A polynomial $f(x)$ has a remainder of $-1$ when divided by $2x + 3$.

1. Apply the Remainder Theorem to find the remainder of $f(x) - 1$ when divided by $2x + 3$

By the Remainder Theorem, the remainder when $f(x)$ is divided by $2x + 3$ is $f\left(-\frac{3}{2}\right) = -1$.

The remainder when $f(x) - 1$ is divided by $2x + 3$ is:
$f\left(-\frac{3}{2}\right) - 1 = -1 - 1 = -2$

1. Determine the polynomial divisible by $2x + 3$

By the Remainder Theorem, since $f(x)$ has a remainder of $-1$ when divided by $2x + 3$, we have $f\left(-\frac{3}{2}\right) = -1$.
For a polynomial to be divisible by $2x + 3$, its value at $x = -\frac{3}{2}$ must be $0$.
Since $f\left(-\frac{3}{2}\right) + 1 = -1 + 1 = 0$, the polynomial $f(x) + 1$ is divisible by $2x + 3$.

1. Formulate an expression for $g(x)$ using its roots and leading coefficient.

Since the roots are $-1$, $m$, and $2m$, and the coefficient of $x^3$ is $2$, we can write:
$g(x) = 2(x + 1)(x - m)(x - 2m)$

2. Apply the Remainder Theorem to set up an equation for $m$.

Given that $g(x)$ has a remainder of $12$ when divided by $x - 1$, by the Remainder Theorem, $g(1) = 12$.
$2(1 + 1)(1 - m)(1 - 2m) = 12$
$4(1 - 3m + 2m^2) = 12$
$1 - 3m + 2m^2 = 3$
$2m^2 - 3m - 2 = 0$

3. Solve the quadratic equation to find the integer value of $m$.

$(2m + 1)(m - 2) = 0$
$m = -\frac{1}{2}$ or $m = 2$
Since $m$ is given to be an integer, $m = 2$.

4. Substitute $m$ back into $g(x)$ and expand into descending powers of $x$.

$g(x) = 2(x + 1)(x - 2)(x - 4)$
$g(x) = 2(x + 1)(x^2 - 6x + 8)$
$g(x) = 2(x^3 - 6x^2 + 8x + x^2 - 6x + 8)$
$g(x) = 2(x^3 - 5x^2 + 2x + 8)$
$g(x) = 2x^3 - 10x^2 + 4x + 16$