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?
