# Polynomial Remainder Theorem

1. Nov 2, 2008

I know how the polynomial remainder theorem works but I can't see how knowing this is useful in any way. So I have f(X). I know that if I divide the statement in f(X) by X - a the remainder will be a. How is this useful knowledge though? What can I discover using this principle that I wouldn't ordinarily be able to find out?

2. Nov 2, 2008

### olliemath

Well it isn't a, it's f(a).
This idea is used (without mention) again and again throughout mathematics. Whenever you find the roots a,b of a quadratic equation you tend to write it in the form (X-a)(X-b). So a question to ask yourself is: how do you know this is always possible? What about cubics and quartics?

3. Nov 2, 2008

### majesticman

Umm well if you are talking about a real world application then i got a hypothetical for you.

Suppose we have to maximize/minimize a function f(x). Now after finding the derivative i.e. f'(x) (i am not sure if you have done calculus yet).

So alternatively think about a function f(x) that predicts the inventory usage in a company. This f(x) is obviously a model so let's say it has a degree of 20 or something big. Remember now that the set of equation for which we can analytically solve the equations are greatly reduced.

Now we can program a computer to start from say x= -80 and end at x = 80 with a increment of 0.001 and then using remainder theorem we can say as long as the value after applying the theorem is less than 0.05 <an arbitrary criterion, we only have to make is as close to zero as we want to the level of accuracy desired>.

So we should be able to find the times the past 80 to the 80 days in future when the inventory is 0.

4. Nov 2, 2008

### HallsofIvy

If you can find x= a that satisfies P(x)= 0, then the remainder theorem tells you that x- a divides P(x) with remainder 0. That means that x-a is a factor of P(x) and you can write P(x)= (x-a)Q(x) where Q(x) is a polynomial of of degree one less than P(x). Since Q(x) has lower degree it may be easier to solve Q(x)= 0 and so continue factoring P(x).

5. Nov 4, 2008

### symbolipoint

I spent some time rethinking about this. Some applications which can lead to second or higher degree polynomial functions can involve voluem calculations for box shapes, and cost modeling. You would possibly be able to use one of these theorems if your model can be approxomated with rational coefficients. Not certain is whether an analyst would actually use remainder or factor theorems. Would it not be easier to simply use a graphing computer program to display a graph for the function model, and directly read the roots from the resulting displayed graph? I imagine the analyst could then apply synthetic division to reach any complex roots or simply accept the quadratic factors without trying to reach complex roots.

Anyone with real-life experience with this to contribute more to this discussion?