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## Algebra 2

### Course: Algebra 2 > Unit 4

Lesson 4: Polynomial Remainder Theorem- Intro to the Polynomial Remainder Theorem
- Remainder theorem: finding remainder from equation
- Remainder theorem examples
- Remainder theorem
- Remainder theorem: checking factors
- Remainder theorem: finding coefficients
- Remainder theorem and factors
- Proof of the Polynomial Remainder Theorem
- Polynomial division: FAQ

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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!

## Want to join the conversation?

- How does the polynomial remainder theorem behaves when you have to deal with polynomials of more than one variable?(37 votes)
- 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(65 votes)

- At1:21, 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.(32 votes)
- 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)(28 votes)

- Can this formula be used when the divisor is a quadratic equation?(7 votes)
- we use x = a in the theorem .............but that suugests we have to divide the polynomial by 0....

as x - a = 0 then(8 votes)- I don't understand your answer but i have the same question! Could you please rephrase?(4 votes)

- Does the theorem only work when the remainder is a constant? What if the remainder is a polynomial having several x terms?(7 votes)
- If the divisor is just (x-a), you can always reduce to a remainder that does not include x. Otherwise, yes it does still apply. You just get f(a) = some function of x.(1 vote)

- Would the same theorem apply to
*f(x)/(x+a)*, with the remainder would be equal to*f(-a)*?(3 votes)- In the polynomial Remainder Theorem, a can be either positive or negative. Since the expressions you referenced in your question are equal to f(x) / (x + (-a) and f(-(-a)), the theorem is true here.(4 votes)

- Just to clarify, The Polynomial Remainder Theorem only works when you are dividing a polynomial by the binomial in the form of x-a?(3 votes)
- Yes. It works for (x + a), because you could rewrite that as (x - (-a)).(3 votes)

- 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)- 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)

- 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)- 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)

- 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)
- 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)

## 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.