Main content

### Course: Algebra (all content) > Unit 10

Lesson 24: Polynomial Remainder Theorem# Remainder theorem: checking factors

Learn how to determine if an expression is a factor of a polynomial by dividing the polynomial by the expression. If the remainder is zero, the expression is a factor. The video also demonstrates how to quickly calculate the remainder using the theorem.

## Want to join the conversation?

- At1:17, why didn't he take the (-) sign? Is there a reason?(15 votes)
- He didn't take the negative sign because you take the number that when you add it to that number, you should get zero. In this case, what plus -3 = 0?

+3 would be the answer(15 votes)

- in School i am told to say this as FACTOR THEOREM(6 votes)
- Factor Theorem is a special case of Remainder Theorem. Remainder Theorem states that if polynomial ƒ(x) is divided by a linear binomial of the for (x - a) then the remainder will be ƒ(a). Factor Theorem states that if ƒ(a) = 0 in this case, then the binomial (x - a) is a factor of polynomial ƒ(x).(30 votes)

- I have a few questions.

If its in the form (x-a), do higher powers of x work? would (x^2-a) work?

Why isn't the negative carried over into the operation? Why does Sal take the absolute value of a?

Does it then matter if I use (x-a) or (x+a)? If I take the absolute of either, it would still give me a right?(4 votes)- In this case, the factor (x-3) is solved as x=3 when you add 3 to both sides to get x alone. It is possible to do this with higher powers of x, namely to the power of 2. When you solve for a factor such as (x²+a), you get x=plus or minus the square root of a. The plus or minus comes from the basic idea that any real number to the power of 2 is positive (though this way of solving DOES also work for factors with imaginary numbers). To solve a factor with this condition, you have to use synthetic division (at least, as far as I know). You can use a graphing calculator if you need to, but you set up for synthetic division and use plus or minus square root of a to divide. You will divide twice, though, once with positive square root of a and THEN with negative square root of a (using the answer from the first division). It does not matter whether you divide with the positive or negative first. Either way, you will either end up with a remainder or you won't; and this will properly prove if the factor is or is not a factor of the polynomial. I apologize if this was difficult to understand; I am far better at explaining thing visually and through solving for proof.(5 votes)

- how do you get 2x^4 to 81?(3 votes)
- We don't.

We evaluate 2𝑥⁴ for 𝑥 = 3, which gives us

2𝑥⁴ = 2 ∙ 3⁴ = 2 ∙ (3 ∙ 3 ∙ 3 ∙ 3) = 2 ∙ 81(4 votes)

- Whats the differencse between remainder theorem v.s factor theorem(3 votes)
- Actually the factor theorem is a special case of the remainder theorem. The remainder theorem states more generally that dividing some polynomial by x-a, where a is some number, gets you a remainder of f(a). The factor theorem is more specific and says when you use the remainder theorem and the result is a remainder of 0 then that means f(a) is a root, or zero of the polynomial. Or I guess more accurately it states that if (x-a) gets a remainder of 0 then you can factor the polynomial into the answer of the division times (x-a)

I hope that made sense.(3 votes)

- What if it was like(2x-30) and you had to see if it was a factor what would you do to the 2x(3 votes)
- The goal is to get the form of (x-a), with the coefficient of x being 1. Divide everything by 2 in the parenthesis only,

2x-30/2 = (x-15)

Another way of thinking about it is setting is equal to 0.

2x-30=0

2x=30

x=15

And then think how we can make x equal to zero.

(x-15)

Hopefully that helps !(3 votes)

- is this also known as the factor theorem?(3 votes)
- yes we can.check if it is a factor using the remainder theorem(1 vote)

- Would the main use of the polynomial remainder theorem be to test if something is a factor of a polynomial?(1 vote)
- I think it is an application of the polynomial remainder theorem to test for factors, but it isn't the "only" use or the "main" use, like many other things in math.(2 votes)

- how do I submit my practice(1 vote)
- As long as you are logged in when you do the practice exercises, the system keeps track of your work automatically when you check your answers.(2 votes)

- How come that here a division with zero is allowed? He divides p(x) with x-3 and assumes x=3.(1 vote)
- The answer ends in 0 but he doesn't divide anything by 0(2 votes)

## Video transcript

- [Voiceover] So we're asked, Is the expression x minus three, is this a factor of this
fourth degree polynomial? And you could solve this by
doing algebraic long division by taking all of this business and dividing it by x minus three and figuring out if you have a remainder. If you do end up with a remainder then this is not a factor of this. But if you don't have a
remainder then that means that this divides fully
into this right over here without a remainder which
means it is a factor. So if the remainder is equal to zero, the remainder is equal to zero, if and only if, it's a factor. It is a factor. And we know a very fast way of calculating the remainder of when you take some polynomial and you divide it by a first
degree expression like this. I guess you could say when you divide it by a first degree polynomial like this. The polynomial remainder theorem, the polynomial remainder theorem tells us that if we take some polynomial, p of x and we were to divide it by some x minus a then the remainder is
just going to be equal to our polynomial evaluated at our polynomial evaluated at a. So let's just see what's a in this case. Well in this case our a is positive three. So let's just evaluate our
polynomial at x equals 3, if what we get is equal to zero that means our remainder is zero and that means that x
minus three is a factor. If we get some other remainder that means well we have
a non-zero remainder and this isn't a factor,
so let's try it out. So, we're gonna have, so I'm just gonna do it all in magenta. It might be a little
computationally intensive. So it's going to be two times
three to the fourth power, three to the fourth, three
to (mumbles), that's 81. 81. Minus 11 Yeah, this is gonna get a
little computationally intensive but let's see if we can power through it. 11 times 27, I probably should have
picked a simpler example, but let's just keep going. Plus 15 times nine. Plus four times three is 12. Minus 12 So lucky for us, at least those
last two terms cancel out. And so this is going to be the rest from here is arithmetic. Two times 81 is 162. Now let's think about what 27 times 11 is. So let's see, 27 times
10 is going to be 270. 270 plus another 27 is minus 297. 297, did I do that, yeah, 270 So 27 times 10 is 270 plus 27, 297 Yep, that's right. And then we have, I'm prone to make careless errors here, see 90 plus 45 is 135. So plus 135. And let's see, if I were to take if I were to take 162 and 135, that's going to give me 297 minus 297. Minus 200, we do it in that green color, minus 297. And we do indeed equal zero. So the remainder, if I were to divide this by this, is equal to zero. So x minus three is indeed a factor of all of this business.