### Instruction: Find a monic quartic polynomial, in $x,$ with rational coefficients such that $2+\sqrt{2}$ and $1-\sqrt{3}$ are roots of the polynomial.### Response: I know that if $2+\sqrt{2}$ is a root of a polynomial with rational coefficients, then so is its conjugate $2-\sqrt{2}.$ Similarly, if $1-\sqrt{3}$ is a root, then so is $1+\sqrt{3}.$
So, the polynomial must have at least four roots: $2+\sqrt{2}, 2-\sqrt{2}, 1-\sqrt{3},$ and $1+\sqrt{3}.$
To find the polynomial, I can use the fact that the product of two monic binomials with rational coefficients is a monic trinomial with rational coefficients.
For example, $(x-(2+\sqrt{2}))(x-(2-\sqrt{2})) = x^2 - 4x + 2.$
Similarly, $(x-(1-\sqrt{3}))(x-(1+\sqrt{3})) = x^2 - 2x - 2.$
So, the polynomial I'm looking for must be the product of these two trinomials: $(x^2 - 4x + 2)(x^2 - 2x - 2).$
To multiply them, I can use the distributive property: $x^4 - 4x^3 + 2x^2 - 2x^3 + 8x^2 - 4x - 2x^2 + 8x - 4.$