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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: 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?

- I have a question. Where did you get the 5 from in the first example in the video? (At1:11)(7 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! ^~^(18 votes)

- how does this help me in real life problems(4 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(8 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?(7 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.(8 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 !(4 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 !(2 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?(4 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)
- Why is it necessary to add missing degree terms?(1 vote)
- Consider the example in the video:

x^2+1/x+2

To begin you would divide x into x^2 right?

And then you would multiply the answer (x) times the divisor (x+2). Your answer would come out as x^2+2x. When you subtract them from the problem, x^2-x^2 cancels out but what would you subtract the 2x from? As in the video, you can imagine that there is a 0x that you would subtract it from in the problem. You don't necessarily*have*to add the missing degree terms but it helps to make the problem more organized.(2 votes)

- Seems pretty similar to completing the square...(1 vote)
- How do we solve it with negatives(1 vote)
- To do Algebraic Long Division with negative numbers you just do it normally but instead of multiplying the divisor by a positive number you multiply it by a negative number. For example, you want to divide x^2+4x+5 by -x-3. It would look something like this.

-x-1 R2 = answer

-x-3/x^2+4x+5 = Problem

-x^2-3x = (-x^2-x)*-x to give you the -x

x+5 = result from subtracting x^2+3x from above

-x-3 = result from (-x-3)*-1 and then subtracting

2 = remainder

basically you are making a positive number so that you when you subtract you are not subtracting a negative number.(1 vote)

## 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.