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

Lesson 24: Polynomial Remainder Theorem# Remainder theorem: finding coefficients

Learn how to find the value of 'c' that makes 'x - 5' a factor of a polynomial 'p(x)'. Understand how setting 'p(5)' equal to zero and solving for 'c' unveils the mystery. Grasp the concept of the Polynomial Remainder Theorem with a real example.

## Want to join the conversation?

- Why did sal khan plug in 5 for x?(28 votes)
- The polynomial remainder theorem states that if you take a polynomial function "f(x)" and divide it by "x-a", then the remainder will be equal to "f(a)". In this case, he was dividing by "x-5," so "a" was equal to 5. Therefore, in order to solve it, he had to plug in "a" for "x" to solve "f(a)".

I hope this helps.(49 votes)

- Where can I find the Factor Theorem tutorials?(6 votes)
- Factor theorem tutorials do not exist on KA, but luckily, Factor Theorem is simply a special case of the Polynomial Remainder Theorem. In fact, with some elementary school intuition, you should be able to figure it out pretty easily:

Polynomial Remainder Theorem tells us that when function ƒ(x) is divided by a linear binomial of the form (x - a) then the remainder is ƒ(a).

Factor Theorem tells us that a linear binomial (x - a) is a factor of ƒ(x) if and only if ƒ(a) = 0.

Which makes since because, if you combine that with Polynomial Remainder Theorem, all Factor Theorem says is that a linear binomial is a factor of a function if and only if the remainder when you divide them is 0.

We learned in elementary school that if you divide two numbers and have a remainder of 0, then the divisor is a factor of the dividend.

Here is a tutorial that is "technically" on Factor theorem, but the video states that it is on Polynomial Remainder Theorem:

https://www.google.com/url?q=https://www.khanacademy.org/math/algebra2/arithmetic-with-polynomials/polynomial-remainder-theorem/v/polynomial-remainder-theorem-to-test-factor&sa=U&ved=0ahUKEwjR7sfv1rPJAhXHqx4KHW2oDpQQFggEMAA&client=internal-uds-cse&usg=AFQjCNE4TdbxCcj0T_Q-bVIxMa_g_hmAGg(24 votes)

- What additional advantage does polynomial remainder theorem have over synthetic division?(8 votes)
- It's useful if you only need to find the remainder or if you need to test factors of polynomials.(9 votes)

- Sal said x-5 is a factor ONLY if p(5) =0, but for this polynomial p(5)=-37. So how is it a factor of x^3+2x^2-37x+10?(6 votes)
- Slight misunderstanding on your part. p(5) DOES NOT equal -37. We are solving for the coefficient c, since the original question is p(x) = x³ + 2x² - cx + 10. We set p(x) equal to zero and solve for c.

If x-5 was a factor, then you could write p(x) as p(x)=(x-5)g(x), where x-5 is the factor we are given and g(x) is the other factor. Now recall when factoring quadratics, you would get, for example x²-6x+5 which would factor into (x-5)(x-1) and that since these were factors, we can now easily see what are the zeros of the original expression x²-6x+5, (those being 5 and 1). So if you put x=1 into x²-6x+5, the answer will be zero. If you put in x=5 into x²-6x+5, the answer will be zero.

It is the same deal here, but we only need to worry about 1 of the factors, this one (x-5).

p(x) can be written as p(x)=x³+2x²+cx+10, and it can also be written with one of its factors like this p(x)=(x-5)g(x). This means that x=5 MUST be a zero for p(x). Since it is, we can calculate p(5), set the result equal to zero and then solve for the missing coefficient, c. When you do that, you get c=-37.

That means p(x)=x³+2x²-37x+10 can be factored in a way that (x-5) is a factor (the other factor is x²+7x-2, that is p(x)=x³+2x²-37x+10 = (x-5)(x²+7x-2).(13 votes)

- If we were asked for example if (x + 5) was a factor, could we use the PRT with (-5)

as (x + 5) is the same as (x - (-5))?(6 votes)- Exactly. Excellent observation. You would indeed use -5.(8 votes)

- if (x-5) is the factor we are solving for shouldn't we have plugged in -5 for x?(3 votes)
- what if (x-5) is not a factor, and there is remainder how would you proceed then?(4 votes)
- If you divide any polynomial and you are left with a non-zero remainder which can't be further divided, then the divisor is not a factor of that polynomial.

Hope you understood! :)(4 votes)

- Finding coefficients still confuses me...for my math assignment I have to solve this polynomial: p(x)= -5x^3-8x^2+2x+c

I have to solve for "c." Can someone please help me and explain the steps?(3 votes)- what do you mean, "solve for c"? perhaps you're given a factor such (x-1) or (x+2) or (x-10) or something?(5 votes)

- Does ''a'' need to be an integer?(4 votes)
- it can be any numerical value like a fraction, integer or decimal as a matter of fact(2 votes)

- How to solve similar questions when the divisor is quadratic?(4 votes)

## Video transcript

- So for what value of c, or values of c, is x minus five? Let me write it this way. For what c is x minus five a factor of p of x? I encourage you to pause the video and try to work through it. That p is actually a lower case p. P of x. All right, so I'm assuming
you have attempted this. We just have to realize,
okay, if x minus five is a factor of p of x,
that means you could write p of x, let me do
this in that green color, that means you could write p of x as being equal to x minus five
times some other business, times some other polynomial. So times some other polynomial. I don't know, let's just
call it g of x maybe, g of x, and this would be
a 2nd degree polynomial. So what would p of five have to be? Well p of five would have to zero. Five would have to be a
root of this polynomial. You see it right over here. If you replace this with the five, then this is going to be a five, and this is going to be a five. It doesn't matter what g of five is. This is going to be zero. So if x minus five is a factor, if and only if p of five is equal to zero. You could say that five is a root of the polynomial p. P of five is equal to zero. So let's just set p of five equal to zero and then try to solve for c. All right, so we're going to get five to the 3rd power. So this right over here, five
to the 3rd, so p of five. P, let me just write it down, p of five is equal to five to the 3rd is 125 plus two times five squared. So that's two times 25, plus 50, plus c times five, plus five c plus 10. That needs to be equal to zero. Now let's see, if we were
to add 125 to 50 to 10, we are going to get 175, 185. So we get 185 plus five
c is equal to zero. Subtract 185 from both sides, you get five c is equal to -185, or c is equal to -185 over five, which is going to be, let's see, it's -185 over five, which is equal to, let's see, five goes into 180, let's see, it goes into, well it's five, let's see, it's going to be 30. Five times 30 is 150. Then we'll have another 35 to go, so it's going to be 37, -37. Is that right? Five times 30 is 150,
five times seven is 35, yep, -37. And we're done. If this was x to the 3rd plus 2 x squared minus 37 x plus 10, then x minus five would be a factor of this polynomial.