(i) Since \((2x + 1)\) is a factor of \(p(x)\), substituting \(x = -\frac{1}{2}\) into \(p(x)\) gives:

\(-\frac{1}{4} + \frac{5}{4} - \frac{1}{2}a + b = 0\)

\(\Rightarrow -\frac{1}{2}a + b = -1\)

When \(x = -2\), the remainder is 9:

\(-16 + 20 - 2a + b = 9\)

\(\Rightarrow -2a + b = 5\)

Solving the equations:

\(-\frac{1}{2}a + b = -1\)

\(-2a + b = 5\)

Subtract the first from the second:

\(-2a + b + \frac{1}{2}a - b = 5 + 1\)

\(-\frac{3}{2}a = 6\)

\(a = -4\)

Substitute \(a = -4\) into \(-\frac{1}{2}a + b = -1\):

\(2 + b = -1\)

\(b = -3\)

(ii) With \(a = -4\) and \(b = -3\), \(p(x) = 2x^3 + 5x^2 - 4x - 3\).

Divide \(p(x)\) by \(2x + 1\) to get a quotient of \(x^2 + 2x - 3\).

Factorise \(x^2 + 2x - 3\) as \((x + 3)(x - 1)\).

Thus, \(p(x) = (2x + 1)(x + 3)(x - 1)\).