##### Like what you saw?

##### Create FREE Account and:

- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- Attend and watch FREE live webinar on useful topics

# Finding Zeros of a Polynomial Function - Concept

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Finding the polynomial function zeros is not quite so straightforward when the polynomial is expanded and of a degree greater than two. One method is to use synthetic division, with which we can test possible **polynomial function zeros** found with the rational roots theorem. Once we find a zero we can partially factor the polynomial and then find the polynomial function zeros of a reduced polynomial.

We're finding the zeros of polynomial functions. Let me show you two examples: f(x)= 2(x+3) and x 1(x+10). If you're given a polynomial like this, it's really easy to find the zeros of the function because each of these factors contributes a 0. So you'll have 3, 1, and 10. You're generally not going to get a problem this easy. What about g(x), which is not factored? This is the kind of problem you'll see: Find the zeros of g(x). Well there are generally three steps we have to think about. First, use the rational roots theorem to find potential zeros. These are potential zeros. They don't necessarily work but we have to check them. And we use synthetic division to test the potential zeros. And finally, we'll factor out whatever factor corresponds to the 0, and we'll get a reduced polynomial that will contain the rest of the zeros.

Let's see how this works with the example, g(x). So I want to find all the zeros of this polynomial function. Now the rational roots theorem says to look at the integer factors of the leading coefficient and the constant. Now the leading coefficient is 1; its integer factors are 1 and 1. For example, 1x1 is 1, and 1x 1 is 1. So those are integer factors of 1.

And for 10, we look at plus or minus1 or plus or minus 2, plus or minus 5, and plus or minus 10. We can take two of these numbers, multiply them, and get 10: 1 and 10 would work. Now the rational roots theorem says that the potential zeros are these integer factors divided by these. So it's plus or minus 1, plus or minus 2, plus or minus 5, and plus or minus 10, all over plus or minus 1. Of course, that's just going to give you the stuff on the numerator, this doesn't really contribute anything.

So what I usually like to, and I suggest you do too, is start with the easy ones. Start with one and 1. So let's start with 1. Now we're going to need synthetic division to test these. The way synthetic division works is I'm using one as my 0, and I write the coefficients of this polynomial down here: 1, 1, 8, and 10. The one comes down, I multiply and write the result here. So 1x1 is 1.

Then I add and I get 0. Then I multiply again, 1x0, and I write the result here. And I add again, 8, and I multiply, 1x8 is eight, and I get 18. This last number is my remainder. Now the remainder's not 0, so one is not a 0. That means that g(1) is actually 18. You remember the remainder theorem. This is actually the value of the function g(1). So one didn't work. Let's try 1.

Same coefficients 1, 1, 8, and 10, and the same process. Bring the one down and multiply. Add and multiply. Add and multiply and that works. The remainder's 0 and that means that 1 is a 0, and that means that x 1 is a factor. And by the way, these are the coefficients of the other factor. So this function can be written g(x)= one factor is (x 1)(x+1). The other factor is x2 2x+10; this is the reduced polynomial.

Now if you want to find the remaining zeros of this function, you've got to look here. This one has a 0 of 1. This has the other two zeros. Now it's a quadratic, so we can use the quadratic formula: a is 1, b is 2, and c is 10. So b, which is +2, Â±b2; 2 squared is 4; 4ac, or 4x1x10 is 40, all over 2a. So this is going to give me Â±2 root negative 36. It's imaginary. Root negative 36 is 6i. So 2(Â±6i)/2 is 1(Â±3i). And that represents two zeros; two imaginary zeros for our polynomial. So the zeros are 1, 1+3i, and 1 3i.

Now remember what we did. First, we used the rational roots theorem to find potential zeros. We had all these potential zeros. We were lucky to find one of them so quickly. But I would always check one and 1 first; the arithmetic is going to be the easiest. Then once you find a 0, you can take the reduced polynomial and looks for the zeros of that. Now with the reduced polynomial, if you start with a fourth degree, the reduced polynomial will be a cubic, and you have to do more of this synthetic division. But when you finally do get a quadratic, use the quadratic formula and find the remaining zeros that way.

Please enter your name.

Are you sure you want to delete this comment?

###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

###### Get Peer Support on User Forum

Peer helping is a great way to learn. Join your peers to ask & answer questions and share ideas.

##### Sample Problems (3)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete