 Let's tie together three important ideas. A root of a polynomial is a value that makes the polynomial equal to zero. Now remember how you speak influences how you think, and so while we use the term root, we also call this the zero of a polynomial. Another important idea is that if x equals a is a root of a polynomial, then x minus a is a factor, and then finally we introduce the idea of the multiplicity of a root. This is the number of times the corresponding factor appears in the factorization of the polynomial. So let's find the roots and multiplicities of this horrifying thing. And so here we'll be relying on our rational root theorem. The potential rational roots of a polynomial with leading coefficient q and constant p are of the form p prime divided by q prime, where p prime is a divisor of our constant term, and q prime is the divisor of our leading coefficient. So we can find the divisors of the constant term 108 and the divisors of the leading coefficient, and so a potential root is one of these divisors of 108 divided by one of the divisors of one. And since we need to impress bands of ferrule mathematicians, we'll start with the biggest numbers first. Well, wait. Since it doesn't matter where we start, we'll start with the easiest numbers first. And we can simultaneously find a root and a factorization by using synthetic division and the remainder theorem. If r is the remainder when f of x is divided by x minus a, then f of a is equal to r. And so we want a remainder of 0, which will give us a factorization and a root. So we'll check our roots one by one. We'll set up our synthetic division table. Don't forget to include that coefficient of one. And we'll try our root x equal to one. So we'll divide by x minus one and get a non-zero remainder, which means that x equals one is not a root. We'll try negative one and get a non-zero remainder. We'll try two and get a non-zero remainder. We'll try negative two and get a non-zero remainder. We'll try three and get a non... wait, oh, no, that worked. And so we find x equals three is a root. And remember the terms of our quotient can be read off the last line of the synthetic division table. And so our quotient will be one x cubed minus two x squared plus nine x minus 36. Now we already know plus or minus one and plus or minus two are not roots, but three might be a repeated root. So we'll try dividing by three again and we find, and since our remainder is zero, x minus three is another factor, and the other factor is x squared plus x plus 12. And since x minus three appears twice as a factor, x equals three is a root of multiplicity two. While we could use synthetic division and the rational root theorem to try and find additional rational roots, since x squared plus x plus 12 is a quadratic, we'll use the quadratic formula. And so the quadratic gives roots of negative one half plus i root 47 halves and negative one half minus i root 47 halves. And since the factor corresponding to these roots was raised to the first power, these roots have multiplicity one.