Lesson Objectives

- Demonstrate an understanding of radicals
- Learn how to multiply radicals using the product rule for radicals
- Learn how to divide radicals using the quotient rule for radicals
- Learn how to simplify a radical

## Multiplying, Dividing, and Simplifying Radicals

In this lesson, we will learn how to multiply, divide, and simplify radicals. Note, for this lesson, we will assume that all variables represent positive real numbers. Let's begin by learning about the product rule for radicals.

Example 1: Simplify each. $$\sqrt{7}\cdot \sqrt{3}$$ We will take the square root of the product of the two radicands (7 and 3). $$\sqrt{7}\cdot \sqrt{3}=\sqrt{7 \cdot 3}=\sqrt{21}$$ Example 2: Simplify each. $$\sqrt[3]{3}\cdot \sqrt[3]{11}$$ We will take the cube root of the product of the two radicands (3 and 11). $$\sqrt[3]{3}\cdot \sqrt[3]{11}=\sqrt[3]{3 \cdot 11}=\sqrt[3]{33}$$ Example 3: Simplify each, assume all variables represent positive real numbers. $$\sqrt[9]{25x^2}\cdot \sqrt[9]{13x^4}$$ We will take the ninth root of the product of the two radicands (25x

Example 4: Simplify each. $$\large{\sqrt{7}\cdot \sqrt[3]{3}}$$ We will first rewrite our problem using rational exponents. $$\large{\sqrt{7}\cdot \sqrt[3]{3}=7^{\frac{1}{2}}\cdot 3^{\frac{1}{3}}}$$ What is the LCD of the fractions? The LCM of 2 and 3 is 6, so let's rewrite our problem as: $$\large{7^{\frac{3}{6}}\cdot 3^{\frac{2}{6}}}$$ We can rewrite the problem as: $$\large{7^{\frac{3}{6}}\cdot 3^{\frac{2}{6}}}=\sqrt[6]{7^3 \cdot 3^2}=\sqrt[6]{3087}$$

Example 5: Simplify each. $$\sqrt{\frac{4}{25}}$$ We will split this up into a quotient of two radicals. $$\sqrt{\frac{4}{25}}=\frac{\sqrt{4}}{\sqrt{25}}=\frac{2}{5}$$ Example 6: Simplify each. $$\sqrt[3]{\frac{64}{729}}$$ We will split this up into the quotient of two radicals. $$\sqrt[3]{\frac{64}{729}}=\frac{\sqrt[3]{64}}{\sqrt[3]{729}}=\frac{4}{9}$$

Example 7: Simplify each, assume all variables represent positive real numbers. $$\sqrt{243x^3}$$ Let's focus on the number and the variable separately for this example. We will first factor 243.

243 = 3 • 3 • 3 • 3 • 3

We know that 3

54 = 3 • 3 • 3 • 2

125 = 5 • 5 • 5

Let's update our problem as: $$\frac{\sqrt[3]{54y^{17}}}{\sqrt[3]{125}}=\frac{\sqrt[3]{27y^{15}}\cdot \sqrt[3]{2y^{2}}}{\sqrt[3]{125}}$$ We know the cube root of 27 is 3, and the cube root of 125 is 5. We broke y

### The Product Rule for Radicals

$$\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{a \cdot b}$$ We may also see this concept presented in exponent form: $$a^{\frac{1}{n}}\cdot b^{\frac{1}{n}}=(a \cdot b)^{\frac{1}{n}}$$ Let's look at a few examples.Example 1: Simplify each. $$\sqrt{7}\cdot \sqrt{3}$$ We will take the square root of the product of the two radicands (7 and 3). $$\sqrt{7}\cdot \sqrt{3}=\sqrt{7 \cdot 3}=\sqrt{21}$$ Example 2: Simplify each. $$\sqrt[3]{3}\cdot \sqrt[3]{11}$$ We will take the cube root of the product of the two radicands (3 and 11). $$\sqrt[3]{3}\cdot \sqrt[3]{11}=\sqrt[3]{3 \cdot 11}=\sqrt[3]{33}$$ Example 3: Simplify each, assume all variables represent positive real numbers. $$\sqrt[9]{25x^2}\cdot \sqrt[9]{13x^4}$$ We will take the ninth root of the product of the two radicands (25x

^{2}and 13x^{4}). $$\sqrt[9]{25x^2}\cdot \sqrt[9]{13x^4}=\sqrt[9]{25x^2 \cdot 13x^4}=\sqrt[9]{325x^6}$$### Multiplying Radicals with Different Indexes

What happens if we want to multiply radicals with different indexes? Let's look at an example.Example 4: Simplify each. $$\large{\sqrt{7}\cdot \sqrt[3]{3}}$$ We will first rewrite our problem using rational exponents. $$\large{\sqrt{7}\cdot \sqrt[3]{3}=7^{\frac{1}{2}}\cdot 3^{\frac{1}{3}}}$$ What is the LCD of the fractions? The LCM of 2 and 3 is 6, so let's rewrite our problem as: $$\large{7^{\frac{3}{6}}\cdot 3^{\frac{2}{6}}}$$ We can rewrite the problem as: $$\large{7^{\frac{3}{6}}\cdot 3^{\frac{2}{6}}}=\sqrt[6]{7^3 \cdot 3^2}=\sqrt[6]{3087}$$

### The Quotient Rule for Radicals

$$\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$$ We may also see this concept presented in exponent form: $$\left(\frac{a}{b}\right)^{\frac{1}{n}}=\frac{a^{\frac{1}{n}}}{b^{\frac{1}{n}}}$$ Let's look at a few examples.Example 5: Simplify each. $$\sqrt{\frac{4}{25}}$$ We will split this up into a quotient of two radicals. $$\sqrt{\frac{4}{25}}=\frac{\sqrt{4}}{\sqrt{25}}=\frac{2}{5}$$ Example 6: Simplify each. $$\sqrt[3]{\frac{64}{729}}$$ We will split this up into the quotient of two radicals. $$\sqrt[3]{\frac{64}{729}}=\frac{\sqrt[3]{64}}{\sqrt[3]{729}}=\frac{4}{9}$$

### Simplifying a Radical

When we simplify a radical, we have four conditions that must be met.- The radicand contains no factor that is raised to a power that is greater than or equal to the index
- The radicand contains no fractions
- No denominator contains a radical
- There can be no common factor between the index of the radical and the exponent in the radicand

Example 7: Simplify each, assume all variables represent positive real numbers. $$\sqrt{243x^3}$$ Let's focus on the number and the variable separately for this example. We will first factor 243.

243 = 3 • 3 • 3 • 3 • 3

We know that 3

^{2}is 9 and the square root of 9 is 3. Let's rewrite our problem as: $$\sqrt{243x^3}=\sqrt{9}\cdot \sqrt{9}\cdot \sqrt{3x^3}$$ Since the square root of 9 is 3, we can just replace this in our problem. $$\sqrt{243x^3}=3 \cdot 3 \cdot \sqrt{3x^3}=9\sqrt{3x^3}$$ For the variable part, it is a bit simpler. We don't really need to factor x^{3}as we know from the exponent, this is just three factors of x. We can rewrite the problem as: $$9\sqrt{3x^3}=\sqrt{x^2}\cdot 9\sqrt{3x}$$ We know the square root of x^{2}is just x, let's replace this in our problem. $$\sqrt{x^2}\cdot 9\sqrt{3x}=9x\sqrt{3x}$$ $$\sqrt{243x^3}=9x\sqrt{3x}$$ Example 8: Simplify each, assume all variables represent positive real numbers. $$\sqrt[3]{\frac{54y^{17}}{125}}$$ Let's first rewrite the problem as: $$\frac{\sqrt[3]{54y^{17}}}{\sqrt[3]{125}}$$ If we factor 54 and 125:54 = 3 • 3 • 3 • 2

125 = 5 • 5 • 5

Let's update our problem as: $$\frac{\sqrt[3]{54y^{17}}}{\sqrt[3]{125}}=\frac{\sqrt[3]{27y^{15}}\cdot \sqrt[3]{2y^{2}}}{\sqrt[3]{125}}$$ We know the cube root of 27 is 3, and the cube root of 125 is 5. We broke y

^{17}up into y^{15}• y^{2}. We are using y^{15}since 15 is the largest number that is smaller than 17 that is divisible by 3. We can just divide 15, the exponent, by 3, the index. The result will be 5 and this will be the exponent on the variable after we simplify. $$\sqrt[3]{y^{15}}=\sqrt[3]{(y^5)^3}=y^5$$ Returning to our problem, we can simplify as: $$\frac{\sqrt[3]{27y^{15}}\cdot \sqrt[3]{2y^{2}}}{\sqrt[3]{125}}=\frac{3y^5\sqrt[3]{2y^2}}{5}$$ $$\sqrt[3]{\frac{54y^{17}}{125}}=\frac{3y^5\sqrt[3]{2y^2}}{5}$$ Example 9: Simplify each, assume all variables represent positive real numbers. $$-5\sqrt[3]{189x^3y^2z^4}$$ 189 = 3 • 3 • 3 • 7 $$-5 \cdot \sqrt[3]{27x^3z^3}\cdot \sqrt[3]{7y^2z}$$ $$-5 \cdot 3xz \cdot \sqrt[3]{7y^2z}$$ $$-5\sqrt[3]{189x^3y^2z^4}=-15xz\sqrt[3]{7y^2z}$$#### Skills Check:

Example #1

Simplify each, assume all variables represent positive real numbers. $$-8\sqrt{32x^{4}y^{4}z^{5}}$$

Please choose the best answer.

A

$$40x^{2}z^{2}\sqrt{5xyz}$$

B

$$-24\sqrt{3xyz}$$

C

$$6x\sqrt{3yz}$$

D

$$-32x^{2}y^{2}z^{2}\sqrt{2z}$$

E

$$-8x^{2}y^{2}z^{2}\sqrt{4x^{2}y^{2}z}$$

Example #2

Simplify each, assume all variables represent positive real numbers. $$-6\sqrt{245xy^{4}z}$$

Please choose the best answer.

A

$$18xy\sqrt{5xz}$$

B

$$-72xyz\sqrt{2xyz}$$

C

$$18x^{2}y^{2}z\sqrt{z}$$

D

$$-9xy^{2}z\sqrt{15xy}$$

E

$$-42y^{2}\sqrt{5xz}$$

Example #3

Simplify each, assume all variables represent positive real numbers. $$3\sqrt{\frac{162p^{3}q^{2}r}{16}}$$

Please choose the best answer.

A

$$\frac{5p^{2}q^{2}r^{2}\sqrt{7pr}}{16}$$

B

$$\frac{81q^{2}p\sqrt{pr}}{4}$$

C

$$\frac{27pq\sqrt{2pr}}{4}$$

D

$$54r^{2}q\sqrt{6pq}$$

E

$$96\sqrt{p^{2}q^{2}r^{2}}$$

Congrats, Your Score is 100%

Better Luck Next Time, Your Score is %

Try again?

Ready for more?

Watch the Step by Step Video Lesson Take the Practice Test