If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

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?

  • piceratops tree style avatar for user Storm
    I have a question. Where did you get the 5 from in the first example in the video? (At )
    (10 votes)
    Default Khan Academy avatar avatar for user
    • aqualine sapling style avatar for user Selena Bean UwU
      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! ^~^
      (20 votes)
  • aqualine ultimate style avatar for user Christie
    how does this help me in real life problems
    (7 votes)
    Default Khan Academy avatar avatar for user
  • old spice man green style avatar for user masinol
    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?
    (8 votes)
    Default Khan Academy avatar avatar for user
    • mr pants teal style avatar for user VincentTheFrugal
      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.
      (9 votes)
  • marcimus pink style avatar for user Ashna
    At why isn't the x+2 also included in the remainder? Shouldn't the remainder be 5/ x+2 ?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • leaf orange style avatar for user A/V
      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 !
      (5 votes)
  • winston baby style avatar for user Noah Colina
    Wait, in the answer of algebraic long division, where do you put the remainder?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • mr pink green style avatar for user David Severin
      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?
      (5 votes)
  • blobby green style avatar for user earoheen
    I noticed a recurring issue with these courses in that they often explain how to solve a problem that you were already asked to solve in the previous exercise without any introduction.
    (3 votes)
    Default Khan Academy avatar avatar for user
  • leaf red style avatar for user euan
    How does the -2x-4 become 2x+4 at ? Is it because of the double negatives?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • piceratops ultimate style avatar for user Goutham Pedinedi
    How is Sal so popular on the street?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Leon Robert C. Lopez
    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)
    Default Khan Academy avatar avatar for user
  • piceratops ultimate style avatar for user lee.jared.p
    Why is it necessary to add missing degree terms?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user NonExistentUser
      Consider the example in the video:


      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)

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.