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## Integrated math 3

### Course: Integrated math 3 > Unit 3

Lesson 2: Dividing quadratics by linear factors- Intro to long division of polynomials
- Dividing quadratics by linear expressions (no remainders)
- Divide quadratics by linear expressions (no remainders)
- Dividing quadratics by linear expressions with remainders
- Dividing quadratics by linear expressions with remainders: missing x-term
- Divide quadratics by linear expressions (with remainders)

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# Dividing quadratics by linear expressions with remainders

Learn how to simplify complex math problems by dividing quadratics by linear factors with remainders. Discover two methods for simplifying these kinds of math problems: factoring the numerator or using algebraic long division. This process helps break down tricky equations, making them easier to solve.

## Want to join the conversation?

- If someone came up to me on the street and told me to do a random math problem, I would probably tell them off.(12 votes)
- In this question, would the remainder just be 2, or would it be (2/x+2)? Because there is a difference between the two, and it seems to me that (2/x+2) would make more sense, but in the video, Sal says 2 is the remainder.(5 votes)
- You're right, I guess sal was assuming everyone knew that when you find the remainder with long division it isknown that you take the result and divide it by the divisor. The first time he got the answer by messing witht he numerator, or dividend if you wanna call it that, and canceled out the terms. He was just showing another way to do it.(8 votes)

- is there a video where he divides by (AX+-B) where a and b are constants and x is the variable. If so, may someone please tell me the title so I can see it( I can't manage to find it. And if not, it would be a good topic to put into video form(5 votes)
- The formula you just wrote is exactly the one of linear equations! you might want to search ''Linear equations'' seach bar. Sal has some REALLY old videos on that, you might start your search from there(3 votes)

- I like how the guy says "Quick" and it takes 4 minutes to do the problem. Anyone agree with me?(6 votes)
- I was wondering about when Sal gets the remainder. Why can't he divide the 2 from the numerator with the 2 in the denominator?(4 votes)
- so in 2/(x+2)? Wellyou would need some multiple of 2 multiplying each part of the denominator. I assume you are saying you would simplify 2/(x+2) to 1/(x+1) Let me know if that is incorrect in my assuming.

before aything else, f you check these two exprressions you will see they are not the same. Canceling out numbers int he numerator and denominator will always lead to the same result. for instance if you had 2/4 you could divide both numerator and denominator by 2 and get 1/2. Both fractions are the same. Or with functions, if you had x/(x^2) this could be rewritten as 1/x. You get the same number no matter what x you plug in.

This is actually something important to know. dividing something fromt he numerator or denominator means yu have to divide the whole numerator and denominator. with 2/(x+2) if you divided a 2 from both you would get 1/(x/2 + 1). Let me know if that didn't help or was not what you were asking.(3 votes)

- At2:18, Sal cancels out the denominator of his fraction with an identical part of the numerator, ex. (x+2)(x+3)/(x+2). However, just a few videos ago, he broke up a fraction so that each part of the numerator was divided by the denominator seperately, ex. (18x^4-3x^2+6x-4)/6x, (18x^4/6x)+(-3x^2/6x)+(-4/6x). When the equation (x+2)(x+3)/(x+2) is seperated this way, a different answer is found from Sal's. I'm just not sure what about the situations differed. Was it maybe that in the first situation, (x+2)(x+3)/(x+2), the numerator is multiplied instead of added, (18x^4-3x^2+6x-4)/6x? If so, why did that alter the method?(3 votes)
- Can you show me how you separated (x+2)(x+3)/(x+2) because I feel as though there may have been a slip up.

Anyway, you are basically 100% correct about the multiplication and addition making a difference. If you expanded out the (x+2)(x+3) you could use the then divide each term.

(x+2)(x+3)

x^2 + 2x + 3x + 6

x^2 + 5x + 6

So then (x^2 + 5x + 6)/(x+2)COULD be rewritten as x^2/(x+2) + 5x/(x+2) + 6/(x+2).

The point here is if you have two terms being multiplied and then divide it by a third term it is actually like a third multiplication, just one over it. So for example

(x*y)/z = x*y*(1/z) or x*y*z^-1

Meanwhile if you have two terms added or subtracted and you divide them all by a third term it distributs just like multiplication.

(x-y)/z = x/z-y/z

Again, because you can rewrite a denominator as multiplication.

(x-y)(z^-1) = xz^-1 - yz^-1 = x/z - y/z(4 votes)

- for the remainder, could i write (x+3)(x+2)+2?(2 votes)
- I'm pretty sure you can. That's what I would do. 😉(4 votes)

- Does math have any other thing besides variables and numbers?(2 votes)
- Mathematics is the concept of using variables and numbers in a mathematical context. There are therefore functions and other concepts that help you use those variables and numbers.(4 votes)

- Can we write things down in the form of mixed fractions? (x+3) and 2/(x+2). After all, we are getting some remainder.(2 votes)
- Yes, that is how Sal originally writes it at2:41. He continues to show how to divide to get this answer.(3 votes)

- Would anyone else be concerned if someone just walked up to you on the street asking you to do math problems?(3 votes)
- Very much. Unless they had a camera man and a microphone, then no.(1 vote)

## Video transcript

- [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.