 Now that we have the notion of what it means to factory a 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 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 3891 Well, that's hard Let's set down some ground rules when we factor integers We look for integer factors to 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 could 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 2 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 a quality 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 zero 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 Four so we could remove a factor of four 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 one and again What we really care about is the remainder because that'll tell us the value of the polynomial when x equals one 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 zero if the remainder isn't zero We don't care what it actually is and here the remainder is not zero So positive one 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 one So we'll clear out our table and find the remainder when we divide by x minus negative one And again since the remainder is not zero. We don't actually care what it is. We know that negative one is not a root So we can try two So we'll check by applying our synthetic division algorithm and finding the remainder So plus two is not a root. We'll check negative two and Use our synthetic division to find the remainder when we're dividing by x minus negative two Still not a root. So we'll try x equals three and Since our remainder is zero that means x equals three 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 three 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 one and plus or minus two So we don't need to check whether these are roots. We already know they aren't So the first root will bother to check is four Since the remainder is zero four is a root and our synthetic division table tells us that the original dividend is x minus four 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