 Because of Gauss' lemma, the question of factoring a polynomial in d adjoin x, where d is a unique factorization domain versus f adjoin x, where f is the field of fractions of d, the question of factoring a polynomial, one of these two rings is in fact the same. We can factor a polynomial in f adjoin x, where we're allowed to divide non-zero coefficients, or we can factor a polynomial in d adjoin x, where we have primes and divisibility. Even if the polynomial is properly contained inside of f adjoin x, we can still factor out the denominators and factor with the resultant polynomial over here in d adjoin x. Therefore, it behooves us to establish a few strategies of identifying irreducible polynomials in either of these two rings. Now, one of these mini-strategies is to search for roots. Because by the factor theorem, a polynomial has a linear factor overfield if and only if it has a root. In particular, if we have a polynomial and it has a root in the field of fractions, then in the field of fractions, it'll have a linear factor. Then in d adjoin x, it'll also have a linear factor. Even if it doesn't have a root in d adjoin x, if it does have a root in the field of fractions, that still gives us something about factorization. It still gives us a factorization over here, even if the root doesn't exist. Because of that, we really are interested in finding roots, and this forces us to go back to the field side, because there might not be the root in d adjoin x here. Like for example, if you take the polynomial 2x minus 1, if you view it as an integer polynomial, then it doesn't have a root because the root should be 1 half, which 1 half belongs to q adjoin x, but not to the integers. But we still do have the linear factor. As we're trying to factor up integer polynomial, maybe one of the factors was 2 minus x. Well, if you're looking for roots, you're not going to find them among the integers, but you could find it with the field elements. Then once you find it amongst the field elements, then you get your linear factor and can bring it back to z adjoin x. For the moment b, at least in this video, we're going to focus on the field where we are going to have the root and consider finding roots. A very useful strategy for finding irreducible polynomials, the following is when your polynomial has degree 2 or 3. If your polynomial has degree 1, then it's a linear polynomial. You can't factor it, at least not over the field, you can't. All linear polynomials over a field are irreducible. But if you have, if you're quadratic or cubic, the following strategy can work. You have some polynomial f of x, which is degree 2 or degree 3. Then f is reducible if and only if it has a root. Now, clearly, if you have a root, then you are reducible. This is a consequence of the factor theorem. But this is an if and only if statement. It goes the other way around. That if it's reducible, then it has to have a root. That's because it has to have a linear factor of its reducible. By the factor theorem, if it has a root, then it has a linear factor, and that's a proper factorization. That's the easy direction. But suppose that f has a factorization, a proper factorization. So g and h are polynomials of smaller degree, but they're non-constant because otherwise, they'd be a unit. And so in that situation, if you look at g or if you look at h, then because of 2 and 3, you can only break it up in one of two ways. When it comes to polynomial degree 2, you're going to have to break it up as two linear polynomials. So actually, you get two roots in that situation. Degree 3, you're going to have to break it up as a linear polynomial and maybe a quadratic polynomial. Or that quadratic polynomial could break up into two linears. But nonetheless, there's always a linear factor. So in degree 2 or degree 3, polynomial is going to be divisible only if it has a root because it would have to have a linear factor in that case. So of course, an important example to look at here is to look at the polynomial f of x equals x squared plus 1 and view this as a rational polynomial. I mean, it's an integer polynomial, but because of the previous theorem, we want to be looking at the rationals here because what if the root was a fraction, right? That's a polynomial. That's a possibility. Before going on, let me make a comment about the so-called rational roots theorem. Imagine you have a polynomial f of x, which could be written as a to the n, x to the n, all the way down to a to the 0. And this belongs to some polynomial ring whose coefficients come from a unique factorization domain, D. If you're looking for a quote unquote rational root of this polynomial, what you mean is you're looking for roots that live inside the field of fractions. So let f be the field of fractions for this unique factorization domain, D. So I'll make a comment about D is a unique factorization domain. Then the roots of little f inside of capital F are of the form. They're going to look like some p over q. So like x equals p over q where p here divides the constant term and q divides the leading coefficient, like so. This is a common fact that students learn about in a college algebra class like Math 1050 at SUU. And so with this polynomial here, we look at the divisors of one divided by the divisors of one. This give us two possible rational roots plus or minus one. And a very quick evaluation shows that neither of those work because in both cases you're going to get two, which is not zero. And so this shows us that by the rational roots theorem, which I leave as an exercise to the viewer here, because of the rational roots theorem, the only possible roots to the polynomial x squared plus one, x squared plus one as a rational polynomial would be plus or minus one. Neither of them works. So it doesn't have a root in the rational field, which then since this is a degree two polynomial, that means this polynomial is irreducible over the rational field. Let's look at another example. Let's look at the polynomial f of x equals x to the fourth minus two x squared plus x plus one. Again, by the rational roots theorem, if there is a, because we're viewing this as a rational polynomial right now, if there is a rational root to this polynomial, it would have to again be x equals plus or minus one. So we can check very quickly that doesn't happen if you plug in one, right? One minus two plus one plus one, you end up with one, that's not zero. And if you do negative one, you're still going to get a positive one there. You're going to get a plus one there. You're going to get a minus one there, plus one. So you end up with three this time, assuming my arithmetic's not incorrect, which is still not, it's not zero. So this polynomial doesn't have any roots to it whatsoever. Now this polynomial is degree four. So there is one other important possibility here. So having no roots means that this polynomial doesn't have a linear factor, but because it's degree four, it is possible that the polynomial could be factored into two irreducible quadratics, two quadratic polynomials with rational coefficients that don't have rational roots, such as x squared plus one. And so let's consider that for a moment. Since f is a monic polynomial here, this is a nice thing about rational factorizations. If it wasn't monic, we could just factor out the leading coefficient from everything. You might get some fractions, but with respect to the rational numbers, that's no biggie whatsoever. And so in particular, if this polynomial was reducible, I could factor it as two irreducible monic quadratic polynomials. So that would look something like x squared plus ax plus b, and x squared plus cx plus d for some coefficients a, b, and cd. For which if you were to multiply this out by the distributive property, this would look like x to the fourth plus acx cubed plus b plus ac plus d x squared plus ad plus bcx plus bd. So if you just look at all the possible combinations combined like terms, that's how those things are going to combine. And this actually gives us a system of equations, right? Because when we compare coefficients, the coefficient of x cubed should be negative two. It's going to be a plus c. So we get the equation a plus c equals negative two. If we look at the coefficient of x, which is one, it's given by the formula b plus ac plus d. So we end up, I'm sorry, that was the linear term, which is ad plus bc equals, that equals one. This function doesn't have a quadratic term, right? So we have b plus ac plus d actually equals zero in that situation. And then the constant term bd, right? This should equal one right here. And so this then gives us the possibilities. But this is a great thing, even though we're working over the rational numbers for which there are a lot of possibilities here. By Gauss's lemma, we can look at the integers here. We can assume, I mean, since f is in fact an integer factorization, we can assume that these factors have to be integer polynomials as well. Gauss's lemma's awesome with this regard. Because over the rational field, there's a lot of possibilities with these equations right here. But with integers, it gets a lot more restrictive. After all, if you have a product to integers that are equal to one, that tells you that the only way you could do that is b and d are both one, b and d are both negative one, like so. In particular, b and d are the same thing, in which case they're equal to plus or minus one. But then when you look at this equation right here, all right? Actually, nope, I wanna look at this one, J.K. If you look at this one, if you plug in one for b and d, you're gonna get a plus c equals one. Or if you plug in negative one for b and c, then you'd end up with negative a minus c equals one. Or alternatively, a plus c equals negative one. So in particular, a plus c is gonna equal one or negative one. So a plus c is gonna equal whatever b and d happen to be. And so we end up with a plus c is equal to plus or minus one. But wait a second, if you look at this coefficient, we have a plus c has to equal negative two. You can't do that over the integers. You can't be negative two and one or negative one. We get a contradiction here. In which case this tells us that the polynomial, let me bring it back to the screen. Our polynomial f here, in fact, is irreducible. It doesn't have a linear factor because it doesn't have a rational root, but it also doesn't have a factorization using irreducible quadratic factors. Now in many situations, this brute force approach to proving the irreducibility of a polynomial is quite possible and could be practical, right? It essentially works well for quartic polynomials because if you're a quartic polynomial degree four, if you're reducible, you have either a linear factor, which corresponds with the rational roots that we just talked about. Or it has to have these two irreducible quadratics and you could try to play this game with the nonlinear system of equations and if you get a contradiction, it means it's irreducible. If you don't get a contradiction, you probably actually found a factorization and voila, Bob's your uncle, you have your factorization. But in general, this technique is not going to do well. Like when you start getting degree five, degree six and beyond, it's going to be very difficult to brute force all the possible factorizations and so it would be nice to have some other criteria. In the next video, we'll talk about Eisenstein's criteria, which is actually a very useful technique for showing that arbitrarily large polynomials can be in fact irreducible.