If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Dividing quadratics by linear expressions (no remainders)

We can divide polynomials similarly to how we divide integers. For example, when we divide (x²+7x+10) by (x+2), we are asking "what can we multiply by (x+2) to get (x²+7x+10)?" We can answer this question in many ways. One is with using factorization, and another one is long division.

Want to join the conversation?

Video transcript

- [Instructor] Let's say someone walks up to you on the street, and they give you this expression. X squared plus 7x plus 10 divided by x plus two. And they say, see if you can simplify this thing. And so, pause this video and see if you can do that. And one way to think about it is, what is x squared plus 7x plus 10 divided by x plus two, what is that going to be? All right now, there's two ways that you could approach this. One way is to try to factor the numerator and see if it has a factor that is common to the denominator. So let's try to do that. So we've done this many, many times. If this looks new to you, I encourage you to review factoring polynomials other places on Kahn Academy. But what two numbers add up to seven, and when you multiply em, you get 10? Well, that would be two and five. So we could rewrite that numerator as x plus two times x plus five. And then of course, the denominator, you still have x plus two. And then we clearly see we have a common factor. And so as long as x does not equal negative two, because if x equals negative two, this whole expression is undefined, because then you get a zero in the denominator. So as long as x does not equal negative two, well then, we can divide the numerator and the denominator by x plus two. Once again, the reason why I put that constraint is that we can't divide the numerator and the denominator by zero. So for any other values of x, this x plus two will be non zero, and we could divide the numerator and the denominator by that, and they would cancel out, and we would just be left with x plus five. So another way to think about it is this expression, our original expression, could be viewed as x plus five for any x that is not equal to negative two. Now the other way that we could approach this is through algebraic long division, which is very analogous to the type of long division that you might remember from, I believe it was, fourth grade. So what you do is you say, all right, I'm gonna divide x plus two into x squared plus seven x plus 10. And in this technique, you look at the highest degree terms. So then, you have an x there and an x squared there. And say, how many times does x go into x squared? Well it goes x times. Now you'd write that in this column, because x is just x to the first power. You could view this as the first degree column. It's analogous to the place values that we talk about when we first learn numbers, or how we group or about place value. But here you could view it as degree places or something like that. And then you take that x and you multiply it times this entire expression. So x times two is 2x. Put that in the first degree column. X times x is x squared. And then what we wanna do is we wanna subtract these things in yellow from what we originally had in blue. So we could do it this way. And then we will be left with 7x minus 2x is 5x. And then x squared minus x squared is just a zero. And then we can bring down this plus 10. And once again, we look at the highest degree term. X goes into 5x five times. That's a zero degree. It's a constant, so I'll write it in the constant column. Five times two is 10. Five times x is five. And then I'll subtract these from what we have up here. And notice, we have no remainder. And what's interesting about algebraic long division, and we'll probably see it in another video or two, you can actually have a remainder. So those are going to be situations where just a factoring technique alone would not have worked. In this situation, this model would have been easier, but this is another way to think about it. You can say, hey, look, x plus two times x plus five is going to be equal to this. Now, if you wanted to rewrite this expression the way we did here, and say, hey, this expression is equal to x plus five, we would have to constrain the domain. You'd say, hey, for all x's not equaling negative two for these to be completely identical expressions.