Carl Horowitz

**University of Michigan**

Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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The rational roots theorem states that all potential roots are in the positive or negative form of the last coefficient s factors divided by the first coefficient s factors. With a large polynomial, solving by factoring is more difficult, and so finding the **rational roots** will give some potential zeros to start with. With these rational roots, the solutions after factoring complicated expressions is narrowed down to a select few answers.

Using the rational roots theorem to try to factor a polynomial, so I'm going to start out explaining this by looking at a fairly straight forward polynomial x squared plus 6x plus 8 equals 0. In order to figure this how we would want to factor our x squared plus 6x plus 8 and said it equal to zero to find out what x is okay.

This particular problem isn't all that hard you will be looking at equal to zero we know that our first 2 terms have to be x because both are positive we know we are adding and then when we factor in this, we think about possible factors of 8 that go here and out of the 6. It could be 1 and 8 but when you think about that we know that we would end up with a 9x here okay that leaves only the other options to be 2 and 4 because the x's are singular leaving us with 2 and 4 order doesn't matter okay? So that's sort of logic we go through when we factor a standard quadratic equation okay and this tells us that x is equal to -2 or -4 okay. What if we're dealing with a larger polynomial? Okay something of degree 3, 4, 5 so on and so forth that we don't know right now where to start factoring okay so what I have here is a general form for a polynomial okay when we first introduced polynomials we always have these a sub n subscripts as the coefficient an a term of degree and or whatever it is so here a sub n minus 1 is the coefficient on x the n minus 1. For this particular example all I've done is I've changed my first coefficient to a p and my last constant term to a q everything else is exactly the same okay.

And what the rational roots theorem says is that all potential factors all potential roots all potential zeros are of form factors of q over factors of p and it could be either positive or negative so we're looking at that the middle terms don't matter at all really all we're concerned with are the factors of q which is our last term the constant term over the factors of p our first coefficient okay and then we can either have positive or negative any of those combinations okay so that's the theorem in letters let's take a look at it in numbers so moving down.

I have a polynomial here okay it's a fourth degree first coefficient is 2 last coefficient is 8, so what I'm looking for is the plus or minus factors of the last term over factors of the first term okay so how I write this out is I just draw a giant fraction line then think about all the factors of the last term factors of q okay 1, 2, 4 and 8 that's the only thing that goes into 8 this is going to be 1, 2, 4 and 8 plus or minus over factors of the first term in this case our first term is just 2 so the only factors we have are 1 and 2. Okay so we now have our weird fraction and basically what this means is any combination of these numbers so let's write these out, so we could have +1 or we could have -1 so that's 1 over 1, we could have the 2 over 1 which will give us 2 or the -2 over 1 -2 and so on and so forth positive 4 over 1, negative 4 over 1, positive 8 over 1 negative 8 over 1 so here are some of our potential roots we also then have to compare it to 2 okay so we can have positive one half, negative one half we could pair the 1 and the 2 then going down the row 2 over 2 is just 1 which we already have listed here 4 over 2, 2 already have listed and 8 over 2 is 4 we already have listed, so those 3 over 2 are already excess so these are the only potential roots for this particular polynomial okay how we actually end up factoring this is a little bit more involved we have to go into synthetic division and figure out which gives us a remainder of 0 but our list of numbers where we can actually start is a lot smaller okay no longer are we just fumbling at numbers randomly we have a list of a few select numbers we have to deal with.

So using the rational roots theorem factors of our last term over factors of the first term is really easy way to at least limit the number of roots we have to consider when factoring a larger polynomial.