Suppose that all four of the numbers \[3 - 2\sqrt{2}, \; -3-2\sqrt{2}, \; 1+\sqrt{7}, \; 1-\sqrt{7}\]are roots of the same nonzero polynomial with rational coefficients. What is the smallest possible degree of the polynomial?
Because the polynomial has rational coefficients, the radical conjugate of each of the given roots must also be roots of the polynomial. However, $1+\sqrt{7}$ and $1-\sqrt{7}$ are each other's radical conjugates, so we only get $2$ more roots. (You might be tempted to think that $3-2\sqrt2$ and $-3-2\sqrt2$ are also a pair of radical conjugates, but the radical conjugate of $3-2\sqrt2$ is $3+2\sqrt2,$ while the radical conjugate of $-3-2\sqrt2$ is $-3+2\sqrt2.$ Therefore, each one of the numbers $3-2\sqrt2$ and $-3-2\sqrt2$ is actually the negation of the radical conjugate of the other one.) In total, the polynomial must have at least $4+2=6$ roots.

Furthermore, the polynomial
\[(x - 3 + 2 \sqrt{2})(x - 3 - 2 \sqrt{2})(x + 3 + 2 \sqrt{2})(x + 3 - 2 \sqrt{2})(x - 1 - \sqrt{7})(x - 1 + \sqrt{7}) = (x^2 - 6x + 1)(x^2 + 6x + 1)(x^2 - 2x - 6)\]has roots $3 \pm 2 \sqrt{2},$ $-3 \pm 2 \sqrt{2},$ and $1 \pm \sqrt{7},$ and has rational coefficients.  Hence, the smallest possible degree is $\boxed{6}.$