 Again, we've laid out our coefficients from our quotient that we got, so we reduced it down to one level and we took the coefficients and laid them out here and we're gonna have to try possible factors of negative 20. Again, possible factors of negative 20 are gonna be plus or minus one, plus or minus two, plus or minus four, plus or minus five, plus or minus 10 and plus or minus 20, right? I know what the possible factors of this are and I'm just gonna pick one of those, right? Now, if you didn't know and if you won't know if you're given a question like this, the best chance that you have of finding a possible factor is starting off with the lowest possible factors first and the more of these you do, again, the more of these you do, you'll get a pretty good idea of what the possible factor is going to be because if you look at this right now, you got a two and an eight, which are positive and you got negative five and negative 20. Now, right away, just looking at this, you should have an idea of what might work, right? Because right now, the positives add up to 10 and the negatives add up to negative 25, right? So you know one is not going to work because if everything's just being multiplied by one, what's gonna happen is the positives don't balance out the negatives. Negative one is not, you know, it probably is not going to work, okay? So you could skip positive one and negative one and just go to positive two and negative two, okay? So what we're gonna do for this is, you know, pick a factor that I already, pick a number that I already know was a factor and we're gonna try out x is equal to negative four, which means x plus four is a possible factor of this. So we're gonna try out x is equal to negative four and if we bring negative four over, it just means the factor is x plus four, right? So what happens is the two's gonna come down, multiply by negative four, it's gonna come up here and so forth. So two times negative four is gonna be negative eight. Add them together, you get zero. Zero times negative four is just gonna be zero. Add them together, you get negative five. Negative five times negative four is gonna be positive 20. Add them up, you get zero. Negative 20 plus 20 is gonna be zero, right? So that means x plus four is a factor of this polynomial, right? And what we have down here is the quotient and one thing to keep in mind is, we have a zero term here. What that means is this was x to the power of three, so we just took out an x from a polynomial from a function that was x to the power of three, that means what's left over, the quotient is gonna be x to the power of two. So this is gonna be two x squared plus zero x and we don't really need that, right? As soon as you get a zero in your synthetic division, in the quotients, you know, you can skip those terms, but make sure you keep track of what you're skipping. If that was x squared, that's x squared, right? X to the power of one goes here and that just becomes a constant. Do not put your x over here because this guy's zero. Everything goes down sequentially, right? Think, think, think. All the powers have to go down in order. So if you have a zero, that uses up one of the x's, okay? So let's write down the quotient here. What we have now is two x squared minus five. Now we're down to a quadratic, right? Something to the power of two and it happens to be two things subtracted from each other, so we can factor this using the difference of squares or we can factor it using the quadratic formula or we can just solve for this by setting it equal to zero, moving the five over, dividing by two and square root in both sides, which is basically what I'm gonna end up doing, right? So what you end up doing for this one, you can factor this one right now and let's just do it over here. So what we do, we go two x squared minus five, you set that equal to zero, bring the five over and then divide both sides by two. So what you end up having is x squared is equal to five divided by two and to solve for this, you gotta get the x by itself so you take the square root of both sides, okay? Let's just write that down here, right? Squared of x squared is just gonna be x, squared of five over two and remember, squared of anything is always plus or minus, so that's gonna be square root of plus or minus the square root of five over two, okay? So those are two other factors because that's two different solutions, right? So our solutions for that is gonna be x is equal to plus or minus five over two. So, you know, going back to our original question we had, you know, the h of x to the power of four, the possible factors or the roots, the zeros, where the h of x, where the function crosses the x-intercept or x is equal to positive square root of five over two, x is equal to negative square root of five over two, x is equal to negative four and x is equal to three. So there's four places where our polynomial crosses the x-axis, okay? And those would be the solutions if that function was equal to zero. But we asked for it to be, you know, we asked, the question would have been if they gave it to you in the function form to factor that polynomial, right? So we're gonna have to write it down and in its factored form. Now, keep in mind, if we're gonna have to write this in this factored form, this is square root of five over square root of two, so we're gonna grab square root of two, cross multiple up here and take the square root of five and bring it over, right? So I'm just gonna write these down, both the factors of it in, you know, in its factored form, that way we're just gonna go to the original function and write down what it looks like in its factored form. So in its factored form, x is equal to plus or minus square root of five over two, it just looks like this, right? Square root of two x and the x is not under square root symbol, right? Square root of two x minus square root of five, square root of two x plus the square root of five. Again, the x is not under the square root symbol, okay? So let's take all of these and write out the function in, you know, where we started off with, basically. So you know what the original function looks like in its factored form. So h of x, this function here in its factored form is gonna look like the following. That function up there could be written as h of x could be this guy, x minus three times x plus four times square root of two x plus square root of five times square root of two x minus the square root of five. If you foiled out this whole thing down here, you would end up with your original function. And all of those points means that each one of those is gonna be an x-intercept. Each one of those points gives you an x-intercept, okay? One thing we are gonna talk about later on when we start graphing polynomials in a lot more detail anyway is whenever you have, you know, a polynomial function to a power here, the highest power decides the maximum possible number of x-intercepts that you can have, okay? So this was x to the power of four and the maximum number of x-intercepts we could have was four and we ended up getting four, okay? You don't necessarily need four or get four. Sometimes you get none, right? But the highest power in any polynomial gives you the maximum number of x-intercepts, okay? And right now we got four x-intercepts. We couldn't have five x-intercepts because this was a power of four, right? So just, you know, just a pointer for the next stuff coming up. The highest power in any polynomial decides the maximum number of x-intercepts you can have. You can have less than that but that's the maximum number you can have, okay? Hopefully this made sense. There's a fair bit of stuff that we did here, a fair bit of calculations but it's all fairly straightforward. You're bringing numbers down, multiplying by whatever they're coming up, adding them together. You gotta know how to use your multiplication table or you're gonna know your multiplication table, you know, and bringing it down. If it's a factor of it, this one wasn't. If that ends up being a zero, that number, there's a factor of that guy. That's your quotient. You continue on this one. You use synthetic division on this one until you kick it down to something to a power of two or something that you can factor using the other techniques. And then you use that one, you factor it if it can and you get all the possible factors of your original question that you start with the original function you start with, okay? We'll end up doing a couple more questions at least and see where it goes from there. Hopefully this makes sense and you know, it looks fairly easy for you even though it's taken a fair bit of work to get it done. Okay.