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Study Guides > Intermediate Algebra

Read: Multiply and Divide Radical Expressions

Learning Objectives

  • Multiply and divide radical expressions
  • Use the product raised to a power rule to multiply radical expressions
  • Use the quotient raised to a power rule to divide radical expressions
You can do more than just simplify radical expressions. You can multiply and divide them, too. 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=axbx {{(ab)}^{x}}={{a}^{x}}\cdot {{b}^{x}} For any numbers a and b and any positive integer x: (ab)1x=a1xb1x {{(ab)}^{\frac{1}{x}}}={{a}^{\frac{1}{x}}}\cdot {{b}^{\frac{1}{x}}} For any numbers a and b and any positive integer x: abx=axbx \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 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 our first example we will work with integers, then we will move on to expressions with variable radicands.

Example

Simplify. 1816 \sqrt{18}\cdot \sqrt{16}

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

1816288\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.

1442 \sqrt{144\cdot 2}

Identify perfect squares.

(12)22 \sqrt{{{(12)}^{2}}\cdot 2}

Rewrite as the product of two radicals.

(12)22 \sqrt{{{(12)}^{2}}}\cdot \sqrt{2}

Simplify, using x2=x \sqrt{{{x}^{2}}}=\left| x \right|.

122122\begin{array}{r}\left| 12 \right|\cdot \sqrt{2}\\12\cdot \sqrt{2}\end{array}

Answer

1816=122 \sqrt{18}\cdot \sqrt{16}=12\sqrt{2}

You may have also noticed that both 18 \sqrt{18} and 16 \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. 1816 \sqrt{18}\cdot \sqrt{16}

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

924433244 \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.

(3)22(4)2 \sqrt{{{(3)}^{2}}\cdot 2}\cdot \sqrt{{{(4)}^{2}}}

Rewrite as the product of radicals.

(3)22(4)2 \sqrt{{{(3)}^{2}}}\cdot \sqrt{2}\cdot \sqrt{{{(4)}^{2}}}

Simplify, using x2=x \sqrt{{{x}^{2}}}=\left| x \right|.

324324\begin{array}{c}\left|3\right|\cdot\sqrt{2}\cdot\left|4\right|\\3\cdot\sqrt{2}\cdot4\end{array}

Multiply.

12212\sqrt{2}

Answer

1816=122 \sqrt{18}\cdot \sqrt{16}=12\sqrt{2}

In both cases, you arrive at the same product, 122 12\sqrt{2}. 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 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, x0 x\ge 0, thus allowing us to avoid the need for absolute value.

Example

Simplify. 12x43x2 \sqrt{12{{x}^{4}}}\cdot \sqrt{3x^2}, x0 x\ge 0

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

12x43x2123x4x2 \sqrt{12{{x}^{4}}\cdot 3x^2}\\\sqrt{12\cdot 3\cdot {{x}^{4}}\cdot x^2}

Recall that x4x2=x4+2 {{x}^{4}}\cdot x^2={{x}^{4+2}}.

36x4+236x6\begin{array}{r}\sqrt{36\cdot {{x}^{4+2}}}\\\sqrt{36\cdot {{x}^{6}}}\end{array}

Look for perfect squares in the radicand.

(6)2(x3)2 \sqrt{{{(6)}^{2}}\cdot {{({{x}^{3}})}^{2}}}

Rewrite as the product of radicals.

(6)2(x3)26x3 \begin{array}{c}\sqrt{{{(6)}^{2}}}\cdot \sqrt{{{({{x}^{3}})}^{2}}}\\6\cdot {{x}^{3}}\end{array}

Answer

12x43x2=6x3 \sqrt{12{{x}^{4}}}\cdot \sqrt{3x^2}=6{{x}^{3}}

 

Analysis of the solution

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 x0 x\ge 0. It is important to read the problem very well when you are doing math.  Even the smallest statement like x0 x\ge 0 can influence the way you write your answer. In our next example we will multiply two cube roots.

Example

Simplify. x5y2358x2y43 \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.

5x5y28x2y4358x5x2y2y4358x5+2y2+4358x7y63\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 x7 {{x}^{7}} is not a perfect cube, it has to be rewritten as x6+1=(x2)3x {{x}^{6+1}}={{({{x}^{2}})}^{3}}\cdot x.

5(2)3(x2)3x(y2)33 5\sqrt[3]{{{(2)}^{3}}\cdot {{({{x}^{2}})}^{3}}\cdot x\cdot {{({{y}^{2}})}^{3}}}

Rewrite as the product of radicals.

5(2)33(x2)33(y2)33x352x2y2x3 \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}

Answer

x5y2358x2y43=10x2y2x3 \sqrt[3]{{{x}^{5}}{{y}^{2}}}\cdot 5\sqrt[3]{8{{x}^{2}}{{y}^{4}}}=10{{x}^{2}}{{y}^{2}}\sqrt[3]{x}

In the following video we present more examples of how to multiply radical expressions. https://youtu.be/PQs10_rFrSM 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 which before you approach the problem, but either order will work for all problems.)

Example

Simplify. 216x94y3481x3y4 2\sqrt[4]{16{{x}^{9}}}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{81{{x}^{3}}y}, x0 x\ge 0, y0 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 44 in each radicand.

2(2)4(x2)4x4y34(3)4x3y4 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(2)44(x2)44x4y34(3)44x3y4 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 44, using the fact that x44=x \sqrt[4]{{{x}^{4}}}=\left| x \right|.

22x2x4y343x3y422x2x4y343x3y4 \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 abx=axbx \sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b} to multiply the radicands.

223x2xy3x3y412x2x1+3y3+14\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 44, 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 x0 x\ge 0, y0 y\ge 0.

12x2x4y4412x2x44y4412x2xy \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}

Answer

216x94y3481x3y4=12x3y,  x0,  y0 2\sqrt[4]{16{{x}^{9}}}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{81{{x}^{3}}y}=12{{x}^{3}}y,\,\,x\ge 0,\,\,y\ge 0

In the following video we show more examples of multiplying cube roots. https://youtu.be/cxRXofdelIM

Dividing 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 abx=axbx \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 (ab)x=axbx {{\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: (ab)1x=a1xb1x {{\left( \frac{a}{b} \right)}^{\frac{1}{x}}}=\frac{{{a}^{\frac{1}{x}}}}{{{b}^{\frac{1}{x}}}}, so abx=axbx \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: (ab)1x=a1xb1x {{\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: abx=axbx \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 more complex expressions like 24xy438y3 \frac{\sqrt[3]{24x{{y}^{4}}}}{\sqrt[3]{8y}}.

Example

Simplify. 4825 \sqrt{\frac{48}{25}}

Answer: Use the rule abx=axbx \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.

4825 \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.

16325or44355 \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.

(4)23(5)2(4)23(5)2 \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.

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

Answer

4825=435 \sqrt{\frac{48}{25}}=\frac{4\sqrt{3}}{5}

Example

Simplify. 640403 \sqrt[3]{\frac{640}{40}}

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

6403403 \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.

64103853 \frac{\sqrt[3]{64\cdot 10}}{\sqrt[3]{8\cdot 5}}

Identify and pull out perfect cubes.

(4)3103(2)353(4)33103(2)33534103253 \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.

4103253 \frac{4\sqrt[3]{10}}{2\sqrt[3]{5}}

Rewrite the numerator as a product of factors.

225323253 \frac{2\cdot 2\sqrt[3]{5}\cdot \sqrt[3]{2}}{2\sqrt[3]{5}}

Identify factors of 11, and simplify.

2253253232123 \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}

Answer

640403=223 \sqrt[3]{\frac{640}{40}}=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? Let’s take another look at that problem.

Example

Simplify. 6403403 \frac{\sqrt[3]{640}}{\sqrt[3]{40}}

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

640403 \sqrt[3]{\frac{640}{40}}

Within the radical, divide 640640 by 4040.

640÷40=16163 \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.

823\sqrt[3]{8\cdot2}

Identify perfect cubes and pull them out.

(2)32323 \begin{array}{r}\sqrt[3]{{{(2)}^{3}}\cdot 2}\\\sqrt[3]{2}\end{array}

Simplify.

2232\cdot\sqrt[3]{2}

Answer

6403403=223\frac{\sqrt[3]{640}}{\sqrt[3]{40}}=2\sqrt[3]{2}

That was a more straightforward approach, wasn’t it? In the next video we show more examples of simplifying a radical that contains a quotient. https://youtu.be/SxImTm9GVNo 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.     Now let’s turn to some radical expressions containing division. Notice that the process for dividing these is the same as it is for dividing integers.

Example

Simplify. 30x10x,x>0\frac{\sqrt{30x}}{\sqrt{10x}},x>0

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

30x10x\sqrt{\frac{30x}{10x}}

Simplify 30x10x \sqrt{\frac{30x}{10x}} by identifying similar factors in the numerator and denominator and then identifying factors of 11.

310x10x310x10x31\begin{array}{r}\sqrt{\frac{3\cdot10x}{10x}}\\\\\sqrt{3\cdot\frac{10x}{10x}}\\\\\sqrt{3\cdot1}\end{array}

Answer

30x10x=3 \frac{\sqrt{30x}}{\sqrt{10x}}=\sqrt{3}

Example

Simplify. 24xy438y3,  y0 \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.

24xy48y3 \sqrt[3]{\frac{24x{{y}^{4}}}{8y}}

Simplify 24xy48y3 \sqrt[3]{\frac{24x{{y}^{4}}}{8y}} by identifying similar factors in the numerator and denominator and then identifying factors of 11.

83xy3y8y33xy318y8y33xy3113\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.

3xy33(y)33x3 \sqrt[3]{3x{{y}^{3}}}\\\sqrt[3]{{{(y)}^{3}}\cdot \,3x}

Simplify.

(y)333x3 \sqrt[3]{{{(y)}^{3}}}\cdot \,\sqrt[3]{3x}

Answer

24xy438y3=y3x3 \frac{\sqrt[3]{24x{{y}^{4}}}}{\sqrt[3]{8y}}=y\,\sqrt[3]{3x}

In our last video we show more examples of simplifying radicals that contain quotients with variables. https://youtu.be/04X-hMgb0tA 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, while you can think of 8y2225y4 \frac{\sqrt{8{{y}^{2}}}}{\sqrt{225{{y}^{4}}}} as equivalent to 8y2225y4 \sqrt{\frac{8{{y}^{2}}}{225{{y}^{4}}}} since both the numerator and the denominator are square roots, notice that you cannot express 8y2225y44 \frac{\sqrt{8{{y}^{2}}}}{\sqrt[4]{225{{y}^{4}}}} as 8y2225y44 \sqrt[4]{\frac{8{{y}^{2}}}{225{{y}^{4}}}}. In this second case, the numerator is a square root and the denominator is a fourth root.

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: abx=axbx \sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}. When dividing radical expressions, the rules governing quotients are similar: abx=axbx \sqrt[x]{\frac{a}{b}}=\frac{\sqrt[x]{a}}{\sqrt[x]{b}}.

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