 So, let's do something simple to start off with on these, you know, factoring higher degree polynomials than just degree 2, right? So let's say we want to factor the following polynomial. So the question would be, they would be asking you to factor this polynomial, x cubed minus 1. Now, there are formulas for this, just like there are formulas for factoring quadratic functions, the quadratic formula, but we're not going to use the formula to factor this, right? Because it's going to work on all possible, all polynomials that have, you know, have factors that exist in the real number set, right? Or at least as part of the rational numbers, integers that we have, right? Things that factor out simply. So what we're going to do to factor this guy is we're going to look at the last term and the coefficient from the highest power, right? The last term, the possible factors of 1 is just plus or minus 1. The possible factors of negative 3, or the 1 here is just plus or minus 1. So possible factors of this is just going to be plus or minus 1 that we're going to try out. Now if those didn't work out, then you're going to have to use a calculator or go to the formula and try to figure this out. But you know, we came up with this question and we already know, I already know one of these is going to work. So let's try possible factors that's just going to be plus or minus 1. And this is x cubed going down to a constant. And we have to put everything in descending order, right? But we're missing the x squared term and we're missing the x term. So what we've got to do is put place markers for the x squared term and the x term. So let's lay out the division statement here, or the synthetic division statement here, right? Or the synthetic division form here, that way you can see how it all lays out. And then we're going to go ahead and do the synthetic division. So the possible factors of this polynomial are going to be plus or minus 1 divided by plus or minus 1, which is just really plus or minus 1. So what we're going to do, we're going to try out x is equal to 1, which means x minus 1. We're going to see if x minus 1 is a possible factor of this, right? And what we did when we laid it out in the synthetic division form, we put in place markers for x to the power of 2 and x to the power of 1, right? Because we're missing those terms. So that's 1 and then 0x squared plus 0x minus 1, right? And what we're going to do is just go through the synthetic division part. So the 1 comes down here, 1 multiplies 1, comes up here. You add those guys, whatever the result is, comes here, multiplied by 1 again, comes up here. So we're just doing this exact thing, thing, thing, thing, thing. So again, what we did was brought the 1, multiply 1 times 1, 1. Add these guys together, 1, multiply 1 times 1, 1. Add these guys together, you get 1, multiply 1 times 1, 1. Negative 1 plus 1, 0, right? That way, we know that x, x minus 1 is a factor of this top guy. So right away, we know that x minus 1 is a factor of this guy because our remainder is equal to 0. What we have as the end result here, we just took an x cubed and divided an x from it, right? So our end result here is that's 1x squared plus 1x plus 1. If any of these terms were 0, it would just be a place marker for the missing x term, right? So if this guy was 0, your factor here would have been x squared plus 1, right? Because this guy would be missing, but it's not. So we're going to fill in the x terms so you know what it looks like. That's the polynomial that we get, the quotient that we get when we take x minus 1 divided into that if we're thinking about a division, right? What we're looking for here, the question for this guy would have been factor this. So what we have right now, this guy factored, right now, we've taken it down to x minus 1 times this whole thing, x squared plus x minus 1. Now we're going to have to try to factor this further, right? We can't use simple, simple, what's it called, simple trinomial factoring because we don't know two numbers that multiply to give you negative 1 and add to give you 1, right? And we can't use complex trinomial factoring because that's just a 1 there, right? We don't need it. So what we're going to have to use for this guy is to use the quadratic formula, right? Because, you know, top of my head, I can't think of two numbers that multiply to give you negative 1 and add to give you 1. So we're going to use the quadratic formula, but before we use the quadratic formula, we're going to look at the discriminant to see if it even has any factors and the discriminant, if you remember from the quadratic formula section, is b squared minus 4ac, b squared minus 4ac, right? So if it's equal to 0, it's got one real root or two identical roots, right? If it's greater than 0, it's got two real roots and if it's less than 0, it's got no real roots, right? So let's take a look at the discriminant before we go ahead and do the whole, you know, layout of the quadratic formula. So the discriminant for this is going to be b squared minus 4ac, right? Just did a correction here because that shouldn't have been negative one. It should have been positive one, right? That's x squared plus x plus 1, right? So our a is equal to 1, our b is equal to 1, our c is equal to 1. If we subbed it in into the discriminant, this is what we're going to end up getting. So we sub in a, b, and c into the discriminant. We're going to get 1 minus 4 times 1 times 1 times 1. So it's going to be 1 minus 4, which is negative 3. Now that means the discriminant is less than 0. So this guy has no real roots. So we're stuck. So possible factors of this guy is just going to be x minus 1 times this. And we've factored as far as we can in the real number realm, right? If we go to complex numbers, imaginary numbers, we can factor this further. But we're not there yet. We're just functioning right now in the real number realm. So this is where we would end up if they said factored this guy. And we can write that down here. So if you're factoring this guy, right? This is as far as you can go. That's it. Because this guy you can't factor in your real number realm, right? So we just factored something that's higher degree than power of 2 using synthetic division. And that's where we're really going with this, because we want to be able to factor large polynomials that way we can get their x-intercepts, right? And once we get their x-intercepts, we can begin to graph them, right? And what this is here is when x is equal to 1, again, what we're talking about here is this is your x-intercept for this function. When x is equal to 1, y is equal to 0, that's where this function crosses the x-axis, right? So if we have your Cartesian coordinate system, if you graph this guy at x equals 1, the function crosses the x-intercept, okay? Let's go do one that's a little bit more complicated than this guy.