(a) Substitute \(x = -3\) into \(p(x)\):

\(p(-3) = (-3)^3 + 5(-3)^2 + 31(-3) + 75 = -27 + 45 - 93 + 75 = 0\).

Since \(p(-3) = 0\), \((x + 3)\) is a factor of \(p(x)\).

(b) Substitute \(z = -1 + 2\sqrt{6}i\) into \(p(z)\):

Calculate \(z^2 = (-1 + 2\sqrt{6}i)^2 = -1 - 4\sqrt{6}i - 24 = -25 - 4\sqrt{6}i\).

Calculate \(z^3 = z \cdot z^2 = (-1 + 2\sqrt{6}i)(-25 - 4\sqrt{6}i) = 25 + 4\sqrt{6}i + 50\sqrt{6}i + 48 = 73 + 54\sqrt{6}i\).

Substitute into \(p(z)\):

\(p(z) = z^3 + 5z^2 + 31z + 75 = 73 + 54\sqrt{6}i + 5(-25 - 4\sqrt{6}i) + 31(-1 + 2\sqrt{6}i) + 75 = 0\).

Thus, \(z = -1 + 2\sqrt{6}i\) is a root of \(p(z) = 0\).

(c) Let \(z^2 = w\), then \(p(w) = w^3 + 5w^2 + 31w + 75 = 0\).

The roots of \(p(w) = 0\) are \(w_1 = 3i, w_2 = -3i, w_3 = -1 + 2\sqrt{6}i\).

For \(w_1 = 3i\), \(z = \pm \sqrt{3}i\).

For \(w_2 = -3i\), \(z = \pm \sqrt{6}i\).

For \(w_3 = -1 + 2\sqrt{6}i\), \(z = -1 \pm 2\sqrt{6}i\).

Thus, the roots are \(z = \pm \sqrt{3}i, \pm \sqrt{6}i, -1 \pm 2\sqrt{6}i\).