We're considering the extension field $\mathbb R$ of $\mathbb Q$
The first element is $2+\sqrt3$ over $\mathbb Q$.
The polynomial I have is $a^4 - 14a^2 + 1.$
The second element is $\sqrt3 + \sqrt5$ over $\mathbb Q$.
The polynomial I have is $a^4 - 16a^2 + 4.$
On
$2 + \sqrt{3}$ is a root of $(x - 2)^2 - 3$. Since $x^2 - 3$ is irreducible, so is $(x - 2)^2 - 3$.
$\sqrt{3} + \sqrt{5}$ is a root of $(x^2 - 8)^2 - 60 = x^4 - 16x^2 + 4$, as you found. You can show that $x^4 - 16x^2 + 4$ is irreducible by trying to factor it explicitly, or you can show that the degree $[\mathbb{Q}(\sqrt{3} + \sqrt{5}):\mathbb{Q}]$ is equal to $4$.
For the first approach: $x^4 - 16x^2 + 4 = 0$ has no rational solutions. By the rational root theorem we only need to check $x = \pm 1, \pm 2, \pm 4$, and there is no need to check negatives because we only have even powers of $x$. A factorization of the polynomial into quadratics has to look like $(x^2 + ax + b)(x^2 + cx + d)$. We get four equations: $a + c = 0, ac + b + d = -16, ad + bc = 0, bd = 4$. The first implies $a = -c$, and the third then implies $b = d$ or $a = c = 0$. If $a = c = 0$, equations 2 and 4 become $b + d = -16$ and $bd = 4$, which has no integer solutions. If $b = d = \pm 2$, equation 2 becomes $-a^2 \pm 4 = -16$, which also has no integer solutions.
For the second approach: We have $(\sqrt{3} + \sqrt{5})^2 = 8 + 2\sqrt{15}$, meaning there is a sequence of field extensions $\mathbb{Q} \subset \mathbb{Q}[\sqrt{15}] \subset \mathbb{Q}[\sqrt{3} + \sqrt{5}]$. Since $\sqrt{15}$ is not rational, the first extension is of degree 2. Then we just need to show $\sqrt{3} + \sqrt{5}$ is not in $\mathbb{Q}[\sqrt{15}]$. If it were, we would have $a + b\sqrt{15} = \sqrt{3} + \sqrt{5}$, and squaring both sides gives $a^2 + 15b^2 = 8$, and $2ab = 2$ Writing $b = 1/a$, we can simplify the first equation to become $a^4 - 8a^2 + 15 = 0$, or $(a^2 - 3)(a^2 - 5) = 0$, which clearly has no rational solutions.
Note that the second approach generalizes easily to any number of the form $\sqrt{p} + \sqrt{q}$, as long as $p,q,pq$ are all not perfect squares.
$2+\sqrt{3}$ is an element of $E=\mathbb{Q}(\sqrt{3})$ which is generated as a $\mathbb{Q}$-vector space by $\{1,\sqrt{3}\}$. Since $E$ is an extension of degree $2$, any element of this field has minimal polynomial of degree at most $2$.
Consider the map $f:E\rightarrow E$ defined as $f(x)=x\cdot(2+\sqrt{3})$. This is a $\mathbb{Q}$-linear map. Let's write it as a matrix :
$f(1) = 2 + \sqrt{3}$ and $f(\sqrt{3})=2\cdot\sqrt{3}+3$.
Hence the matrix is $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$ which has the characteristic polynomial $f(x)=det(A-x\cdot I)=(2-x)^2-3=x^2-4\cdot x +1$. Check that this polynomial admits $2+\sqrt{3}$ as a root. $f$ is the minimal polynomial we are looking for.
Can you apply the same method for $\sqrt{3}+\sqrt{5}$?
Edit : The minimal polynomial of $2+\sqrt{3}$ has coefficients in $\mathbb{Q}$ by definition. You can show that in fact this polynomial must be $(x-2-\sqrt{3})\cdot (x-2+\sqrt{3})=x^2-4\cdot x +1$ since if $2+\sqrt{3}$ is a root then the conjugate $2-\sqrt{3}$ also must be a root. Similarly compute the minimal polynomial of $\sqrt{3}+\sqrt{5}$.
Edit 2 : There is an easier way to show that the minimal polynomial of $\sqrt{3}+\sqrt{5}$ is $a^4-16\cdot a^2+4$. Note that $\mathbb{Q}(\sqrt{3},\sqrt{5})$ has extension degree $4$ hence the degree of the minimal polynomial of $\sqrt{3}+\sqrt{5}$ is either $2$ or $4$. If the minimal polynomial is $x^2-d\cdot x -c$ then we could write $(\sqrt{3}+\sqrt{5})^2=c+d\cdot(\sqrt{3}+\sqrt{5})$. Note that $(\sqrt{3}+\sqrt{5})^2=8+2\cdot\sqrt{15}$ which is not of this form. Hence the minimal polynomial has degree $4$. Now check that the given polynomial admits $\sqrt{3}+\sqrt{5}$ as a root.