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### Course: Integrated math 3 > Unit 3

Lesson 1: Dividing polynomials by x# Polynomial division introduction

When we divide the polynomial p(x) by q(x) we are basically asking "what should we multiply by q(x) to get p(x)?" If this sounds familiar, it's because it's very similar to dividing numbers! In this introduction we see how some quotients end up as a polynomial, while other times we have a remainder and cannot express the quotient as a polynomial. This is very similar to quotients of integers!

## Want to join the conversation?

- hi everyone,you CAN DO THIS!(37 votes)
- This is supposed to be in Tips and Thanks.(0 votes)

- I am confused by the conclusion that x cannot = -1. Lets say we STARTED the function f(x) = x+2. There would be no reason to not use -1. I understand that in this example we arrive at x-2 by simplifying a function that WOULD have a problem with x = -1, BUT whether or not we start with the unsimplified function and reduce it to x+2 or just come up with x+2 to start with, the end resulting graph of f(x) = x+2 would look the same, right?(6 votes)
- Not exactly. This is where the concept of "holes" comes in. Basically a single point on a graph that doesn't exist. Here is a super simple example.

x/x simplifies to 1. Of course tht is just a horizontal line at y = 1. since we started with x/x thee is actually a hole at x=0. So it is actually pretty important what you start with. The starting expression determines your domain, even if things cancel out. If you have a graphing calculator you will be able to see that if you graph x/x and try to tell what x=0 is you will get that it is undefined, while everywhere else it is 1.(17 votes)

- what grade do most students learn this in?(4 votes)
- I think in 10th, maybe 11th this is alg. 2(7 votes)

- Still don’t fully understand this, any ‘better/easier to understand’ formulas for finding the answers with polynomial division, maybe an easier formula.(6 votes)
- i am answering a problem(devide each polynomial by the given binomial.

(64x³-192x²+5x-15)÷(x-2)

how do i solve this?(3 votes) - So is this similar to factoring polynomials?(5 votes)
- The first step he shows you, factoring would be needed to solve it and im a bit late sry(0 votes)

- he used long division to solve it, will we be learning that later? I don't really understand it.(3 votes)
- Yes, in lesson 2.(3 votes)

- at1:12how is it factorized(3 votes)
- Am I in the right place for Mrs. Green’s class? I can’t tell if I’m doing the right lessons and it’s giving me a lot of anxiety.(3 votes)
- How do we know we can assume x is not 0?(2 votes)
- We can't, x can be 0 but the expression should not be. Since dividing by 0 is a challenging concept - see here

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:foundation-algebra/x2f8bb11595b61c86:division-zero/v/why-dividing-by-zero-is-undefined(2 votes)

## Video transcript

- [Instructor] We're already familiar with the idea of a polynomial and we've spent some
time adding polynomials, subtracting polynomials, and multiplying polynomials, and factoring polynomials. And what we're going to
think about in this video and really start to
think about in this video is the idea of polynomial division. So, for example. If I had the polynomial, and this would be a quadratic polynomial, let's say x squared plus three x plus two, and I wanted to divide it by x plus one. Pause this video and think about what would that be? What would I have to
multiply x plus one by to get x squared plus three x plus two? Well, one way to approach it is we could try to factor x squared plus three x plus two and we've done that
multiple times in our lives. We think about well, what
two numbers add up to three and if I were to multiply them I get two. And the one that might jump, or the ones that might jump out at you are two and one. And so we could express x squared plus three x plus two as x plus two times x plus one, and then all of that is going to be over x plus one. And so, if you were to take x plus two times x plus one, and then divide that by x plus one, what is that going to be? Well, you're just going to be left with an x plus two. This is going, you don't
have to put parentheses, this is going to be an x plus two. And if we want to be really
mathematically precise we would say hey, this would be true as long as x does not equal, x does not equal negative one, because if x equals negative one in this expression or this expression we're gonna be dividing by zero. And we know that leads to all sorts of mathematical problems. But as we see, for any other x as long as we're not
dividing by zero here, this expression is going to be the same thing as x plus two, and that's because x plus two times x plus one is equal to what we have in this numerator here. Now, as we go deeper
into polynomial division, we're going to approach things that aren't as easy to do just purely through factoring. And that's where we're
gonna have a technique called polynomial long division. Polynomial long division, sometimes known as
algebraic long division. And if it sounds familiar, because you first learned
about long division in fourth or fifth grade, it's because it's a very similar process where you would take your x plus one and you would try to divide it into your x squared plus three x plus two. And you do. Something very, and I'm
gonna do a very quick example right over here, but we're gonna do much more detailed examples in future videos, but you look at the highest degree terms. You say okay, I have a first degree term, I have a second degree term here. How many times is x going to x squared? Well, it goes x times. So you put the x in
the first degree column and then you multiply
your x times x plus one. X times x is x squared. X times one is x. And then you subtract this from that. So you might already start to see some parallels with the long division that you first learned
in school many years ago. So when you do that, these cancel out, three x minus x. We are left with a two x. And then you bring down that two. So two x plus two. And you say how many times does x go into two x? Well, it goes two times. So you have a plus two here. Two times x plus one. Two times x is two x. Two times one is two. You can subtract these and then you are going to be left with nothing. Two minus two is zero. Two x minus two x is zero. So in this situation it divided cleanly into it and we got x plus two, which is exactly what we had over there. Now, an interesting scenario that we are also going to approach in the next few videos, is what if things don't divide cleanly? For example, if I were to add one to x squared plus three x plus two, I would get x squared
plus three x plus three. And if I were to try to divide that by x plus one, well, it's not going to divide cleanly anymore. You could it either approach. One way to think about it, if we know we can factor x squared
plus three x plus two is say hey, this is the same thing as x squared plus three x plus two plus one, and then all of that's
going to be over x plus one. And then you could say hey, this is the same thing as x squared plus three x plus two over x plus one, over x plus one, plus one over x plus one. Plus one over x plus one. And we already figured out that this expression on the left, as long as x does not equal negative one, this is going to be equal to x plus two. So this is going to be
equal to x plus two, but then we have that one that we were able to
divide x plus one into, so we're just left with
the one over x plus one. And we'll study that in a lot more detail in other videos. What does this remainder mean and how do we calculate it if we can't factor part of what we have in the numerator? And as we do our polynomial long division we'll see that the remainder will show up at the end when we are done dividing. We'll see those examples in future videos.