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### Course: Get ready for AP® Calculus > Unit 6

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: missing x-term

An interesting case in polynomial division is when one of the terms is missing. The video explains how to divide a quadratic expression, like (x²+1), by a linear one, such as (x+2). It shows two methods: re-expressing the numerator and using algebraic long division. Both methods lead to the same answer.

## Want to join the conversation?

- how does this help me in real life problems(14 votes)
- Knowing how to divide polynomials is essential in calculus. A lot of unsolvable problems become solvable just because it's possible to divide and reduce them(18 votes)

- I have a question. Where did you get the 5 from in the first example in the video? (At1:11)(10 votes)
- The closest expression we can get to x^2+1 that is also a multiple of x+2 is x^2-4 since x^2-4 has a degree of 2 and doesn't have an "x" term. However, x^2-4 is not exactly the same as x^2+1, so we have to do something to make them the same. We can add 5 to x^2-4 so it is equal to x^2+1. That way, we can manipulate the expression (x^2-1)/(x+2) to look like (x^2-4+5)/(x+2). We can then factor x^2-4 in the numerator as (x+2)(x-2). Now the expression should look something like this:

((x+2)(x-2)+5)/ x+2

You can rewrite this expression as ((x+2)(x-2))/(x-2) + 5/(x+2)

In the first term, the x+2 cancels out so you are just left with x-2

You can't really go any further with the second term, so the answer should be

x-2 + 5/(x+2)

Hope this helped! ^~^(24 votes)

- How is Sal so popular on the street?(17 votes)
- He helps a lot of people (like us) learn math, that's why XD.(2 votes)

- Surely x-2+5/(x+2) is identical to (x^2+1)/(x+2) without declaring it not defined at x=-2, as the former also contains an expression with (x+2) as the denominator.

I understand that you'd want to restrict the domain in the case there was no remainder, but here it's just redundant, or am I mistaken here?(9 votes)- Good catch! Yes, in that case it would be redundant.

He included it here because it's a good habit to get into. If you always restrict the domain, you don't need to think about whether you have to or not. I'm with you though, I wouldn't have included it because the remainder already excludes "x = -2" from being a possibility.(11 votes)

- At2:34why isn't the x+2 also included in the remainder? Shouldn't the remainder be 5/ x+2 ?(2 votes)
- Sal is placing emphasis on what is the number that was left over from the division, which is — your remainder.

If you were writing it down however as an expression overall , yes you are right , you are writing it as 5/(x+2).

Hopefully that helps !(7 votes)

- Wait, in the answer of algebraic long division, where do you put the remainder?(1 vote)
- If you try to divide 28/5, you get 5 with a remainder of 3, so you would either get a decimal (5.6) or like in the video, you get a fraction 5 3/5. So remainder goes on top and what you divided by goes on bottom. Of you are dividing by x + 2 and get a remainder of 5, the last part would be 5/(x+2). Is this what you are asking?(6 votes)

- How does the -2x-4 become 2x+4 at4:00? Is it because of the double negatives?(2 votes)
- Yup ! When doing long division, you subtract to go to the next step.

-(-2x-4) = 2x+4 through distributing. hopefylly tht helps !(3 votes)

- Why do people keep walking up!? Where did they come from!?(3 votes)
- they saw this video and they decided to be a part of the video, so they went back in time, and walked up to you and asked you the question.

anyway, it is just an excuse to present the problem(1 vote)

- What would you do if instead of x+2 it was x-2(2 votes)
- so an easy way of memorizing which way it is: if the dividend has a positive (I.E. x+4) the answer to the polynomial will always be a negative (I.E. x-3), and vice versa.(2 votes)

- do we have to use the difference on squares every time we know we will be diving polynomials with a remainder, and can we use it even if the equation has 2nd degree terms?(2 votes)

## Video transcript

- [Instructor] This
polynomial division business is a little bit more fun than we expected. So let's keep going. So let's say that, I guess again, someone walks up to you in the street and says "What is x squared
plus one divided by x plus two." So pause this video and have a go at that. And I'll give you a
little bit of a warning. This one's a little bit more
involved than you might expect. All right, so there's two
ways to approach this. Either we can try to
re-express the numerator where it involves an x plus two somehow, or we could try to do
algebraic long division. So let me do the first way. So x squared plus one, it's not obvious that
you can factor it out. But can you write something
that has x plus two as a factor, and interestingly enough
has no first degree terms? 'Cause we don't want
some first degree weird first degree terms sitting up there. And the best thing that
I could think of is, constructing a different of
squares using x plus two. So we know that x plus two
times x minus two is equal to x squared minus four. So what is we were to write
x squared minus four up here, and then we would just have to
add five to get to plus one. So what if we were to
write x squared minus four and then we write plus five. This expression and that
expression up there, those are completely equivalent. But why did I do that? Well, now I can write
x squared minus four as x plus two times x minus two. And so then I could rewrite
this entire expression as x plus two times x
minus two, all of that over x plus two plus five,
plus five over x plus two. And now as long as x does
not equal negative two, then we could divide the
numerator and the denominator by x plus two. And then we would be left
with x minus two plus five over x plus two, and I'll
put that little constraint, if I wanna say that this
expression is the same as that first expression, for x does not equal, for x not equaling negative two. And so here, we'd say "Hey!
X squared plus one divided by x plus two is x minus two,"
and then we have a remainder of five, remainder of five. Now let's do the same question,
or try to rewrite this using algebraic long division. We'll see that this is actually a little bit more straightforward. So we are going to divide
x plus two into x squared plus one. Now when I write things out I
like to be very careful with my, I guess you could say,
with my different places for the different degrees. So x squared plus one
has no first degree term, so I'm gonna write the one out here. So second degree, no first
degree term, and then we have a one, which is a zero degree term, or constant term. And so, we do the same
drill, how many times does x go into x squared. We're just looking at the
highest degree terms here. X goes into x squared x
times, that's first degree so I put it in the first degree column. X times two is two x. X times x is x squared. And now we wanna subtract. And so what is this gonna be equal to? We know the x squared's cancel out. And then I'm gonna be
subtracting negative two x from, you could do this
as plus zero x up here plus one, and so you're
left with negative two x. And then we bring down that one plus one. X goes into negative two
x, negative two times. Put that in the constant column. Negative two times two is negative four. And then negative two
times x is negative two x. Now we have to be very
careful here because we want to subtract the negative
two x minus four from the negative two x plus one. We could view it as this
or we could just distribute the negative sign. And then this will be
positive two x plus four. And then, the two x's, the two x and the negative two x cancels out. One plus four is five and
there's no obvious way of dividing x plus two
into five so we would call that the remainder,
exactly what we had before. When we divided with
algebraic long division, we got x minus two, x minus
two with a remainder of five.