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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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# Remainder theorem: finding remainder from equation

CCSS.Math: ,

Sal finds the remainder of (-3x^3-4x^2+10x-7) divided by (x-2) using the PRT (Polynomial Remainder Theorem).

## Want to join the conversation?

- Where did the 8 come from?(53 votes)
- He was just plugging in
*a_ (in this case, 2) for _x*.

The first term of the equation is -3_x_^3. When _x_ is plugged in, it is -3*2^3, which is also -3*8.(74 votes)

- Does the polynomial theorem only work for x-a and not x+a?(23 votes)
- If you have (x+a), then the "a" value will actually be "-a" because (x+(-a))=(x-a).(14 votes)

- This question was asked a few comments down but I did not find the answer very satisfying. What is the significance of the remainder? In what situation would I need the remainder of a polynomial instead of both its quotient and remainder?(8 votes)
- If the remainder is 0, then you know that the divisior is a factor of the dividend (they are divisble).(23 votes)

- Would this work if the the coefficient on the x term was greater than 1?(5 votes)
- I think that it could work. Another way to find the remainder is to set the x - a to term equal to 0 and then solve for x. After this, you just plug it back in to find the remainder. Correct me if I am wrong.(11 votes)

- How can a remainder be negative?(6 votes)
- With polynomial division, you can get remainders that are negative.

In numeric long division, you would not have a negative remainder.(7 votes)

- what do you mean when someone says f(x)?(3 votes)
- Function of x is basically a processing machine. You give it an input, and it gives out one output and one only. For example if f(x) = x+1, you can substitute any number for x and find the output. So f(5)= 5+1 = 6. In some ways they are similar to equations and in some ways they are different. I don't know too much about it myself, but here's the link: https://www.khanacademy.org/math/algebra/algebra-functions(8 votes)

- After all this process, should the remainder be expressed as -27/(x-2)??(5 votes)
- It depends on the context of the problem (if it asks for remainder
*term*); but usually, and in this problem, it should be expressed as -27, according to the definition of remainder.(5 votes)

- Would the resulting remainder be just -27 or would it be (-27/x-2)? I'm really confused about this.(2 votes)
- Think about when you're dividing normal numbers that don't go into each other easily, like 9 / 4. In this problem, the remainder would only be 1. However, if you were to write the quotient out, it would be 2 + 1/4. It's the same way with polynomials. When talking about the quotient, you have to write the remainder as a fraction, but when just asked to identify the remainder, you can just give the remainder itself.(5 votes)

- How can I interpret a negative remainder?(3 votes)
- um, what would happen if "a" is a positive number? like "x + 2" instead of "x - 2"? Should I then treat "a" like a negative number?(2 votes)
- If the sign is positive, like in "x +2", then a
*is*a negative number. The formula is "x - a". So, if a is positive, the sign just stays negative like in "x - 2". But if a is negative, then it becomes "x - (-2)" which is the same as "x + 2". So, yes. When plugging it in, a is a negative number in the case of "x + 2"(5 votes)

## Video transcript

- So we have a polynomial here. What I'm curious about
is what is the remainder if I were to divide this polynomial by, let's just say, x minus,
I want the remainder when I divide this
polynomial by x minus two? You could do this. You could figure this out with algebraic long division, but I'll give you a hint. It is much simpler and
much less computation intensive and takes much less space on your paper if you use the polynomial remainder theorem. If that's unfamiliar to
you, there's other videos that actually cover that. So why don't you have a go at it. All right, so now let's
work through this together. The polynomial remainder theorem tells us that when I take a polynomial, p of x, and if I were to divide
it by an x minus a, the remainder of that is just going to be equal to p of a. Is just going to be equal to p of a. So in this case, our p of x is this. What is our a? Well our a is going to be positive two. Remember it's x minus a. So let me do this. Our a is equal to positive two. So to figure out the remainder, we just have to evaluate p of two. So let's do that. So the remainder in this case is going to be equal to p of two, which is equal to, so let's see, it's going to be, I'll
just do it all in magenta, negative three times
eight minus, let's see, minus four times four plus 20 minus seven. So let's see, this is -24 minus 16 plus 20 minus seven. So that gives us, let's see, -24 minus 16, this is -40. All right, I'm just doing it step by step. This is equal to negative,
actually I can do this in my head. All right, here we go. So this is -40 plus 20 is
-20 minus seven is -27. That was pretty neat
because if we attempted to do this without the
polynomial remainder theorem, we would have had to do a bunch of algebraic long division. Now if we did the algebraic long division, we would have gotten the
quotient and all of that, but we don't need the quotient, we don't need to know. So if we did all the
algebraic long division, you know, we would have taken our p of x and then we would have
divided the x minus a into it, and we would have
gotten a quotient here, q of x, and we would have done all this business down here, all this
algebraic long division. Probably wouldn't have
even fit on the page. But eventually we would
have gotten to a point where we got an expression that has a lower degree than this. It would have to be a
constant because this is a 1st degree, so it
would have be essentially a zero degree. So we would have eventually
gotten to our -27. But this was much, much, much, much easier then having to go through
this entire exercise. Hopefully you appreciated that.