Algebra I (2018 edition)
Roots are nice, but we prefer dealing with regular numbers as much as possible. So, for example, instead of √4 we prefer dealing with 2. What about roots that aren't equal to an integer, like √20? Still, we can write 20 as 4⋅5 and then use known properties to write √(4⋅5) as √4⋅√5, which is 2√5. We *simplified* √20. Created by Sal Khan.
Let's see if we can simplify 5 times the square root of 117. So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So clearly it's an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explain why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3, so 117 is going to be divisible by 3. Now, let's do a little aside here to figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11, three times. 3 times 3 is 9. Subtract, you got a remainder of 2. Bring down a 7. 3 goes into 27 nine times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. So we can factor 117 as 3 times 39. Now 39, we can factor as-- that jumps out more at us that that's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as-- and we know this from our exponent properties-- 5 times the square root of 3 times 3 times the square root of 13. Now, what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those-- well, that's just going to give you 3. So this is just going to simplify to 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. I'm actually going to put 26 in yellow, like I did in the previous problem. Well, 26 is clearly an even number, so it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this any more. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't, so we've simplified this about as much as we can. We would just leave this as 3 times the square root of 26.