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Proof of the Polynomial Remainder Theorem

The PRT (Polynomial Remainder Theorem) may seem crazy to prove, but Sal shows how you can do it in less than six minutes!

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  • leaf blue style avatar for user maxsisco
    How does the polynomial remainder theorem behaves when you have to deal with polynomials of more than one variable?
    (34 votes)
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    • aqualine seed style avatar for user Sobhan.Bihan
      This theorem has not been extended to divisions involving more than one variable. A more general theorem is:
      If f(x) is divided by ax + b (where a & b are constants and a is non-zero), the remainder is f(-b/a).
      Proof:
      Let Q(x)be the quotient and R the remainder.
      f(x) = Q(x)*(ax+b) + R
      Substituting the solution of 0 = ax + b we have
      f(-b/a) = Q(x)*(a*(-b/a)+b) + R
      f(-b/a) = Q(x)*(-b + b) + R
      f(-b/a) = Q(x)*0 + R
      f(-b/a) = R
      (62 votes)
  • leafers ultimate style avatar for user Goss, Jerry
    At , why isn't the remainder placed over the divisor as it is in other videos? Is there a difference between the two answers or the way they're are written? Such as having added (6/(3x-1)) or having added 6.
    (29 votes)
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    • spunky sam blue style avatar for user Jack
      It's because if you move (x - 1) from the right side of the formula to the left by dividing, you also have to divide 6 by that monomial. The equation then changes to:
      (3x^2 - 4x + 7)/(x - 1) = (3x - 1) + 6/(x - 1)
      (24 votes)
  • leafers seed style avatar for user Ritikraj Arya
    Can this formula be used when the divisor is a quadratic equation?
    (7 votes)
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  • mr pink red style avatar for user Faiz iqbal
    we use x = a in the theorem .............but that suugests we have to divide the polynomial by 0....
    as x - a = 0 then
    (7 votes)
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  • starky sapling style avatar for user Hodorious
    Does the theorem only work when the remainder is a constant? What if the remainder is a polynomial having several x terms?
    (7 votes)
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  • starky ultimate style avatar for user Levi  Jesus
    Would the same theorem apply to f(x)/(x+a) , with the remainder would be equal to f(-a)?
    (3 votes)
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  • starky ultimate style avatar for user Alessandro V. Santoro
    I just thought of something else.

    In Wikipedia it says that the polynomial remainder theorem states that the remainder of the division of a polynomial f(x) by a linear polynomial x - r...

    That's my question. Does the polynomial remainder theorem only work when the divisor is a linear polynomial? Is it possible for it to be a quadratic or something else? Or the dividend has to change (for example be of higher degree) for the theorem to work for quadratic polynomials (or other higher-degree polynomials).

    Thanks again in advance for any answers and for the work of finding/thinking of an answer ^^
    (3 votes)
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    • leaf orange style avatar for user Nayef
      I remember when I first started learning remainder theorem, these were the initial questions that used to wander in my mind. So here's the answer:
      The reminder theorem is only true when the divisor is a linear polynomial. That means it cannot be utilized when the divisor is something else and if the degree of the divisor polynomial is more than 1 , the sole way to find the remainder is polynomial long division. However if you are able to reduce the divisor polynomial to linear polynomial. Then you can use the remainder theorem.
      (3 votes)
  • starky ultimate style avatar for user Alessandro V. Santoro
    Hello everyone!
    I have already asked many questions here but I really want to know the reason for things. I appreciate every answer I get from you guys, thanks ^^

    My question, which I can already guess must be pretty obvious, is why in the x - r the r has to be a negative r? I don't understand why. Probably it would be understandable with something that was taught in some lesson I haven't seen.

    If you can answer me this question, I'd really appreciate it. Or if you cant tell me on what lesson - if there is one - he explains the reason for this.

    Thanks again, and everyone have a great weekend ^^
    (3 votes)
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    • mr pink green style avatar for user David Severin
      It does not HAVE to be negative, it could be either one. There is a good reason that we often choose the negative. If we have x-a=0, we end up with x=a. If it were positive, we would end up with x=-a which is not as neat. We see this all the time such as in vertex form of quadratic y=a(x-h)^2+k or in general with functions such as given a function f(x), g(x)=f(x-2) shifts the function two units to the right (positive direction) and g(x)=f)(x+2) shifts it to the left.
      (2 votes)
  • blobby green style avatar for user Alexander Wang
    For the long division, when you got 6 as a remainder, isn't it supposed to be 6/x-1 for the remainder?
    (3 votes)
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    • leaf orange style avatar for user A/V
      In equation form, yes you are correct ! But the theorem only account for the number on the numerator (otherwise the remainder of the division), and you have to add the denominator by yourself.
      (2 votes)
  • primosaur tree style avatar for user Collin Bluethmann
    I did not understand what the proof was. Will someone please explain?
    (1 vote)
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    • aqualine tree style avatar for user rex.xiaowen
      f(x)=q(x)*(x-a)+r.
      f is a polynomial,q is the quotient when f is divided by (x-a), (x-a) is what it is being divided by, and r is the remainder.
      We plug in a as x:
      f(a)=q(a)*(a-a)+r
      (a-a) is 0, so that leads us to:
      f(a)=q(a)*0+r
      Anything multiplied by 0 is 0, so that leads us to:
      f(a)=0+r
      Anything added to 0 is that thing, and the home stretch!
      f(a)=r
      (5 votes)

Video transcript

- [Voiceover] Let's now do a proof of the polynomial remainder theorem. Just to make the proof a little bit tangible, I'm going to start with the example that we saw in the video that introduced the polynomial remainder theorem. We saw that if you took three x squared minus four x plus seven and you divided by x minus one, you got three x minus one with the remainder of six. When we do polynomial long division, how do we know when we've got to our remainder? Well when we get to an expression that has a lower degree than the thing that is the divisor, the thing that we're dividing into the other thing. So in this example we could have re-written what we just did right over here as our f of x. Let me just write it right over here. So we could have said three x squared minus four x plus seven is equal to x minus one times the quotient right over here, or I could say the quotient times x minus one. So it's going to be equal to this business. It's going to be equal to three x minus one times the divisor, times x minus one. When you multiply these two things, you're not going to get exactly this. You still have to add the remainder. So plus the remainder. Actually, let me write the actual remainder down. So plus six. The analogy here is exactly the analogy to when you did traditional division. If I were to say, actually, let me just show you the analogy. If I were to say 25 divided by four, you would say, okay, well four goes into 25 six times, six times four is 24. You would subtract, and then you would get a remainder, one. Or another way of saying this is you could say that 25 is equal to six times four plus one. So we just did the exact same thing here, but we just did it with expressions. So once again, I haven't started the proof yet, I just wanted to make you feel comfortable with what I just wrote right over here. If I divided this expression into this polynomial, and I were to get this quotient, that's the same thing as saying this polynomial could be equal to three x minus one times x minus one plus six. Now this is true in general. Let's abstract a little bit. This is our f of x. So that is f of x. So f of x is going to be equal to whatever the quotient is. Let me call that q of x. So do this in a different color. So I'm going to call that q of x. This right over here is q of x. So f of x is going to be equal to the quotient, q of x, times, this is our x minus a, in this case a is one, but I'm just trying to generalize it a little bit. So x minus a, and then plus the remainder. We know that the remainder is going to be a constant because the remainder is going to have a lower degree then x minus a. X minus a is a first degree. So in order to have a lower degree, this has to be zeroth degree. This has to be a constant. So this is true in general. This is true for any polynomial f of x divided by any x minus a. So this is just true. So this is true for any f of x and x minus a. Now what is going to happen if we evaluate f of a? Well if f of x can be written like this, well we could write f of, let me do a in a new color so it sticks out. We could write f of a would be equal to q of a times, I think you might see where this is going, times a minus a plus r. Well what's that going to be equal to? What's all this business going to be equal to? Well a minus a is zero, and q of a, I don't care what q of a is, if you're going to multiply it by zero all of this is going to be zero. So f of a is going to be equal to r. And so you're done. This is the proof of the polynomial remainder theorem. Any function, if when you divide it by x minus a you get the quotient q of x and the remainder r, it can then be written in this way. If it's written in this way and you evaluated at f of a and you put the a over here, you're going to see that f of a is going to be whatever that remainder was. That is the polynomial remainder theorem. And we're done. One of the simpler proofs that exists for something that at first seems somewhat magical.