College Algebra: Find Zeros of Polynomials

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So I thought we'd take a look at one of these quest ions from start to finish. Someone hands us a polynomial and wants us to find the roots. I'll make this one a cubic, so let's consider f(x) = x³ + 3x² - 6x - 8. Okay, so there is the polynomial and I want to find all the zeros, which of course, means I want to set that equal to zero and solve for x. Well, it's a cubic, so I don't know how to factor that directly. So one technique to try after thinking about factoring techniques that would all basically fail, as far as I can tell, is to see if I can find a rational root. So let's look for candidates of rational roots. Now, where do you look? Well, I use the rational zero theorem, and the rational zero theorem says that if you're going to have rational zeros, their numerators will be factors of -8, and the denominators, in this case, will be factors of ± 1.
So the candidates would be what? Well, look at all the factors at the bottom of the -8, which would be ± 1, ± 2, ± 4, and ± 8. And now look at all the factors of 1. That's pretty easy, actually, ± 1. So now I look at all the quotients. And if you think about it, any of these numbers divided by ± 1 is still just these numbers. So, in fact, in this case, since this was a 1 in front, all of the candidates are right in front of you. So, in fact, there are only 8 candidates, if this is going to have a rational root, and it may not have one, by the way. It may not have one, but if it does, it's going to be one of these eight numbers. So let's just trial and error and see what happens.
If I plug in a 1 in, so let's look at f(1). Is that zero? If I put in a 1 I see 1 + 3 is 4 and then 4--and then here I'm going to have a -6, -8, and -6 - 8 + 4 is certainly not zero. It's negative. So, in fact, this is not zero.
What about the next one? How about f(-1)? IF I put a -1 in here that's a -1 here, this becomes a +3, so -1 + 3 is +2. +2, and I put a minus sign here and it becomes a +6. So +2 + 6 is actually 8, -8 is zero. So that looks pretty good. Did I do that correctly, by the way? Let's see, this is going to be a 6 and then I'm going to have a... Yeah, I'm going to have a +2, which is 8, so this is zero. Wow! I found one. Sort of surprising. I didn't expect that. So there's a root right there--boom, I've got one. What can I do now? Well, I can keep going on my list and see if there's others or I can just say if -1 is actually a zero, that means that x + 1 is a factor, and then I can actually use the division algorithm, or I could use the synthetic division method to actually see what's left over. So either one's fine. Let's keep going on our list though and see if we can maybe detect some other ones.
So let's look at 2. If I plug in 2 I see 8, and this is going to be 4 and 12, so we have an 8 and a 12 and I subtract a 12, so these cancel out, and then I see an 8 and I see an 8, so again I see zero. Wow! I'm seeing all the factors here.
And what about a -2? Well, you could try a -2 and stuff, but I've already found two roots. That seems like an awful lot to me. So let's see what happens when we actually start to get these roots out and see what's left over.
So I'll use this one right here. I see that 2 is a zero. If 2 is a zero what that means is that x - 2 must be a factor. So if that's a factor what that means is I should be able to synthetic divide this into that and see what's left. So I switch the signs, so I get now 2, make a little house, put down the coefficients--1, 3, -6, -8--the remainder should be zero if this is correct, because otherwise it's not going to divide in evenly. I bring down the 1, I get 1, multiply it by 2, I get 2, add, get 5. I combine these guys, I get 10. If I combine these things I get 4, 4 times 2 is 8, 8 - 8 is zero. Cool! So that's a good check. This is zero. And what's left over here is an x², because I have an x³ and I'm taking out an x, so this is x², this is the x, and this is the constant.
So what I see here is that f(x) can be factored as (x - 2)(x² + 5x + 4). And now the question is, can I factor this? Well, let's see if I can factor this in some way. Well, yeah, I think I actually can factor that. X + 4 and x + 1. So I want to find the zeros of this. I set it equal to zero and now I can read it off since it's factored. I see x = 2, that was the one we already saw. I see x = -4, and I see x = -1, and notice the x = -1 was also the one we already found. So, in fact, we found two of them. The one we didn't find was the x = -4. But notice something--if we would have kept going on my list I would have come across -4 and I would have uncovered that one, too. So there are a lot of different ways of finding the roots, especially if a root is rational, you can really uncover it in this finite list, check using synthetic division, see what's left over, factor and find the zeros. Try these on your own now.
Polynomial and Rational Functions
Zeros of Polynomials
Finding the Zeros of a Polynomial from Start to Finish Page [1 of 1]