Lesson Objectives
• Demonstrate an understanding of radical expressions
• Learn how to multiply two or more radicals
• Learn how to simplify a radical using the product rule for radicals
• Learn how to simplify a radical using the quotient rule for radicals
• Learn how to simplify a radical with variables
• Learn how to simplify higher-level roots

The product rule for radicals tells us the product of two square roots is the square root of the product. Let's suppose we saw the following problem: $$\sqrt{3} \cdot \sqrt{5}$$ Based on our rule, we could place the 3 and 5 under one square root symbol and find their product: $$\sqrt{3} \cdot \sqrt{5} = \sqrt{3 \cdot 5} = \sqrt{15}$$ Let's look at a few examples.
Example 1: Find each product $$\sqrt{13} \cdot \sqrt{2}$$ We can write one square root symbol and place the two radicands (13 and 2) underneath. These two numbers will be multiplied together: $$\sqrt{13} \cdot \sqrt{2} = \sqrt{13 \cdot 2} = \sqrt{26}$$ Example 2: Find each product $$\sqrt{6} \cdot \sqrt{7}$$ We can write one square root symbol and place the two radicands (6 and 7) underneath. These two numbers will be multiplied together: $$\sqrt{6} \cdot \sqrt{7} = \sqrt{6 \cdot 7} = \sqrt{42}$$

### Simplifying Square Roots

A square root is simplified when no perfect square factor remains under the radical sign. Let's suppose we saw the following problem: $$\sqrt{8}$$ If we factor 8, we get 23 or 2 • 2 • 2. We know that 4 is a perfect square (2 • 2). Let's use our product rule for radicals to rewrite our problem: $$\sqrt{8} = \sqrt{4 \cdot 2}$$ We can now break this problem up into: $$\sqrt{4} \cdot \sqrt{2}$$ Since the square root of 4 is 2, we can just replace the square root of 4 with a 2: $$2 \cdot \sqrt{2}$$ We can just place the number outside of the square root symbol to imply multiplication: $$2\sqrt{2}$$ Another way to think about this is to factor the radicand (number under the radical symbol): $$\sqrt{8} = \sqrt{2 \cdot 2 \cdot 2}$$ Since the square root of a number is a number that multiplies by itself to give the number, each time we have a pair (two) of factors, we can remove the two factors from the inside of the square root symbol and place one outside: $$\require{color} \sqrt{8} = \sqrt{\colorbox{yellow}{2} \cdot \colorbox{yellow}{2} \cdot 2} = \colorbox{yellow}{2} \sqrt{2}$$ Let's look at some examples.
Example 3: Simplify each $$\sqrt{72}$$ We will begin by factoring the radicand (number under the radical symbol):
72 = 2 • 2 • 2 • 3 • 3
Each pair of factors can be removed from the inside of the square root symbol and we can place one outside: $$\sqrt{72}$$ $$\sqrt{\colorbox{yellow}{2} \cdot \colorbox{yellow}{2} \cdot 2 \cdot \colorbox{yellow}{3} \cdot \colorbox{yellow}{3}}$$ $$\colorbox{yellow}{2} \cdot \colorbox{yellow}{3} \cdot \sqrt{2}$$ $$6\sqrt{2}$$ Another way to think about this: $$\sqrt{72} = \sqrt{4} \cdot \sqrt{9} \cdot \sqrt{2}$$ We know the square root of 4 is 2 and the square root of 9 is 3. We can replace each: $$\sqrt{4} \cdot \sqrt{9} \cdot \sqrt{2} = 2 \cdot 3 \cdot \sqrt{2}$$ $$6\sqrt{2}$$ Example 4: Simplify each $$\sqrt{26} \cdot \sqrt{39}$$ We can write this problem as: $$\sqrt{26} \cdot \sqrt{39} = \sqrt{26 \cdot 39}$$ Let's factor each number and look for pairs: $$\sqrt{2 \cdot \colorbox{yellow}{13} \cdot 3 \cdot \colorbox{yellow}{13}}$$ $$\colorbox{yellow}{13} \sqrt{2 \cdot 3}$$ $$13 \sqrt{6}$$ Another way to think about this: $$\sqrt{26} \cdot \sqrt{39} = \sqrt{169} \cdot \sqrt{6}$$ We know the square root of 169 is 13. We can replace this: $$\sqrt{169} \cdot \sqrt{6} = 13\sqrt{6}$$

When working with division, we have a similar rule. The quotient rule for radicals tells us the square root of the quotient is the quotient of the square roots. Let's suppose we saw the following problem: $$\sqrt{\frac{100}{81}}$$ We can rewrite this as: $$\frac{\sqrt{100}}{\sqrt{81}}$$ We can then simplify. We know the square root of 100 is 10 and the square root of 81 is 9: $$\frac{\sqrt{100}}{\sqrt{81}} = \frac{10}{9}$$ Let's look at a few examples.
Example 5: Simplify each $$\sqrt{\frac{90}{5}}$$ Let's rewrite our problem as: $$\sqrt{\frac{90}{5}} = \frac{\sqrt{90}}{\sqrt{5}}$$ Now let's simplify each part: $$\frac{\sqrt{90}}{\sqrt{5}} = \frac{\sqrt{\colorbox{yellow}{3} \cdot \colorbox{yellow}{3} \cdot 2 \cdot 5}}{\sqrt{5}}$$ $$\frac{\sqrt{\colorbox{yellow}{3} \cdot \colorbox{yellow}{3} \cdot 2 \cdot 5}}{\sqrt{5}} = \frac{3 \sqrt{2 \cdot 5}}{\sqrt{5}}$$ Square root of 5 over square root of 5 can be canceled: $$\frac{3 \sqrt{2} \cdot \sqrt{5}}{\sqrt{5}}$$ $$\require{cancel}\frac{3 \sqrt{2} \cdot \cancel{\sqrt{5}}}{\cancel{\sqrt{5}}} = 3\sqrt{2}$$

Example 6: Simplify each $$\sqrt{44} \cdot \sqrt{55x^2}$$ $$\sqrt{\colorbox{yellow}{2} \cdot \colorbox{yellow}{2} \cdot \colorbox{yellow}{11} \cdot \colorbox{yellow}{11} \cdot 5 \cdot \colorbox{yellow}{x} \cdot \colorbox{yellow}{x}}$$ $$\colorbox{yellow}{2} \cdot \colorbox{yellow}{11} \cdot \colorbox{yellow}{x} \sqrt{5}$$ $$22x \sqrt{5}$$
The product and quotient rules for radicals also works for higher-level roots. As an example, if we want to simplify a cube root, we will look for factors that are perfect cubes. Let's suppose we ran into a problem such as: $$\sqrt{16}$$ We can factor the radicand and look for 3 copies of the same number. When we have a cube root, we look for 3 copies. If we had a fourth root, we would look for four copies, and so on and so forth. Once we find our 3 copies, we can remove these from under the radical symbol and place one outside of the radical. $$\sqrt{16} = \sqrt{\colorbox{yellow}{2} \cdot \colorbox{yellow}{2} \cdot \colorbox{yellow}{2} \cdot 2}$$ $$\sqrt{\colorbox{yellow}{2} \cdot \colorbox{yellow}{2} \cdot \colorbox{yellow}{2} \cdot 2} = \colorbox{yellow}{2} \sqrt{2}$$ Another way to think about this: $$\sqrt{16} = \sqrt{8} \cdot \sqrt{2}$$ We know the cube root of 8 is 2, so we can replace this: $$\sqrt{8} \cdot \sqrt{2} = 2 \sqrt{2}$$ Let's take a look at an example.
Example 7: Simplify each $$\sqrt{96x^7}$$ Let's begin by factoring 96:
We could also rewrite x7 as: x5 • x2 $$\sqrt{96x^7} = \sqrt{2^5} \cdot \sqrt{x^5} \cdot \sqrt{3x^2}$$ We know the fifth root of 32 is 2 and the fifth root of x5 is x. Let's replace each: $$\sqrt{2^5} \cdot \sqrt{x^5} \cdot \sqrt{3x^2} = 2x\sqrt{3x}$$