 This video will talk about zeroes of functions. Then we have the fundamental theorem of algebra which says every complex polynomial with a degree greater than or equal to one has at least one complex zero, a plus bi, two is a complex number, it's two plus zero i. And then we have the linear factorization theorem that says you have this polynomial, degrees n greater than or equal to one, then it has exactly n linear factors and can be written as times x minus c one times x minus c two, so on. As long as a is not equal to zero and c one, c two, c three, these numbers that are being subtracted from x are not necessarily distinct, but they are complex numbers, or the product of non zero constant and exactly n linear factors. That's another way to word that. Let's try it. Rewrite the polynomial q of x is equal to x to the fourth plus 21x squared minus 100 as a product of linear factors and find the zeroes. Then it gives us this hint. It says let u equal x squared, u is equal to x squared, and that means that if I have u squared, that's equal to x squared squared or x to the fourth. Instead of x to the fourth, we can say that that's really u squared. And then instead of x squared, we can say that that's u and then we have our minus 100. So now it's just a quadratic that we can factor factors of u squared are going to be u and u and factors of negative 100 that will add up to 21 would be minus four and plus 25. But now I need to reassure that I go back while you was equal to x squared. So both of these factors start with x squared and then it's x squared minus four and x squared plus 25. Those are the linear factors. But if I wanted to find the zeroes, I could say that x minus four x squared minus four was equal to zero. So x squared would be equal to positive four and take the square root of both sides. X would be equal to plus or minus four. There's two of my zeros. And then I have x squared plus 25 is equal to zero and take it to the other side x squared is equal to negative 25. And if I take the square root of both sides, I find out that x is equal to plus or minus square root of negative one is I and the square root of 25 is five. So those are my two factors. And actually these are not linear factors because they've got quadratics in them. So I need to rewrite them as these are like my C's. Okay, these are like the C's. Remember it was x minus C was the factors. So I have x minus four x plus four. And then here I have again this is C. So x plus five I and x minus five I. So if I want it to be linear, I'm going to have to have the eyes in my factorization. Now I want us to write this one as a product of linear factors and find the zeros. We can look and see if we could just factor it just the way it is. We've got four terms here. So let's see if we take a common factor of x squared out would be left with x minus three. And we take a common factor here out as negative nine and we'd have x minus three. So that leaves us with the common factor of x minus three and x squared minus nine as my other factor. And this is linear, but this one is not. So we have to go one step further and this would be x minus three and x plus three. So we found the linear factors and now we just have to find the zeros. Well, we have x minus three. We actually had that more than once. So we just have to, so x would be equal to three. And then we have the x plus three and that will give us x equal negative three. So we had this zeros of multiplicity. That's basically if you have a degree of polynomial that's greater than one and you've got these linear factors that occur more than once, like in the previous problem. Here we had x minus three that happened twice. So that's what we're talking about. Then it says that that zero, which is C, has a multiplicity of m. So again, going back here, we could have written this as x minus three quantity squared. So the zero is three and the multiplicity is two. This exponent right here is going to tell me how many factors of that I actually have. So I only see on a graph one point, but I know that there were two there because the exponent tells me that. So this says factor completely, name the zeros and their multiplicities. So this would be x and x and factors of 36 that added to 12 would be plus six and plus six. And then in here, I'm going to have x and x and factors of negative 24 that add up to two would be plus six and minus four. And then I have this x minus four. So all three of them are completely factored now. And I want to think about this. I have x plus six. So x is going to be equal to negative six and it has a multiplicity of how many of them do I have? One, two, three. The multiplicity is equal to three. I could write that as x plus six cubed. And then I have this x minus four. So x would be equal to positive four and it has a multiplicity and that's equal to two of them. Or we could simplify this and say x minus four quantity squared. So find the zeros of this and then write the function in factored form and assume a and it says try one and negative one first. That means synthetic division. That's what they're talking about. You can find zeros from synthetic division. We learned that. So I'm going to try one. See if that works. It's on the outside. On the inside we have four, eight, negative three and negative nine. Nothing's missing. If I bring down the four and then four times one would be four. And remember if we do that, that means that this one is a zero because the remainder was zero. And this also means that now I have four x squared plus 12x plus nine. We could do a couple of different things with this one. We could just try to factor it like we see it or we could try synthetic division again. But actually come find out that we have rational zeros. So let's just factor it from this point. It's a quadratic. We could use a calculator, quadratic formula if we really wanted to. But this one's not too bad. Two x and two x. Factors of nine that add up to with those two x is to 12. Well that looks like three and three because then I'd have six plus six. And then that tells me that my zero is x is equal to negative three over two. So here's one of my zeros. And here's one of my zeros. And they want it in completely factored form. So I would write it as that first zero x equal one would be x minus one. And the other zero is two x plus three, but we have two factors of those. There's our completely factored form. The complex zeros. If a polynomial has real coefficients, the complex zeros must occur in conjugate pairs. That's the important part right there. So you'd have a plus bi and a minus bi. So it's asking us here to find the polynomial having real coefficients degree three. So that means that all my multiplicities have to add up to my degree of the polynomial. So degree three and zero is at negative five and negative three i. This tells us that we're also going to have to have plus three i because of this definition here. So writing that x minus c. So it'd be x plus five. And then we have x plus three i and its conjugate x minus three i. It says find the polynomial having real coefficients. We don't have a polynomial here. We're just factored. So we have to keep going and we can say that this is x plus five. I'm just going to leave that one. And this one is x squared and then that minus three i and plus three i cancel out. And then we have minus nine i squared almost there. So x plus five and this is x squared plus nine converting the i squared. So now we can multiply a little easier. So x times x squared is x cubed x times nine is plus nine x five times x squared and then five times nine. So if we simplify the polynomial that we're looking for is x cubed plus five x squared plus nine x plus forty five if we put it in standard form. And you'll notice that we have a degree of three which is exactly what it said that we needed to have. Find the polynomial in factored form and find the rational in complex zero. So both kinds assuming the leading coefficient is one. All right. So here we go. When you look at this and you say I really don't know how to factor that there's a couple of things you could do. We could do synthetic division again and you can always start with one and negative one. Those are always good places to start or you could also and we may do this from both ways use the calculator. So I'm going to put y equal and I'm just going to do a standard window because all I really care to see are my intersection points here. Or my zeros. So second trace I could just do the zero to and ask for the left bound. So I need to be down the left hand side of my graph. So that would be on top above this press enter. And then I need to be below that X intercept and we find out that X is equal to X is equal to negative one. And then I'll write that later. Let's find the other one. Second trace two for the zero. And I want to keep going through beyond this side now. And now the left is going to be below the X axis in this case. Enter and then keep going and go above it. Press enter and then enter again. And we find that it's one point five. So we said the X was negative one and X is equal to a positive one point five. So we make that into a fraction. It's equal to three over two. So we have two factors. And that's X plus one and two X minus three. What I did was three over two equals was equal to X. And then I bring the two over by multiplying both sides. So three is equal to two X. And then we bring the three to the other side and we have zero is equal. So zero is equal to two X minus three. So that's how I got that factor. But that's only two. There might be multiplicities on it. I really don't. But I know for a fact there aren't. So what do we do about this? Well then we definitely have to do the combination of synthetic division. But this one then tells me that I needed the negative one. I might have tried the one and been in trouble. So I have the two and a negative one and three and negative three and negative nine. I could try to factor my four terms. But I also know that I have a zero at three halves. So bring down the two. And if you don't believe me, three halves times two over one is just going to be three. And negative three plus three is going to be zero. And zero times anything is zero. So I get to bring down the six and then the three halves times the six. Well two goes into six three times. So three times three would be nine and it's a positive. So there's my remainder. And now I'm ready to say that my other factor is two X squared plus six. It's now in factored form. I know two of the zeros because we grafted to find those zeros. But I have to find the other two, these complex ones. I'm going to change colors here and say that two X squared plus six equals zero. So two X squared will equal negative six and X squared will equal negative three. And then that tells me that X is equal to plus or minus I and then the square root of three. Because we don't know if that's not a perfect square. So there are my other two. There's one, two, and then three, four right here.