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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?
• 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.
• Where can I find the Factor Theorem tutorials?
• 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:
• What additional advantage does polynomial remainder theorem have over synthetic division?
• It's useful if you only need to find the remainder or if you need to test factors of polynomials.
• 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?
• 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).
• 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))?
• Exactly. Excellent observation. You would indeed use -5.
• if (x-5) is the factor we are solving for shouldn't we have plugged in -5 for x?
• ``(-5 - 5) = -10``
• what if (x-5) is not a factor, and there is remainder how would you proceed then?
• 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! :)
• Finding coefficients still confuses me...for my math assignment I have to solve this polynomial: p(x)= -5x^3-8x^2+2x+c