Package of problems

## Suggestion

• Can you find a quadratic polynomial with integer coefficients which has $x=1+\sqrt{2}$ as a root?

What is the other root of this polynomial?

I know that the quadratic factorises as $(x-(1+\sqrt{2}))$ times something. So perhaps I could write the quadratic as $(x-(1+\sqrt{2}))(ax+b)$ and work out what $a$ and $b$ could be.

Alternatively, perhaps I could think about the relation between $x=\sqrt{2}$ and $x^2-2=0$, and see where that leads me to if I start with $x=1+\sqrt{2}$ instead.

As a third possible approach, I could think about the graph of $y=x^2-2$, which has $x=\sqrt{2}$ as a solution, and wonder what I would have to do to get $x=1+\sqrt{2}$ as a solution instead.

As a fourth approach, since $x$ involves square roots, I could try squaring it to get $x^2=3+2\sqrt{2}$. Can I somehow use this together with $x$ to get rid of the $\sqrt{2}$ part?