Find a polynomial with rational coefficients 
Let $γ$ be a root of $x^5 − x + 1 = 0$ in an algebraic closure of $\mathbb{Q}$. Find a polynomial with rational coefficients of which $γ +\sqrt2$ is a root

Is it possible to directly modify the polynomial itself so that its root is $\gamma + \sqrt2$ to get the polynomial?
 A: You’ll use $\gamma$’s polynomial, yes. If $\alpha = \gamma + \sqrt2$, then $\alpha-\sqrt2$ solves the polynomial; substitute. Open the powers with the binomial theorem. You’ll have terms with $\alpha$ and $\sqrt 2$. Group terms with $\sqrt 2$, leave them alone in one side of the equation and square. Finally, this polynomial over $\alpha$ has rational coefficients and we found what we wanted.
A: $1$, $\gamma$, $\gamma^2$,  $\gamma^3$,  $\gamma^4$,  $\sqrt{2}$, $\gamma\sqrt{2} $, $\gamma^2\sqrt{2} $,  $\gamma^3 \sqrt{2} $,  $\gamma^4 \sqrt{2} $ form a basis for the $\mathbb{Q}$ vector space $\mathbb{Q}(\gamma, \sqrt{2})$.
This means that $\{1, (\gamma + \sqrt{2}), (\gamma + \sqrt{2})^2, ... ,  (\gamma + \sqrt{2})^{10} \} $ is not linearly independent over $\mathbb{Q}$. Write these numbers down in terms of the basis, and use linear algebra to find a linear combination of these things that is $0$, which will be your polynomial.
A: You can compute the resultant of  $x^5 − x + 1$ and $(y-x)^2-2$ to eliminate $x$:
$$
y^{10} - 10 y^8 + 38 y^6 + 2 y^5 - 100 y^4 + 40 y^3 + 121 y^2 + 38 y - 17
$$
The resultant can be computed by hand but it is a large determinant and is best found with a computer, for instance with Resultant in Mathematica and WolframAlpha.
