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## Algebra 2

### Course: Algebra 2 > Unit 4

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 (no remainders)

CCSS.Math: ,

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?

- I was wondering if Khan Academy had a video on the foil method. It is confusing me.

Thanks(5 votes)- FOIL is a method for multiplying binomials. It stands for First Outside Inside Last.

For example:

(x+1)(x+1)

= (x)(x) [First] + (x)(1) [Outside] + (1)(x) [Inside] + (1)(1) [Last]

= x^2+2x+1(14 votes)

- Did anyone else notice he spelled Khan wrong at0:41?(7 votes)
- Correct, but Sal did not spell anything wrong, it was Youtube. They auto generate those based on the sound, so people who cannot hear, will be able to watch his videos, or other videos. Because it is automated, it does not always have the correct word.(6 votes)

- I just found a really cool way to solve this!! Bc of the binomial therom, we know that the answer is going to be a binomial. We set our equation up like this: (x+2)(a+-b), and then work from there. x multiplied by what = x^2? x. So a = x. our equation is now (x+2)(x+-b), or when distributed, x^2+-bx+2x+-2b. Because the last term in our trinomial was ten, we know that b has too equal to 5, as 2*5 = 10. We can even check our work by putting in 5 for b in bx, and you get 5x+2x = 7x! Has this method been discovered before?(7 votes)
- Why did you assume that we multiply (x+2) with a polynomial of two terms (a+b)? why not (a+b+c) for example?(0 votes)

- Is factoring the expression faster/easier?(5 votes)
- Really it's whichever you prefer, and some may be easier one way for you and others would be easier the other.(3 votes)

- Incredibly helpful video. I was just wondering why the answer at0:58is a positive, and if it was a negative, why not a negative?(4 votes)
- I'm not quite sure what you mean. Would you be able to further explain?(1 vote)

- i dont understand the -2 rule.what does he mean a1:44(2 votes)
- Do not know about what you mean by rule, but it had to do with the whole conversation before that. From the original expression, x cannot equal -2 because that would create a place where the expression would be undefined because it would mean we are dividing by 0 (there are always discontinuities/or vertical asymptotes if this were a function y=) where the denominator is equal 0. Just because we can factor and cancel that out later, it does not change the fact that this discontinuity is present even with the expression x+5.(2 votes)

- Why do people keep walking up to me asking me these questions. Like, I don't know you, go away please.(2 votes)
- How do we determine when to cancel and when to distribute the division?

Ok I have maybe a silly question but I want to clarify this. Sal said that we can distribute a division to all of the terms in the numerator but this this particular video we cancel the`x+2`

with the`x+2`

denominator.

Why aren't we destributing like the previous time?

Is it because we have the same terms both in the denominator and numerator?(2 votes) - why below 7x is 2x and not 1x2:41(1 vote)
- The division x^2+7x+10 : x+2 returns

1. x, for the first term -> = x

2. to subtract x^2 one also has to take the whole

term into consideration

3. x*(x+2) = x^2 + 2x

Sal subtracts the dividend to get the result of the remainder, that luckily for us turns out to be a nice factorization without a rational remainder.(1 vote)

- Well it all seems very straightforward to add a domain, because you cannot divide by 0, but if that even would be the case, wouldn't that make the first fraction undefined in the first place? Cause it also contains x+2 in the denominator?(1 vote)

## 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.