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### Course: Algebra 2>Unit 4

Lesson 4: Polynomial Remainder Theorem

# Remainder theorem examples

The polynomial remainder theorem says that for a polynomial p(x) and a number a, the remainder on division by (x-a) is p(a). This might not be very clear right now, but you will understand this much better after watching these examples.

## Want to join the conversation?

• At , may i know why P(-3)=k?
• According to the polynomial remainder theorem, when you divide the polynomial function, P(x), by x-a, then the remainder will be P(a). In this case, we are dividing P(x) by x+3. x+3 can be thought of as x-(-3) and since the value "a" in the polynomial remainder has to be the constant that is being subtracted from x, our "a" value would be -3.

After that, you would plug in -3 into P(x) to find the remainder, so it would be P(-3)=k, where k is the remainder, or the y value in the function in the graph when x=-3

Hope this helps ^~^
• "we are having so much fun" had me
• My response to his comment was, "Oh, no, you are the only one having fun, you schadenfreude."
• I have a question. At , why do we want (x-2) = 0, that is, we are essentially dividing by 0? Isn't anything divided by 0 undefined?
• If you divide a polynomial by (x-a) the remainder to that is the same as solving for f(a), where f is the polynomial. So in other words you plug a in for all xs. In other words, f(a) = the remainder of f(x)/(x-a) = r, so r is both the answer to f(a) and the remainder to that division problem.

I do think Sal mentions (x-2)=0 somewhat mistakenly, though it may have some reason to do with the reason f(a) equals the remainder of f(x)/(x-a), though it is a fact that if you plug in a for x in (x-a) you do get (a-a) which is 0, so maybe he was just making that link to help better remember it.
• if i were to divide a polynomial by 2 values, like (x-2) and (x-1) how would you find the remainder then?
• When you divide by a polynomial of degree n, you get a remainder of degree n-1 or less. So here, your remainder would be of the form ax+b.

If you're dividing a polynomial f(x) by (x-2)(x-1), then f(2)=a·2+b and f(1)=a·1+b. You just need to solve the system to find a and b, and you have your remainder.
• what if we were to divide a polynomial p(x) by a linear equation such as 2x+5??
• A polynomial 𝑝(𝑥) divided by (𝑎𝑥 − 𝑏) will have a remainder of 𝑝(𝑏∕𝑎).
• Why do we have to find the value of x to make (x+3) equal to ZERO (0.40)?
• P(x) = ((divisor)*(quotient)) + (remainder)
P(x) = (x-a)*(q) + r

Now, if we take x = a,

P(a) = (a-a)*q + r
P(a) = 0 + r = r

If you had taken some other value for x like b such that b ≠ a, then you would've gotten some other answer.

Hope that helps..
• Why is it that when (x-3), it's p(3).
And while is (x+3), it's (p-3)?
• In the last example at how do we know that (x-0) or (x+0) will evaluate to 0?
I would have expected the answer to be -x because that would cancel the x.
• At around , he says that p(2)/(x-2) gives a remainder of 1. How is this so? Aren't we substituting x for p(x)?
• That is correct that for that given example p(2)/(x-2) does give a remainder of 1. However, the remainder represents what the p(a) should evaluate to when we substitute "x" for "a".
(1 vote)
• x+0 or x-0 cuz I don't see how x+0 works
(1 vote)
• 0 is neither positive nor negative, so x + 0 and x - 0 both works.

## Video transcript

- [Instructor] So we have the graph here of Y is equal to P of x, I could write it like this, Y is equal to P of x. And they say, what is the remainder when P of x is divided by x plus three? So pause this video and see if you can have a go at this and they tell us your answer should be an integer. So, as you might have assumed, this will involve the polynomial remainder theorem and all that tells us is that hey if we were to take P of x and divide it by x plus three, whatever the remainder is here, so we'll say the remainder is equal to k. That value k is what we would have gotten if we took our polynomial and we evaluated it at the value of x that would have made x plus three equal zero or just what would have happened if I evaluated our polynomial as x equals negative three. You have to be very careful there sometimes people get confused. They see a positive three and then they evaluate the polynomial at the positive three to figure out the remainder. No! If you saw a positive three there, you would evaluate the polynomial at negative three but this should be equal to k as well. And so what is the remainder when P of x is divided by x plus three? Well it's going to be equal to P of negative three. P of negative three it looks like it is equal to negative two. It is equal to negative two, so our remainder is equal to negative two in this situation. Let's do another example, actually let's do several more examples. Here we're told that P of x is equal to all of this business where k is an unknown integer, very interesting. P of x divided by x minus two has a remainder of one. What is the value of k? So pause this video again, see if you can work it out. All right, well this second sentence that P of x divided by x minus two has a remainder of one, that tells us that P not of negative two but P of positive two, whatever x value would make this expression equal zero. That P of two is equal to one. And then we could use this top information to figure out what P of two would be. It would be two to the fourth power minus two times two to the third power plus k times two squared. So, times two squared minus 11, and so all of that, that's P of two right over here that's going to be equal to one. Two to the fourth is 16 and then two times two to the third that's two to the fourth again, so it's minus 16 plus four k minus 11 is equal to one, these cancel out. Now let's see we can add 11 to both sides of this equation and we get four k is equal to 12. Divide both sides by four and we get k is equal to three and we're done. Let's do another example, in fact let's do two more because we're having so much fun. So this next question tells us, P of x is a polynomial and they tell us what P of x divide by various things are, what the remainder would be when you divide P of x by these various expressions. Find the following values of P of x, P of negative four and P of one. Pause this video and see if you can have a go at it. All right, so P of negative four, this is going to be equal to the remainder when P of x divided by what. You might be tempted to say x minus four but they're trying to trick you intentionally. This would be the remainder when P of x is divided by x plus four. And so they tell us right over here P of x divided by x plus four has a remainder of three. So, it's going to be three right over there. And similarly P of one, this is going to be the remainder, this is the remainder when P of x divided by not x plus one, but x minus one. So when P of x is divided by x minus one, the remainder is zero. Let's do one last example. So once again P of x is a polynomial and then they give us a few values of P of x. And they say what is the remainder when P of x is divided by x minus three? Pause the video and try to think about that. Well, we've gone over this multiple times, the remainder when P of x is divided by x minus three that would be P of not negative three, P of positive three. Whatever value of x makes this entire expression equal zero. So P of positive three is equal to five. And similarly what is the remainder, actually not so similarly, this is interesting. What is the remainder when P of x is divided by x? I know what you're thinking it's like wait what number am I dealing with? But if I were to rewrite this instead of saying divided by x, if I were to say divided by x plus zero then you'd be like oh now I get it. Or if I wrote divided by x minus zero, you'd be like oh, now I get it. This is going to be P of and it doesn't matter whether I take a positive or a negative zero, it's going to be P of zero. And P of zero, they tell us, is negative one and we're done.