Let $f = (x^4 - 1)(x + 2)$ and let $g = (x^2 - 4)(x + 1)$. Find a single polynomial that generates the ideal $I = (f,g)$ of $\mathbb{R}[x]$ The Problem:
Let $f = (x^4 - 1)(x + 2)$ and let $g = (x^2 - 4)(x + 1)$. Find a single polynomial that generates the ideal $I = (f,g)$ of $\mathbb{R}[x]$.
My Approach:
This seemed a bit too easy, which made me think I'm misunderstanding the problem...
Since $f = (x+1)(x-1)(x^2 +1)(x+2)$ and $g= (x+2)(x-2)(x+1)$, it follows that $f$ and $g$ are both in the ideal generated by $p = x+1$ (or, for that matter, the ideal generated by $q = x-1$). Thus $p$ generates $I$.
Is that all there is to it?
EDIT: I had misdefined $g$.
 A: No: you have shown that $(f,g)$ is contained in $(p)$ (or $(q)$), but you need to show $(f,g)$ is equal to the ideal generated by some single polynomial.  In other words, you would need to show the reverse inclusion as well, that $(f,g)$ contains $(p)$.  You have not proved that, and in fact it is not actually true for the $p$ you have chosen.
A: It is well-known that $\Bbb R[x]$ is a principal ideal domain; thus the ideal
$(f(x), g(x)) = (d(x)) \tag 1$
for some
$d(x) \in \Bbb R[x]; \tag 2$
it immediately follows that
$d(x) = r(x)f(x) + s(x)g(x) \tag 3$
for some 
$r(x), s(x) \in \Bbb R[x]; \tag 4$
furthermore, it is clear from (1) that
$d(x) \mid f(x), \; d(x) \mid g(x); \tag 5$
it is equally clear via (3) that any
$p(x) \in \Bbb R[x] \tag 6$
such that
$p(x) \mid f(x), \; p(x) \mid g(x) \tag 7$
will also satisfy
$p(x) \mid d(x); \tag 8$
(5) and (8) together show that $d(x)$ by definition is in fact given by
$d(x) = \gcd(f(x), g(x)); \tag 9$
that is, $d(x)$ is the greatest common divisor of $f(x)$, $g(x)$ in $\Bbb R[x]$; this fact gives us a tool by which $d(x)$ may be had: calculating the irreducible, hence prime, factors of $f(x)$ and $g(x)$; we find that
$f(x) = (x^4 - 1)(x + 2) = (x^2 + 1)(x + 1)(x - 1)(x + 2), \tag{10}$
$g(x) = (x^2 - 4)(x + 1) = (x - 2)(x + 2)( x + 1); \tag{11}$
inspection of (10), (11) reveals that every factor on the right-hand sides of these equations is a prime (irreducible) in $\Bbb R[x]$, being either linear or the quadratic $x^2 + 1$ which has no roots in $\Bbb R$; the factors they have in common are $x + 1$ and $x + 2$; therefore, since $\Bbb R[x]$ is a unique factorization domain (as is any principal ideal domain), the unique prime factors of $d(x)$ must be these two; that is
$d(x) = (x + 1)(x + 2). \tag{12}$
$OE\Delta$.
