Dividing quadratics by linear expressions with remainders

- [Instructor] So if you've been watching these videos, you know that we have a lot of scenarios where people seem to be walking up to us on the street and asking us to do math problems. And I guess this will be no different. So let's say someone walks up to you on the street and says, "Quick, you, "x squared plus five x plus eight over x plus two, "what can this be simplified to? "Or what is x squared plus five x plus eight "divided by x plus two?" Pause this video and see if you can work through that. So there's two ways that we can approach this. We can try to factor our numerator, and see if we have a common factor there, or we can try to use algebraic long division. Let's first try to factor this numerator. And we would ideally want x plus two to be one of the factors. So let's see, what two numbers can add up to five, and when I multiply 'em I get to eight, and ideally two is one of them. So I can think of two and three. But two times three is going to be equal to six, not eight. And I can't think of anything else. But that still gives us some progress. Because, what if we did say, all right, let's rewrite part of it. What if we were to write x squared plus five x, and we wanna write a plus six, because that actually would be divisible by x plus two. So I'm gonna write a plus six. But of course we have an eight here, so then we're going to have an extra two right over there. And then all of that is divisible by x plus two. And now I can rewrite this part, up here in orange. That is x plus two times x plus three. So let me write it here. x plus two times x plus three, I still have that plus two sitting out there in the numerator, plus two. And then all of that over x plus two. Or I could write this as being over x plus two. And this being over x plus two. All I did is I said, hey, if I have something plus something else over x plus two, that could be the first something over x plus two plus the second something over x plus two. And then here, we can say, hey, look, this first part, as long as x does not equal negative two, because then we would be changing the domain, then these two would cancel out. What you could say, hey, I'm just dividing the numerator and the denominator by x plus two. And so this would be equal to x plus three, plus, and I don't necessarily even have to put parentheses there, plus two over x plus two. And I would have to constrain the domain so this is for x does not equal negative two. So in this situation, we had a remainder. And people will refer to the two as the remainder. We divide it as far as we can, but it still remains to be done to divide the two by x plus two. And so we would refer to the two as the remainder. Now that wasn't too difficult, but it also wasn't too straightforward and we'll see that this is a situation where the algebraic long division is actually a little bit more straightforward. So let's try that out. Once again, pause this video and see if you can figure out what this is through algebraic long division. So we're trying to take x plus two and divide it into x squared plus five x plus eight. Look at the highest degree terms, the x and the x squared. x goes into x squared x times, put it in the first degree column. x times two is two x, x times x is x squared, subtract these from x squared plus five x and we get five x minus two x is three x. x squared minus x squared, that's just zero. Bring down that eight. Look at the highest-degree term. And we get x goes into three x three times. Put that in the constant column, or the zeroth degree column, so plus three. Three times two is six, three times x is three x. Subtract ds, and we are left with, let me scroll down a little bit, you're left with, those cancel out, and you're left with eight minus six, which is, indeed, equal to two. And we could say, hey, we don't really know how to divide x plus two into two for an arbitrary x, so we will say, hey, this is going to be equal to x plus three with a remainder of two. Now once again, if you wanted to rewrite that original expression, and you wanted it to be completely the same, including the domain, you would have to constrain, you would have to constrain the domain, just like that.