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

(i) To find the value of \(a\), we perform polynomial division of \(4x^3 - 4x^2 + 3x + a\) by \(2x^2 - 3x + 3\). The division should result in a remainder of zero since \(p(x)\) is divisible by \(2x^2 - 3x + 3\).

Performing the division, we find that the remainder is zero when \(a = 3\).

(ii) With \(a = 3\), the polynomial becomes \(4x^3 - 4x^2 + 3x + 3\). We need to solve the inequality \(p(x) < 0\).

Since \(2x^2 - 3x + 3\) is never zero and always positive, we analyze the sign of \(p(x)\) by considering the roots of the polynomial.

The inequality \(p(x) < 0\) holds when \(x < -\frac{1}{2}\), as justified by the behavior of the polynomial and its derivative.

Since \(x^2 + x + 2\) is a factor of \(p(x) = x^4 + 3x^2 + a\), we can perform polynomial division of \(p(x)\) by \(x^2 + x + 2\).

1. Divide \(x^4 + 3x^2 + a\) by \(x^2 + x + 2\) to get a quotient of \(x^2 - x + 2\).

2. The remainder must be zero for \(x^2 + x + 2\) to be a factor.

3. Equate the constant remainder to zero and solve for \(a\):

\(a = 4\).

Thus, the other quadratic factor is \(x^2 - x + 2\).

(i) Since \((x + 2)\) is a factor of \(p(x)\), substituting \(x = -2\) into \(p(x)\) should yield zero:

\((-2)^3 - 2(-2) + a = 0\)

\(-8 + 4 + a = 0\)

\(a = 4\)

(ii) With \(a = 4\), the polynomial becomes \(x^3 - 2x + 4\). To find the quadratic factor, divide \(x^3 - 2x + 4\) by \(x + 2\):

Perform polynomial division:

\(x^3 - 2x + 4 = (x + 2)(x^2 - 2x + 2)\)

Thus, the quadratic factor is \(x^2 - 2x + 2\).

(i) To find \(a\), perform polynomial division of \(x^4 + 5x + a\) by \(x^2 - x + 3\). The quotient is \(x^2 + x\) and the remainder is \(3x + a\). Equating the remainder to zero gives:

\(3x + a = 0\)

Since \(x^2 - x + 3\) is a factor, the remainder must be zero for all \(x\). Thus, \(a = -6\).

Now, factorise \(x^2 + x - 2\) as \((x + 2)(x - 1)\). Therefore, \(p(x) = (x^2 - x + 3)(x + 2)(x - 1)\).

(ii) The quadratic \(x^2 + x - 2 = 0\) has two real roots, \(x = -2\) and \(x = 1\). The quadratic \(x^2 - x + 3 = 0\) has no real roots (discriminant \(b^2 - 4ac = 1 - 12 = -11\)). Therefore, \(p(x) = 0\) has two real roots.