(i) To verify that \(u = 1 + \sqrt{2}i\) is a root of \(p(x) = 0\), substitute \(x = 1 + \sqrt{2}i\) into \(p(x)\):

\(p(1 + \sqrt{2}i) = (1 + \sqrt{2}i)^4 + (1 + \sqrt{2}i)^2 + 2(1 + \sqrt{2}i) + 6\).

Calculate \((1 + \sqrt{2}i)^2 = 1 + 2\sqrt{2}i - 2 = -1 + 2\sqrt{2}i\).

Calculate \((1 + \sqrt{2}i)^4 = ((1 + \sqrt{2}i)^2)^2 = (-1 + 2\sqrt{2}i)^2 = -1 - 4 + 8i^2 = -5 - 8 = -13\).

Substitute back: \(-13 + (-1 + 2\sqrt{2}i) + 2 + 2\sqrt{2}i + 6 = 0\).

Thus, \(u = 1 + \sqrt{2}i\) is a root. The second complex root is \(1 - \sqrt{2}i\).

(ii) To find the other two roots, factor \(p(x)\) using the quadratic factor \(x^2 - 2x + 3\) obtained from the roots \(1 \pm \sqrt{2}i\).

Divide \(p(x)\) by \(x^2 - 2x + 3\) to get a quotient \(x^2 + kx + c\).

Perform polynomial division to find \(x^2 - 2x + 2\) as the other factor.

Find the roots of \(x^2 - 2x + 2 = 0\) using the quadratic formula: \(x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{2 \pm \sqrt{-4}}{2} = 1 \pm i\).

Thus, the other two roots are \(-1 + i\) and \(-1 - i\).