrt[3]{2}(\sqrt[3]{4}-2\sqrt[3]{28})=\sqrt[3]{2}\cdot \sqrt[3]{4}-\sqrt[3]{2}\cdot 2\sqrt[3]{28}$

$=\sqrt[3]{8}-2\sqrt[3]{56}$

Prime Factor each radicand

$=\sqrt[3]{2\cdot 2\cdot 2}-2\sqrt[3]{2\cdot 2\cdot 2\cdot 7}$

Identify groups of three identical factors

$=\sqrt[3]{2^3}-2\sqrt[3]{2^3\cdot 7}$

$=2-2\cdot 2\sqrt[3]{7}$

$=2-4\sqrt[3]{7}$

Solution:

$\sqrt[3]{2}(\sqrt[3]{4}-2\sqrt[3]{28})=2-4\sqrt[3]{7}$

Multiply binomials containing radicals

We will multiply binomials containing radicals in much the same way we multiplied binomials in Section 5.6. We will use the Distributive Property, and we can also use the FOIL Method.

Example 7

Multiply $(5+2\sqrt{2})(\sqrt{5}+3\sqrt{7})$

Show Answer

Use the FOIL Method to multiply the two binomials

$(5+2\sqrt{2})(\sqrt{5}+3\sqrt{7})$

$\begin{array}{l}\text{First}:\,\,\,\,\,5\cdot \sqrt{5}\\\text{Outer}:\,\,\,5\cdot 3\sqrt{7}\\\text{Inner}:\,\,\,2\sqrt{2}\cdot \sqrt{5}\\\text{Last}:\,\,\,\,\,2\sqrt{2}\cdot 3\sqrt{7}\end{array}$

$=5\cdot \sqrt{5} + 5\cdot 3\sqrt{7} + 2\sqrt{2}\cdot \sqrt{5} + 2\sqrt{2}\cdot 3\sqrt{7}$

$=5\sqrt{5}+15\sqrt{7}+2\sqrt{10}+6\sqrt{14}$

Since none of these radicals simplify further and there are no like radicals, this is the final answer.

Solution:

$(5+2\sqrt{2})(\sqrt{5}+3\sqrt{7})=5\sqrt{5}+15\sqrt{7}+2\sqrt{10}+6\sqrt{14}$

Example 8

Multiply $(3\sqrt{7}+2\sqrt{5})(2\sqrt{7}-4\sqrt{5})$

Show Answer

Use the FOIL Method to multiply the two binomials

$(3\sqrt{7}+2\sqrt{5})(2\sqrt{7}-4\sqrt{5})$

$\begin{array}{l}\text{First}:\,\,\,\,\,3\sqrt{7}\cdot 2\sqrt{7}\\\text{Outer}:\,\,\,3\sqrt{7}\cdot -4\sqrt{5}\\\text{Inner}:\,\,\,2\sqrt{5}\cdot 2\sqrt{7}\\\text{Last}:\,\,\,\,\,2\sqrt{5}\cdot -4\sqrt{5}\end{array}$

$=3\sqrt{7}\cdot 2\sqrt{7} - 3\sqrt{7}\cdot 4\sqrt{5} + 2\sqrt{5}\cdot 2\sqrt{7} - 2\sqrt{5}\cdot 4\sqrt{5}$

$=6\sqrt{49}-12\sqrt{35}+4\sqrt{35}-8\sqrt{25}$

Simplify each radical, if possible

$=6\cdot 7-12\sqrt{35}+4\sqrt{35}-8\cdot 5$

$=42-12\sqrt{35}+4\sqrt{35}-40$

Combine like terms

$=2-8\sqrt{35}$

Solution:

$(3\sqrt{7}+2\sqrt{5})(2\sqrt{7}-4\sqrt{5})=2-8\sqrt{35}$

Simplify the square of a sum or difference of radicals

Example 9

Simplify: $(\sqrt{3}-\sqrt{2})^2$

Show Answer

Expand

$(\sqrt{3}-\sqrt{2})^2=(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2})$

Use the FOIL Method to multiply the two binomials

$(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2})$

$\begin{array}{l}\text{First}:\,\,\,\,\,\sqrt{3}\cdot \sqrt{3}\\\text{Outer}:\,\,\,\sqrt{3}\cdot -\sqrt{2}\\\text{Inner}:\,\,\,-\sqrt{2}\cdot \sqrt{3}\\\text{Last}:\,\,\,\,\,\sqrt{2}\cdot \sqrt{2}\end{array}$

$=\sqrt{3}\cdot \sqrt{3} - \sqrt{3}\cdot \sqrt{2} - \sqrt{2}\cdot \sqrt{3} + \sqrt{2}\cdot \sqrt{2}$

$=\sqrt{9}-\sqrt{6}-\sqrt{6}+\sqrt{4}$

Simplify each radical, if possible

$=3-\sqrt{6}-\sqrt{6}+2$

Combine like terms

$=5-2\sqrt{6}$

Solution:

$(\sqrt{3}-\sqrt{2})^2=5-2\sqrt{6}$

Divide radical expressions

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 $\displaystyle {{\left( \frac{a}{b} \right)}^{x}}=\frac{{{a}^{x}}}{{{b}^{x}}}$ . This rule is applicable for roots as well:

Quotient Rule for Radicals

For any real numbers $\sqrt[n]{a}$ and $\sqrt[n]{b}$ , $b\neq0$ ,

$\displaystyle \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

As you did with multiplication, you will start with some examples featuring integers. (Moving on to radicals with variables will be in a future course).

Example 10

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

Show Solution

Use the rule

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

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

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

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

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

Identify and pull out perfect squares.

$\displaystyle=\frac{\sqrt{{{4}^{2}}\cdot 3}}{\sqrt{{{5}^{2}}}}$

Simplify.

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

Answer

$\displaystyle \sqrt{\frac{48}{25}}=\frac{4\sqrt{3}}{5}$

Sometimes radicals within both numerator and/or denominator will simplify completely, getting rid of the radical as seen in the following example.

Example 11

Simplify $\sqrt[3]{\frac{27}{8}}$

Show Answer

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

$\displaystyle\sqrt[3]{\frac{27}{8}}= \frac{\sqrt[3]{27}}{\sqrt[3]{8}}$

Both the numerator and denominator are perfect cubes.

$\displaystyle \frac{\sqrt[3]{27}}{\sqrt[3]{8}}=\frac{3}{2}$

If you didn’t notice that the numerator and denominator were perfect cubes, you could also factor and then simplify each radical.

$\displaystyle \frac{\sqrt[3]{27}}{\sqrt[3]{8}}=\frac{\sqrt[3]{3\cdot3\cdot3}}{\sqrt[3]{2\cdot2\cdot2}}=\frac{\sqrt[3]{3^3}}{\sqrt[3]{2^3}}=\frac{3}{2}$

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 this last video, we show more examples of simplifying a quotient with radicals.

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

Module 7: Radicals