Given that $f(x) = 2x^3 + px^2 + 7x + q$. Find the value of $p$ and of $q$ for which $(x-3)$ and $(x-2)$ are factors of $f(x)$. Hence solve $f(x) = 0$.

1. Apply the Factor Theorem for $(x-3)$ to form the first equation.

Since $(x-3)$ is a factor, $f(3) = 0$.
$2(3)^3 + p(3)^2 + 7(3) + q = 0$
$54 + 9p + 21 + q = 0$
$9p + q = -75 \quad \text{--- (1)}$

2. Apply the Factor Theorem for $(x-2)$ to form the second equation.

Since $(x-2)$ is a factor, $f(2) = 0$.
$2(2)^3 + p(2)^2 + 7(2) + q = 0$
$16 + 4p + 14 + q = 0$
$4p + q = -30 \quad \text{--- (2)}$

3. Solve the simultaneous equations to find $p$ and $q$.

Subtracting (2) from (1):
$(9p + q) - (4p + q) = -75 - (-30)$
$5p = -45$
$p = -9$

Substitute $p = -9$ into (2):
$4(-9) + q = -30$
$-36 + q = -30$
$q = 6$

4. Find the remaining linear factor of $f(x)$.

The polynomial is $f(x) = 2x^3 - 9x^2 + 7x + 6$.
Since $(x-3)$ and $(x-2)$ are factors, their product $(x-3)(x-2) = x^2 - 5x + 6$ is a quadratic factor of $f(x)$.
Let the third linear factor be $(ax + b)$.
$2x^3 - 9x^2 + 7x + 6 = (x^2 - 5x + 6)(ax + b)$
Comparing the coefficients of $x^3$: $a = 2$.
Comparing the constant terms: $6b = 6 \implies b = 1$.
So, the third factor is $(2x + 1)$.

5. Solve $f(x) = 0$.

$f(x) = (x-3)(x-2)(2x+1) = 0$
$x - 3 = 0 \implies x = 3$
$x - 2 = 0 \implies x = 2$
$2x + 1 = 0 \implies x = -\frac{1}{2}$