If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Integrated math 3>Unit 3

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 ^~^
• 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.
• "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."
• 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)
• Anyone give me the link on how to read graphs? Like I can't understand what or where the remainder is