# express the product

#### shaochen

##### New member
Express the product of 3?20(2?5-7) in simplet radical form.
I didnt get the question. And i dont know how to express this "root" expression.
Can people help me solving this question?

#### mmm4444bot

##### Super Moderator
Staff member
shaochen said:
Express the product of 3?20 * (2?5 - 7) in simplest radical form.
I'm thinking that it is less work to reduce the radical factor 3?20 first, and then do the multiplications.

The number 20 is called the "radicand" because it's inside the radical sign. In order to reduce this square root, we need to factorize the radicand, followed by taking the square root of any square factors of 20.

Well, 20 is the product of factors 4*5. The number 4 is a square number, so we take the square root of 4.

Here's the property that allows us to do this:

$$\displaystyle \sqrt{a \cdot b} \;=\; \sqrt{a}\cdot\sqrt{b}$$

This property says that the square root of a product is the same thing as the product of square roots of the individual factors.

So, if either a or b are square numbers, their square root can be simplified. Like so:

$$\displaystyle \sqrt{20} \;=\; \sqrt{4\cdot5} \;=\; \sqrt{4}\cdot\sqrt{5} \;=\; 2\cdot\sqrt{5}$$

Since the square root of 4 is 2, I hope that you understand how the property above allows us to simplify ?20 to 2?5 .

So, now we know that the factor 3?20 in your exercise equals (3)(2)(?5), which is clearly 6?5.

Next, we multiply the expression 2?5 - 7 by the expression 6?5 . We use the Distributive Property for that.

$$\displaystyle 6\sqrt{5} \cdot (2\sqrt{5} \;-\; 7)$$

$$\displaystyle (6\sqrt{5})(2\sqrt{5}) \;-\; (6\sqrt{5})(7)$$

Can you finish now?

If it helps you, the order of the factors can be rearranged, using the Commutative Property of Multiplication.

$$\displaystyle (6)(2)(\sqrt{5})(\sqrt{5}) \;-\; (6)(7)(\sqrt{5})$$

Please show your work, or explain what you're thinking, if you want more help finishing this.

Of course, if I wrote anything that you do not understand, reply with specific questions.

Cheers ~ Mark