Given that $z = 2 - \text{i}$ is a root of the equation $4z^4 - 12z^3 + 17z^2 + pz + q = 0$, where $p$ and $q$ are real, find $p$ and $q$. Using the values of $p$ and $q$ found, find the other roots of the equation $4z^4 - 12z^3 + 17z^2 + pz + q = 0$ in exact form.

1. Identify the conjugate root and form the corresponding quadratic factor.

Since the equation has real coefficients ($p, q \in \mathbb{R}$), the complex conjugate $z = 2 + \text{i}$ is also a root.
The quadratic factor containing these two roots is:
$(z - (2 - \text{i}))(z - (2 + \text{i})) = ((z - 2) + \text{i})((z - 2) - \text{i})$
$= (z - 2)^2 - \text{i}^2 = (z^2 - 4z + 4) - (-1) = z^2 - 4z + 5$

2. Set up a polynomial identity with an unknown quadratic factor.

Let the quartic polynomial be identically equal to the product of this quadratic factor and another quadratic factor. By observing the leading coefficient, the second factor must start with $4z^2$:
$4z^4 - 12z^3 + 17z^2 + pz + q \equiv (z^2 - 4z + 5)(4z^2 + cz + d)$

3. Expand the right-hand side and compare the coefficients of $z^3$ and $z^2$ to find $c$ and $d$.

Expanding the right-hand side:
$= 4z^4 + cz^3 + dz^2 - 16z^3 - 4cz^2 - 4dz + 20z^2 + 5cz + 5d$
$= 4z^4 + (c - 16)z^3 + (d - 4c + 20)z^2 + (5c - 4d)z + 5d$

Comparing the coefficients of $z^3$:
$c - 16 = -12 \implies c = 4$

Comparing the coefficients of $z^2$:
$d - 4c + 20 = 17$
Substitute $c = 4$:
$d - 4(4) + 20 = 17 \implies d + 4 = 17 \implies d = 13$

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

Comparing the coefficients of $z$:
$p = 5c - 4d$
Substitute $c = 4$ and $d = 13$:
$p = 5(4) - 4(13) = 20 - 52 = -32$

Comparing the constant terms:
$q = 5d$
Substitute $d = 13$:
$q = 5(13) = 65$

1. Concisely derive the values of $p$ and $q$ required from part (a) by substituting the given root.

Substitute $z = 2 - \text{i}$ into $4z^4 - 12z^3 + 17z^2 + pz + q = 0$:
$4(2-\text{i})^4 - 12(2-\text{i})^3 + 17(2-\text{i})^2 + p(2-\text{i}) + q = 0$
$4(-7-24\text{i}) - 12(2-11\text{i}) + 17(3-4\text{i}) + 2p - p\text{i} + q = 0$
Equating the real and imaginary parts to zero:
$-32 - p = 0 \implies p = -32$
$-1 + 2p + q = 0 \implies q = 65$

2. Identify the complex conjugate root and form the corresponding quadratic factor.

Since the polynomial has real coefficients, the complex conjugate $z = 2 + \text{i}$ is also a root.
The quadratic factor corresponding to these two roots is:
$(z - (2-\text{i}))(z - (2+\text{i})) = (z-2)^2 - \text{i}^2 = z^2 - 4z + 5$

3. Compare coefficients to find the remaining quadratic factor.

Using $p = -32$ and $q = 65$, the polynomial is $4z^4 - 12z^3 + 17z^2 - 32z + 65$.
Let the remaining quadratic factor be $4z^2 + Bz + C$:
$4z^4 - 12z^3 + 17z^2 - 32z + 65 = (z^2 - 4z + 5)(4z^2 + Bz + C)$
Comparing the constant terms:
$5C = 65 \implies C = 13$
Comparing the $z^3$ coefficients:
$B - 16 = -12 \implies B = 4$
Thus, the other factor is $4z^2 + 4z + 13$.

4. Solve the remaining quadratic equation to find the remaining roots.

$4z^2 + 4z + 13 = 0$
$z = \frac{-4 \pm \sqrt{16 - 4(4)(13)}}{8}$
$z = \frac{-4 \pm \sqrt{-192}}{8} = \frac{-4 \pm 8\text{i}\sqrt{3}}{8} = -\frac{1}{2} \pm \text{i}\sqrt{3}$
The other roots are $2+\text{i}$, $-\frac{1}{2} + \text{i}\sqrt{3}$, and $-\frac{1}{2} - \text{i}\sqrt{3}$.