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## Algebra (all content)

### Course: Algebra (all content)>Unit 10

Lesson 8: Multiplying binomials by polynomials

# Multiplying binomials with radicals (old)

An old video Where Sal multiplies and simplifies (x²-√6)(x²+√2). Created by Sal Khan and Monterey Institute for Technology and Education.

## Video transcript

We're asked to multiply and to simplify. And we have x squared minus the principal square root of 6 times x squared plus the principal square root of 2. And so we really just have two binomials, two two-term expressions that we want to multiply, and there's multiple ways to do this. I'll show you the more intuitive way, and then I'll show you the way it's taught in some algebra classes, which might be a little bit faster, but requires a little bit of memorization. So I'll show you the intuitive way first. So if you have anything-- so let's say I have a times x plus y-- we know from the distributive property that this is the same thing as ax plus ay. And so we can do the same thing over here. If you view a as x squared-- as this whole expression over here-- x squared minus the principal square root of 6, and you view x plus y as this thing over here, you can distribute. We can distribute all of this onto-- let me do it this way-- distribute this entire term onto this term and onto that term. So let's do that. So we get x squared minus the principal square root of 6 times this term-- I'll do it in yellow-- times x squared. And then we have plus this thing again. We're just distributing it. It's just like they say. It's sometimes not that intuitive because this is a big expression, but you can treat it just like you would treat a variable over here. You're distributing it over this expression over here. And so then we have x squared minus the principal square root of 6 times the principal square root of 2. And now we can do the distributive property again, but what we'll do is we'll distribute this x squared onto each of these terms and distribute the square root of 2 onto each of these terms. It's the exact same thing as here, it's just you could imagine writing it like this. x plus y times a is still going to be ax plus ay. And just to see the pattern, how this is really the same thing as this up here, we're just switching the order of the multiplication. You can kind of view it as we're distributing from the right. And so if you do this, you get x squared times x squared, which is x to the fourth, that's that times that, and then minus x squared times the principal square root of 6. And then over here you have square root of 2 times x squared, so plus x squared times the square root of 2. And then you have square root of 2 times the square root of 6. And we have a negative sign out here. Now if you take the square root of 2-- let me do this on the side-- square root of 2 times the square root of 6, we know from simplifying radicals that this is the exact same thing as the square root of 2 times 6, or the principal square root of 12. So the square root of 2 times square root of 6, we have a negative sign out here, it becomes minus the square root of 12. And let's see if we can simplify this at all. Let's see. You have an x to the fourth term. And then here you have-- well depending on how you want to view it, you could say, look, we have to second degree terms. We have something times x squared, and we have something else times x squared. So if you want, you could simplify these two terms over here. So I have square root of 2 x squareds and then I'm going to subtract from that square root of 6 x squareds. So you could view this as square root of 2 minus the square root of 6, or the principal square root of 2 minus the principal square root of 6, x squared. And then, if you want, square root of 12, you might be able to simplify that. 12 is the same thing as 3 times 4. So the square root of 12 is equal to square root of 3 times square root of 4. And the square root of 4, or the principal square root of 4 I should say, is 2. So the square root of 12 is the same thing as 2 square roots of 3. So instead of writing the principal square root of 12, we could write minus 2 times the principal square root of 3. And then out here you have an x to the fourth plus this. And you see, if you distributed this out, if you distribute this x squared, you get this term, negative x squared, square root of 6, and if you distribute it onto this, you'd get that term. So you could debate which of these two is more simple. Now I mentioned that this way I just did the distributive property twice. Nothing new, nothing fancy. But in some classes, you will see something called FOIL. And I think we've done this in previous videos. FOIL. I'm not a big fan of it because it's really a way to memorize a process as opposed to understanding that this is really just from the common-sense distributive property. But all this is is a way to make sure that you're multiplying everything times everything when you're multiplying two binomials times each other like this. And FOIL just says, look, first multiply the first term. So x squared times x squared is x to the fourth. Then multiply the outside. So then multiply-- I'll do this in green-- then multiply the outside. So the outside terms are x squared and square root of 2. And so x squared times square root of 2-- and they are positive-- so plus square root of 2 times x squared. And then multiply the inside. And you can see why I don't like it that much is because you really don't know you're doing. You're just applying an algorithm. Then you'll multiply the inside. And so negative square root of 6 times x squared. And then you multiply the last terms. So negative square root of 6 times square root of 2, that is-- and we already know that-- that is negative square root of 12, which you can also then simplify to that expression right over there. So it's fine to use this, although it's good, even if you do use this, to know where FOIL comes from. It really just comes from using the distributive property twice.