(a) For \( g(x) = 2x^3 – 7x^2 + dx + e \), the sum of roots \(\alpha + \beta + \gamma = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} = -\frac{-7}{2} = \frac{7}{2}\).

(b) Since \( h(z) = 2z^5 – 11z^4 + rz^3 + sz^2 + tz – 20 \) has roots \( p + 3i \), its conjugate \( p – 3i \) is also a root (as coefficients are real). The other roots are \(\alpha, \beta, \gamma\). Sum of roots for \( h(z) \):
\( (p + 3i) + (p – 3i) + \alpha + \beta + \gamma = -\frac{-11}{2} = \frac{11}{2} \).
Since \(\alpha + \beta + \gamma = \frac{7}{2}\) from (a), we have:
\( p + p + \frac{7}{2} = \frac{11}{2} \implies 2p = \frac{11}{2} – \frac{7}{2} = 2 \implies p = 1 \).

(c)
(i) Given \( h\left(\frac{1}{2}\right) = 0 \), \(\frac{1}{2}\) is a root, so \(\gamma = \frac{1}{2}\). Product of roots for \( h(z) \):
\( (1 + 3i)(1 – 3i) \cdot \frac{1}{2} \cdot \alpha \beta = -\frac{-20}{2} = 10 \).
Compute: \( (1 + 3i)(1 – 3i) = 1 + 9 = 10 \), so \( 10 \cdot \frac{1}{2} \cdot \alpha \beta = 10 \implies 5 \alpha \beta = 10 \implies \alpha \beta = 2 \).
(ii) Since \(\alpha, \beta \in \mathbb{Z}^+\), \(\alpha < \beta\), and \(\alpha \beta = 2\), the only pair is \(\alpha = 1\), \(\beta = 2\).

Markscheme
(a) Sum of roots: \(\alpha + \beta + \gamma = -\frac{-7}{2} \quad \mathbf{M1} \)
Result: \(\frac{7}{2} \quad \mathbf{A1} \)
(b) Recognize \( p – 3i \) as a root and sum: \( 2p + \frac{7}{2} = \frac{11}{2} \quad \mathbf{M1A1} \)
Result: \( p = 1 \quad \mathbf{A1} \)
(c)
(i) Product of roots: \( 10 \cdot \frac{1}{2} \cdot \alpha \beta = 10 \implies \alpha \beta = 2 \quad \mathbf{A1} \)
(ii) \(\alpha = 1\), \(\beta = 2 \quad \mathbf{A1} \)
[7 marks]