In general, you can find the minimum polynomial for an algebraic number $\alpha$ by determining the smallest power of $\alpha$ for which $\{1,\alpha,\alpha^2,\ldots,\alpha^n\}$ is linearly dependent over $\mathbb{Q}$.
For example, let $\alpha=\sqrt{2} + \sqrt{3}$. Then
\begin{align*}
\alpha^2 &= 5+2\sqrt{6} \\\
\alpha^3 &= 11\sqrt{2}+9\sqrt{3} \\\
\alpha^4 &= 49+20\sqrt{6}
\end{align*}
Since $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ are linearly independent over $\mathbb{Q}$, we can think of the powers of $\alpha$ as vectors:
$$
1 = \begin{bmatrix}1\\\ 0\\\ 0\\\ 0\end{bmatrix},
\qquad
\alpha = \begin{bmatrix}0\\\ 1\\\ 1\\\ 0\end{bmatrix},
\qquad
\alpha^2 = \begin{bmatrix}5\\\ 0\\\ 0\\\ 2\end{bmatrix},
\qquad
\alpha^3 = \begin{bmatrix}0\\\ 11\\\ 9\\\ 0\end{bmatrix},
\qquad
\alpha^4 = \begin{bmatrix}49\\\ 0\\\ 0\\\ 20\end{bmatrix}
$$
As you can see, $\{1,\alpha,\alpha^2,\alpha^3\}$ is linearly independent, so $\alpha$ is not the root of any cubic polynomial. However, $\{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}$ is linearly dependent, with
$$
\alpha^4 - 10\alpha^2 + 1 \;=\; 0
$$
It follows that $x^4-10x^2+1$ is the minimum polynomial for $\alpha$.
This technique depends on being able to recognize a useful set of linearly independent elements like $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$. Depending on how much Galois theory you know, it may be hard to prove that elements like this are linearly independent over $\mathbb{Q}$. In this case, you can use this technique to guess the minimum polynomial, and then prove that you are correct by proving that the polynomial you found is irreducible.
For example, if $\alpha = i\sqrt[4]{2}$, then
$$
\alpha^2 = -\sqrt{2},\qquad \alpha^3 = -i\sqrt[4]{8},\qquad \alpha^4 = 2.
$$
It seems clear that $ \{1,\alpha,\alpha^2,\alpha^3\} $ is linearly independent, while $\{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}$ satisfies $\alpha^4-2 = 0$. Thus $x^4-2$ should indeed be the minimum polynomial over $ \mathbb{Q} $, though the easiest way to prove it is to show that $x^4 -2$ is irreducible. This can be done using Eisenstein's criterion, or by checking that $ x^4-2 $ has no roots and does not factor into quadratics modulo $4$.
Best Answer
We can simplify the expression $\alpha^3 - 14 \alpha$. This occurs because $\alpha^2$ satisfies $\alpha^4 = 16\alpha^2 - 4$, therefore $$\alpha^3 = 16\alpha - 4\alpha^{-1} $$
and so $\theta = \alpha^3 - 14\alpha = 2\alpha - 4\alpha^{-1}$. (note that $\alpha \neq 0$ so is invertible)
Let us square this term : $$ \theta^2 = 4\alpha^2 - 16\alpha^{-2} -16 = \frac{4\alpha^4 - 16}{\alpha^2} - 16 = 4\frac{\alpha^4 - 4}{\alpha^2} -16 $$
but $\alpha^4 - 4 = 16 \alpha^2$! So substitute that in and get $4 \times 16 - 16 = 48 = \theta^2$.
Thus, the minimal polynomial should be $x^2-48$ as long as you see that is irreducible.
We are lucky that such a manipulation worked out. In general, however, we'd either have to visit higher powers, or need good manipulative intuition to find the minimal polynomial. In some cases, knowing a root can be helpful, so you can directly compute the expression for that root.
For example, the polynomial given is in fact the minimal polynomial of $\sqrt 3 + \sqrt 5$, so one can try the manipulation with this as $\alpha$ to see that $\alpha^3 - 14 \alpha = 4\sqrt 3$ (in case that no intelligent techniques work, this is brute force).