Assuming the polynomial has real coefficients. Sort of implied. Rational coefficients are real, unless you want to extend the definition to gaussian rationals.
In that case, because all coefficients are rational, all irrational roots of the form $a+\sqrt{b}$, for rational $a,b$ and irrational $\sqrt{b}$ come in conjugate pairs. We can conclude straight away that there are also roots $x=2+3\sqrt{2}$ and $x=1+\sqrt{3}$.
This should be easy enough for you to figure out from there.
Note: It would be incorrect to say that every irrational root has a conjugate pair. Only irrational roots of the form $a+\sqrt{b}$ have a conjugate pair. Its a subtlety, but its a common misconception Ive seen far too often. A quartic could conceivably have one linear factor and one cubic factor, and cubics typically have irrational roots of the form $a+\sqrt[3]{b+\sqrt{c}}+\sqrt[3]{b-\sqrt{c}}$. The rule would not apply.
A helpful tidbit: If you have irrational conjugate roots $a\pm\sqrt{b}$ then you have a single quadratic factor of the form $x^2 -2ax +(a^2 - b)$.
Suppose that $f(a + \sqrt[n]{b}) = 0$. Suppose that $f(x) \neq 0$. Instead of $f(x)$, I would like to consider $g(x) = f(a+x)$, which is still a polynomial with rational coefficients and whose degree is the same as that of $f(x)$. Note that $\beta := \sqrt[n]{b}$ is a root of $g(x)$.
I denote the set of all polynomials (in $x$) with rational coefficients as $\mathbb{Q}[x]$. I define a subset:
$$
S = \{\, h(x) \in \mathbb{Q}[x] \mid \text{$h(\beta) = 0$ and $h(x) \neq 0$} \,\}.
$$
It is not empty, since $g(x) \in S$. Among the members of $S$, there exists a unique polynomial $m(x) \in S$ such that:
- the coefficient of the highest-degree term in $m(x)$ is $1$;
- the degree of $m(x)$ is not higher than that of any member of $S$.
Existence: Consider
$$
T = \{\, \text{degree of $h(x)$} \mid h(x) \in S \,\},
$$
which is a nonempty subset of $\mathbb{N}$. Hence the minimal member exists, which means that there exists $M(x) \in S$ such that
$$
\text{degree of $M(x)$} \leq \text{degree of $h(x)$}
$$
for any $h(x) \in S$. Suppose that
$$
M(x) = a_k x^k + a_{k-1} x^{k-1} + \dots + a_0
$$
with $a_k \neq 0$. Let $m(x) = \frac{1}{a_k} M(x)$. It is not hard to see that $m(x)$ satisfies the two requirements above.
Uniqueness: Suppose that $m(x)$, $z(x)$ both satisfy the two requirements above but are unequal. By the second point, the degree of $m(x)$ has to be equal to that of $z(x)$. By the first point, $d(x) := m(x) - z(x)$ is a polynomial with lower degree than that of $m(x)$.
Since $d(\beta) = 0$ and $d(x) \neq 0$, $d(x)$ is a member of $S$. This is a contradiction, since the degree of $d(x)$ is lower than that of $m(x)$.
I call $m(x)$ the minimal polynomial of $\beta$.
$m(x)$ is irreducible. I prove the claim by assuming that there exist two polynomials $m_1 (x)$, $m_2 (x) \in \mathbb{Q}[x]$ such that $m(x) = m_1 (x) m_2 (x)$ and that neither of $m_1(x)$, $m_2 (x)$ is a polynomial of degree zero. Since $0 = m(\beta) = m_1 (\beta) m_2 (\beta)$, one of $m_1 (\beta)$, $m_2 (\beta)$ must be zero. Without loss of generality, suppose that $m_1 (\beta) = 0$. $m_1 (x)$ is not a zero polynomial; $\beta$ is a root of $m_1 (x)$. Hence $m_1 (x) \in S$. Hence the degree of $m_1 (x)$ is not lower than that of $m(x)$. Note that the degree of $m(x)$ is not lower than that of $m_1 (x)$. Hence the degree of $m_1 (x)$ is equal to that of $m(x)$. Hence $m_2 (x)$ is a polynomial of degree zero, which is a contradiction.
I claim that $m(x)$ divides $g(x)$. I use "reductio ad absurdum" again. Suppose that it is not the case. Then $m(x)$ and $g(x)$ are coprime. Hence there exist two polynomials $r(x)$, $s(x) \in \mathbb{Q}[x]$ such that
$$
m(x) r(x) + g(x) s(x) = 1.
$$
Hence
$$
m(\beta) r(\beta) + g(\beta) s(\beta) = 1,
$$
which is a contradiction.
Let me summarize what I have got:
Let $f(x)$ be a polynomial with rational coefficients. Suppose that $f(a + \sqrt[n]{b}) = 0$, in which $a$, $b$ are rational numbers. Let $g(x) = f(a + x)$. Let $\beta = \sqrt[n]{b}$. Let the minimal polynomial of $\beta$ be $m(x)$. Then $m(x)$ divides $g(x)$.
If $\beta$ is a rational number, what I can learn is not interesting, since the minimal polynomial of $\beta$ is simply $x - \beta$.
If $\beta$ is not a rational number, what I can learn is not boring, since the minimal polynomial of $\beta$ cannot be of degree $1$.
Example. It can be verified that the minimal polynomial of $\sqrt[3]{7}$ is $m(x) = x^3 - 7$. It can be verified that $4 + \sqrt[3]{7}$ is a root of $f(x) = x^3 - 12x^2 + 48x - 71$. Hence $m(x)$ must divide $g(x)$, in which
$$
g(x) = f(x + 4) = x^3 - 7,
$$
which is just the same as $m(x)$. The roots of $m(x)$ are well known, so it can be learned what other roots $g(x)$ must have, from which it can be learned what other roots $f(x)$ must have.
And what about nested radicals?
You are invited to explore this yourself.
Best Answer
$$ \begin{aligned} x&= 3 - i \sqrt[4]{2} \\ x-3&= -i \sqrt[4]{2} \\ (x-3)^4&= 2 \\ (x-3)^4-2&=0 \\ x^4 - 12 x^3 + 54 x^2 - 108 x + 79&=0 \end{aligned} $$
EDIT. In fact, you can even factor the given polynomial fully (if you desired) because examining the construction closely you can see the roots of $x^4 - 12 x^3 + 54 x^2 - 108 x + 79$ are precisely, $3 \pm i \sqrt[4]{2}$ and $3 \pm \sqrt[4]{2}$, i.e. $3 + \sqrt[4]{2} e^{2\pi ki/4}$ for $k=0,1,2,3$.