# Multiplying and Distributing Radical Expressions - Problem 4

###### Transcript

Here I have two binomials being multiplied together. So I'm going to use the FOIL process; first, outers, inners, last. When I multiply the first, I have 2 root 10 times regular root 10. That's the same thing as 2 times 10, because root 10 times root 10 is 10.

So 2 times 10 gives me 20, that's my first. Outers, I have -6 from these guys, and then square root of 50. Inners, I have 3 root 50, and then lasts, I have -9 in my outside numbers. And then square root of 5 times square of 5 is regular 5. What I have is -9 times 5 which is -45.

I did a couple steps in my head there, so if I lost you, you might want to go back, and rewatch the start of this video. These firsts and lasts, are regular integers, so they're going to be combined. And then these are like term radicals. So I can combine those.

Let's start with this. 20 take away 45 is -25. Here I have -6 plus 3 which is -3 root 50. A lot of students will box this answer, and think they're done . In fact, I need to simplify further. Square root of 50 could be written as square root of 25 times square root of 2. Square root of 25 is a whole number 5, so really what I'm working with is -3 times regular 5 times root 2. That came with the -25 piece.

Combine all that together, I'll have -25 take away 15 root 2. That might be the answer in the back of your text book. Your textbook might also have factored out a common factor of either positive, or -5. I'm going to show you what it will look like. If I factored out -5, then I would have 5 minus 3 root 2. Your textbook might also have given you this answer. Those are equivalent statements, only this guys has the -5 factored out.

So the first step is the most tricky, the foiling. Make sure you're careful with what's inside the radical, or what's a radical, and what's outside. Then I used a shortcut, knowing that if I was doing 2 times square root of 10, times square root of 10, that was just like 2 times 10 or 20. That's how I went through to find this in this term, then combining like terms, simplifying to get to there.