(i) Substitute \(z = -1 + i\) into \(p(z)\):

\(p(-1+i) = ((-1+i)^4) + 3((-1+i)^2) + 6(-1+i) + 10\).

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

Calculate \((-1+i)^4 = ((-2+2i)^2) = 4 - 8i - 4 = -4 - 8i\).

Substitute back: \(p(-1+i) = (-4 - 8i) + 3(-2 + 2i) + 6(-1+i) + 10\).

\(= -4 - 8i - 6 + 6i - 6 + 6i + 10\).

\(= 0\).

Thus, \(u = -1 + i\) is a root.

(ii) Since \(-1 + i\) is a root, its conjugate \(-1 - i\) is also a root.

Find a quadratic factor with roots \(-1+i\) and \(-1-i\):

\((z + 1 - i)(z + 1 + i) = z^2 + 2z + 2\).

Divide \(p(z)\) by \(z^2 + 2z + 2\):

\(z^4 + 3z^2 + 6z + 10 = (z^2 + 2z + 2)(z^2 - 2z + 5)\).

Solve \(z^2 - 2z + 5 = 0\) using the quadratic formula:

\(z = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}\).

\(= \frac{2 \pm \sqrt{4 - 20}}{2}\).

\(= \frac{2 \pm \sqrt{-16}}{2}\).

\(= \frac{2 \pm 4i}{2}\).

\(= 1 \pm 2i\).

The other roots are \(-1 - i\), \(1 + 2i\), and \(1 - 2i\).