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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. For example, (x²+1) divided by (x+2). Learn how to avoid any mistakes in such cases.

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  • 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 )
    (7 votes)
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    • 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! ^~^
      (13 votes)
  • 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?
    (7 votes)
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    • 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.
      (7 votes)
  • 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)
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  • 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)
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    • 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 !
      (2 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)
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    • 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?
      (2 votes)
  • piceratops ultimate style avatar for user lee.jared.p
    Why is it necessary to add missing degree terms?
    (1 vote)
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    • leaf red style avatar for user NonExistentUser
      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)
  • sneak peak blue style avatar for user Stephen Earley
    Seems pretty similar to completing the square...
    (1 vote)
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  • blobby green style avatar for user 4067557
    How do we solve it with negatives
    (1 vote)
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    • piceratops tree style avatar for user Adam N. Chickering
      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)
  • aqualine ultimate style avatar for user Simum
    Why does the difference of squares make sense in reverse?
    (1 vote)
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  • blobby green style avatar for user 100049
    how can I do it in other equation at
    (1 vote)
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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.