Find a polynomial $f(x)$ with integer coefficients such that $f(\sqrt2+i)=0$ 
Given that $α=\sqrt2+i$, construct a polynomial $f(x)$ with
  integer coefficients such that $f(α)=0$.

My try: I squared and got $α^2=-1+2\sqrt2+2$. Here I don't know how to proceed further…
 A: From $\alpha^2-1=2i\sqrt{2}$ we get $(\alpha^2-1)^2=-8$ and $\alpha^4-2\alpha^2+9=0$. Finally, $$f(x)=x^4-2x^2+9$$
A: $f$ is real, so its complex roots come in conjugate pairs. Accordingly:
$$(x-\alpha)(x-\overline\alpha)=(x-(\sqrt2+i))(x-(\sqrt2-i))=x^2-2\sqrt2x+3$$
Now rearrange and square to eliminate the square root:
$$x^2+3=2\sqrt2x$$
$$x^4+6x^2+9=8x^2$$
$$x^4-2x^2+9=0$$
This is the desired $f$ with $\alpha$ as a root.
A: Compute powers of $\alpha$ until you can find a rational linear combination that cancels out the square root.
$$1,\sqrt2+i,1+2\sqrt2i,-\sqrt2+5i,-7+4\sqrt2i,\cdots$$
Now observe that $2\alpha i=-2+2\sqrt2i$ so that
$$2\alpha i=\alpha^2-3$$ and
$$-4\alpha^2=(\alpha^2-3)^2.$$
A: Starting with $\alpha=\sqrt{2}+i$, get rid of the square root by isolating and squaring, so $\alpha-i=\sqrt{2}$ yields $\alpha^2-2\alpha i-1=2$. Now you still have an imaginary part, so isolate it and square, that is, $\alpha^2-3=2\alpha i$ yields $(\alpha^2-3)^2=-4\alpha^2$, so finally
$$
f(x) = (x^2-3)^2+4x^2 = x^4 -2x^2+9.
$$
A: More generally, if $a$ and $b$ are integers, set $r=\sqrt{a}+\sqrt{b}$. Then
$$
(r-\sqrt{a})^2=b
$$
which can be rearranged to $r^2+a-b=2r\sqrt{a}$ and squaring again yields
$$
r^4+2(a-b)r^2+(a-b)^2=4ar^2
$$
so you know that $r$ is a root of
$$
X^4-2(a+b)X^2+(a-b)^2
$$
(note the symmetry between $a$ and $b$).
In your case $a=2$ and $b=-1$, so the polynomial is
$$
X^4-2X^2+9
$$
Further exercise: find a sufficient condition so that $r$ has degree $2$ over $\mathbb{Q}$ by looking at the discriminant of $Y^2-2(a+b)Y+(a-b)^2$. Is it also necessary?
