Do not use a calculator in answering this question. The equation $2z^3 + 5z^2 + pz + q = 0$ has a root $z = -1 + \mathrm{i}$, where $p$ and $q$ are real constants.
Find $p$ and $q$ and all the other roots of the equation.

1. Identify the complex conjugate root using the property of real coefficients.

Since the coefficients $p$ and $q$ are real constants, the complex conjugate of the given root must also be a root of the equation.
Therefore, $z = -1 - \mathrm{i}$ is another root.

2. Form the quadratic factor corresponding to the two complex roots.

The quadratic factor having these roots is:
$(z - (-1 + \mathrm{i}))(z - (-1 - \mathrm{i}))$
$= ((z + 1) - \mathrm{i})((z + 1) + \mathrm{i})$
$= (z + 1)^2 - \mathrm{i}^2$
$= z^2 + 2z + 1 - (-1)$
$= z^2 + 2z + 2$

3. Set up an identity to find the remaining linear factor.

Since $2z^3 + 5z^2 + pz + q = 0$ is a cubic equation with leading coefficient $2$, it can be expressed as the product of the quadratic factor and a linear factor $(2z + c)$:
$2z^3 + 5z^2 + pz + q = (z^2 + 2z + 2)(2z + c)$
Expanding the right side gives:
$= 2z^3 + cz^2 + 4z^2 + 2cz + 4z + 2c$
$= 2z^3 + (c + 4)z^2 + (2c + 4)z + 2c$
Comparing the coefficient of $z^2$:
$5 = c + 4 \implies c = 1$

4. Compare the remaining coefficients to solve for $p$ and $q$.

Substitute $c = 1$ into the expanded expression and compare coefficients.
Comparing the coefficient of $z$:
$p = 2c + 4$
$p = 2(1) + 4 = 6$
Comparing the constant term:
$q = 2c$
$q = 2(1) = 2$

5. Solve the linear factor to find the third root.

The linear factor is $2z + 1$. Setting this to zero gives the final root:
$2z + 1 = 0$
$z = -\frac{1}{2}$
The other roots of the equation are therefore $-1 - \mathrm{i}$ and $-\frac{1}{2}$.