 Now that we have the notion of what it means to factory polynomial, let's talk about the complete factorization of a polynomial. Now fair warning, factoring is the hardest easy problem in mathematics. What we mean by that is that it's easy to explain what the problem is, but it's often very hard to actually solve the problem. Now from your previous experience factoring numbers, you might think that's not the case, factoring's easy. But the truth is, it only seems to be easy because we've only ever asked to solve easy problems. So for example, factor 12, well that's easy. On the other hand, factor 3,891, well that's hard. Let's set down some ground rules. When we factor integers, we look for integer factors, two integers that multiply to a number. So 20 is equal to 5 times 4, and not even though it's true, one-third times 60. And that's because one-third is not an integer factor. We'll make a similar restriction for factoring polynomials. When we factor polynomials with rational coefficients, we look for factors with rational coefficients. So we can factor x squared minus x minus 20 as, well, I don't know, x minus 5 times x plus 4. So here we see that both of our factors have rational coefficients, and this is a factorization. Well, how do we know when to stop? We say a polynomial is irreducible over the rationals if it cannot be written as a product of lower-degree polynomials with rational coefficients. So 3x plus 5 is irreducible. x squared plus x is not irreducible because x squared plus x is x times x plus 1. And so we say that a polynomial has been completely factored when it is written as a product of irreducible polynomials. Now, because our definition of factorization talks about writing a polynomial as a product of lower-degree terms, this does introduce the following special case. What if a polynomial has a constant factor? So suppose I begin with this product 4x plus 8 times 2x minus 3. This is a factorization. It's a product of two first-degree polynomials, neither of which can be written as a polynomial of lower degree. But we can do a little bit more work here. For example, this 4x plus 8, we can actually factor a 4 out of that first term, and because multiplication is associative and commutative, I can move that 4 and migrate it into the second term by using the distributive property. Since I've maintained equality throughout, I can say that 4x plus 8 times 2x minus 3 is the same thing as x plus 2 and 8x minus 12. And this means that 4x plus 8 times 2x minus 3 and x plus 2 times 8x minus 12 are different factorizations of the same polynomial. To avoid this, it's useful to remember if you can remove a constant factor, do so. So the proper final factorization of this polynomial would be this line, where the common factor has been removed, where possible. So let's try to factor something like this. Our ability to factor completely relies on two important ideas. First, there's the factor theorem. If x equals a is a root of a polynomial, in other words, it's something that's going to make the polynomial equal to 0, then x minus a will be a factor. Now in general, finding the root of a polynomial is hard, but we only care about the rational roots because they lead to the rational factors. And so we can use the rational root theorem, which tells us that those rational roots will be of the form a divisor of the constant over a divisor of the leading coefficient. Now these two form the theoretical basis for our factorization process. The practical implementation of the factorization process relies on the remainder theorem. And remember that says if I want to evaluate a polynomial at x equal to a, find the remainder when you divide by x minus a. So let's try and factor this mess. The first thing to remember is that if you can remove a constant factor, do so. It'll make your life much easier. So here we see that every term has a factor of 4, so we can remove a factor of 4. So now we'll use our rational root theorem. The rational root theorem tells us that a rational root, if it exists, must be among a rather long list of possibilities. And unfortunately, there's nothing we can do but to try every single possibility until we find a root. So again, we'll use synthetic division and find the remainder when we divide by x minus a potential root. So we'll set up our synthetic division table and list the roots that we're going to test. And we'll try them out from easiest to hardest. So we'll try out plus 1. And again, what we really care about is the remainder because that'll tell us the value of the polynomial when x equals 1. So using our synthetic division algorithm, at this point there is a useful shortcut. Since we're looking for a root, we want the remainder to be 0. If the remainder isn't 0, we don't care what it actually is. And here the remainder is not 0. So positive 1 is not a root. Well, as the saying goes, if at first you don't succeed, give up and quit. Oh wait, no one says that. We'll keep trying. The next thing we could try is negative 1. So we'll clear out our table and find the remainder when we divide by x minus negative 1. And again, since the remainder is not 0, we don't actually care what it is. We know that negative 1 is not a root. So we can try 2. So we'll check by applying our synthetic division algorithm and finding the remainder. So plus 2 is not a root. We'll check negative 2. And use our synthetic division to find the remainder when we're dividing by x minus negative 2. Still not a root, so we'll try x equals 3. And since our remainder is 0, that means x equals 3 is a root. And because we've actually done the division, we can read this as a factorization. Our original dividend will be x minus 3 times something. So if it's not written down, it didn't happen. Let's record that factorization. Equals means replaceable. So where we had this original dividend, we can replace it with the factored form. And now we have a quadratic polynomial. We could factor this using the quadratic formula to find the roots, but just for practice, let's use our rational root theorem again. The rational root theorem tells us that a rational root, if it exists, must be among. And again, we have a very long list of possibilities. However, here's a useful idea. Remember, arithmetic is bookkeeping and we've already determined that some of these are non-roots. We've already checked plus or minus 1 and plus or minus 2. So we don't need to check whether these are roots. We already know they aren't. So the first root we'll bother to check is 4. Since the remainder is 0, 4 is a root, and our synthetic division table tells us that the original dividend is x minus 4 times something. Which gives us the factorization. And again, if it's not written down, it didn't happen. And since all of our factors are first degree polynomials, we can't write any of them in terms of lower degree polynomials, and our factorization is complete.