## InsanelyChaotic 3 years ago Can you help me simplify a radical below?

1. InsanelyChaotic

$\sqrt 80 - 3 \sqrt 63 - \sqrt 125 - \sqrt 175$

2. KingGeorge

Let's go through these one at a time. First, let's simplify $$\sqrt{80}$$. If we factor 80, we get $$80=2\cdot2\cdot2\cdot2\cdot5$$. Since there are four 2's, we can say that $\sqrt{80}=\sqrt{2^4\cdot5}=2^2\sqrt{5}=4\sqrt{5}$Did this all make sense?

3. 03453660

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4. InsanelyChaotic

@KingGeorge Where did the 2's come from?

5. InsanelyChaotic

When I see myself factoring, the number two doesn't even come to mind, do you just pick any number?

6. 03453660

|dw:1351701840947:dw|

7. 03453660

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8. KingGeorge

There are certain numbers that you should think about when you're factoring this kind of number. In particular, they're the "prime" numbers. The first few prime numbers are 2,3,5,7,11,13. When you're trying to simplify radicals, see if you can divide the number by these "prime" numbers. Usually, we start at the smallest one. So for 80, you can divide by 2 to get $$80=2\cdot 40$$. Again, we notice that 2 divides 40, so we get $$80=2\cdot2\cdot20$$. Keep doing this, and we get $80=2\cdot2\cdot2\cdot10$$80=2\cdot2\cdot2\cdot2\cdot5$Since 5 is on our list of prime numbers, we stop, because we can't divide any more.

9. InsanelyChaotic

That's incredible, thank you for taking the time to explain that

10. KingGeorge

So if you follow this same process, you should get almost what 03453660 did (175 is a bit different). In particular,$80=2^4\cdot 5$$63=3^3\cdot7$$125=5^3$$175=5^2\cdot 7$Now, can you take the square roots of the factored numbers?

11. InsanelyChaotic

Can I what?

12. InsanelyChaotic

I'm sorry, that question caught me off guard

13. KingGeorge

So we already said that $$\sqrt{2^4\cdot5}=4\sqrt{5}$$. Now, what is $\large \sqrt{3^2\cdot 7}?$Also, I've got go now, feel free to call in someone else to help you finish.

14. InsanelyChaotic

12x7?