Rationalizing denominator containing a root of a polynomial - why is this possible / why does it work? The problem is to express the number
$$1\over x^5 + 2x^4 + 3x^3 + 3x^2 + 2$$
using only rational numbers in the denominator, knowing that $x^5 + 2 = -2(x + 1)(x^3 + x^2 + x)$.
(This is an example from a final exam of an algebra course from the past year (which is freely available to students).)
I know how to solve this problem, but I do not understand why this is possible. The trick comes from knowing that this number can be expressed using addition and multiplication of $x$ and rational numbers, ie. as
$$r_0 + r_1x + r_2x^2 + r_3x^3 + r_4x^4 + r_5x^5 + \dots$$
My solution follows, skip to questions if you are not interested.
From the polynomial equation it follows that $x^5$ equals $-2 - 2x - 4x^2 - 4x^3 - 2x^4$, therefore an expression containing $x$ with exponent 5 or greater can be rewritten as an expression with only $x^0$ to $x^4$.
So my number can be expressed as
$$ax^4 + bx^3 + cx^2 + dx + e$$
Putting an equals sign between these two expressions and multiplying by the denominator gives
$$1 = (x^5 + 2x^4 + 3x^3 + 3x^2 + 2)(ax^4 + bx^3 + cx^2 + dx + e)$$
I can multiply the right hand side and "normalize" it (by replacing $x^{\ge5}$) to get
$$1 = (e - f)x^4 + \dots + (2e - 2d)$$
From this formula I can create a system of linear equations, because the coefficient of $x^n$ has to be one if n equals zero, and zero otherwise. Solving this system will let me find the values of the coefficients.
Question 0: Why does this problem even have a solution of the form I used?
Question 1: Is there a solution like this for any expression $1 \over f(x)$, where f is a polynomial of $x$, knowing $x$ is a root of another polynomial? Will my method always work? Why?
Apologies if this is a duplicate, but everything I could find about rationalizing the denominator was about simple fractions like $1 \over 2 + \sqrt 6$.
I'd appreciate even just comments directing me towards what I can study because I am out of keywords.
 A: In short one possible way to learn why this works is via Algebraic Number Theory, but it could also come from other areas.  

The given equation means $x$ is an algebraic number, which are complex numbers that satisfies an irreducible (cannot be factored over $\mathbb Z[x]$) integer polynomial with finite degree. In this case it's an algebraic integer since leading coefficient of the determining polynomial is $1$ (since $1\cdot x^5$).  
Usually we fix one of the roots of
$$
x^5 + 2 = -2(x + 1)(x^3 + x^2 + x),
$$
say $\alpha$, and work using it (so all $\alpha$'s instead of $x$'s), but we'll continue with $x$ here since it's largely similar for intuition purposes.  
Now the key part is if you studied some basics of Algebraic Number Theory, one of the first things you'll learn is that sum and product of algebraic integers are still algebraic integers (same for algebraic numbers). This means that the new arbitrary polynomial you formed, which we denote as $\beta$:
$$
\beta:x^5 + 2x^4 + 3x^3 + 3x^2 + 2
$$
is an algebraic integer (sum and product of $x$) and hence must satisfy some irreducible integer polynomial $g(y)$ with leading coefficient $1$ (That is the definition of algebraic integers). So you have
$$
g(y):=y^k+a_{k-1}y^{k-1}+\cdots+a_0,\;\;\;\; g(\beta)=0
$$
Note that the $a_i$'s are integers and $y$ is a variable. Now in order for $g(y)$ to be irreducible, $a_0$ cannot be $0$. Otherwise we can factor out $y$. Then since $a_0\neq 0$ we can now form
$$
\begin{align*}
0 &= g(\beta) = \beta^k+a_{k-1}\beta^{k-1}+\cdots + a_1\beta + a_0\\
\frac{-(\beta^{k-1}+a_{k-1}\beta^{k-2}+\cdots + a_1)}{a_0} &= \frac{1}{\beta} 
\end{align*}
$$
So your $1/\beta$ always can be turned into an expression in terms of rational coefficients. Notice that the only requirement is that $\beta$ is an algebraic integer (works similarly for algebraic numbers), which provides you the equation to work with.  

On a related note, using your linear algebra arguments you can also argue in an elementary way that such a $g(y)$ exists. i.e. solve for such a $g(y)$.
