The polynomial $\mathrm{f}(x) = Ax^3 - (2A + B)x^2 + 2x + B$, where $A$ and $B$ are constants, is exactly divisible by $(x - 1)$.
When $\mathrm{f}(x)$ is divided by $x$, the remainder obtained is 3.

1. Use the Remainder Theorem to find $B$.

Since the remainder is 3 when $\mathrm{f}(x)$ is divided by $x$, we have $\mathrm{f}(0) = 3$.
$\mathrm{f}(0) = A(0)^3 - (2A + B)(0)^2 + 2(0) + B = 3$
$B = 3$

2. Use the Factor Theorem to set up an equation for $A$.

Since $\mathrm{f}(x)$ is exactly divisible by $(x - 1)$, we have $\mathrm{f}(1) = 0$.
$\mathrm{f}(1) = A(1)^3 - (2A + B)(1)^2 + 2(1) + B = 0$
$A - (2A + B) + 2 + B = 0$

3. Solve for $A$.

$A - 2A - B + 2 + B = 0$
$-A + 2 = 0$
$A = 2$

1. Determine the values of $A$ and $B$ from the conditions in part (a) to write the full polynomial.

From part (a), $\mathrm{f}(0) = 3 \implies B = 3$.
Since $(x-1)$ is a factor, $\mathrm{f}(1) = 0 \implies A(1)^3 - (2A + 3)(1)^2 + 2(1) + 3 = 0 \implies -A + 2 = 0 \implies A = 2$.
Thus, $\mathrm{f}(x) = 2x^3 - 7x^2 + 2x + 3$.

2. Find the quadratic factor by dividing $\mathrm{f}(x)$ by $(x - 1)$.

Let $2x^3 - 7x^2 + 2x + 3 = (x - 1)(2x^2 + cx - 3)$.
Comparing the coefficients of $x^2$:
$-2 + c = -7 \implies c = -5$.
So, the quadratic factor is $2x^2 - 5x - 3$.

3. Factorise the quadratic factor completely.

$2x^2 - 5x - 3 = (2x + 1)(x - 3)$

4. State the completely factorised form of $\mathrm{f}(x)$.

$\mathrm{f}(x) = (x - 1)(2x + 1)(x - 3)$

1. Recall the factorised form of $f(x)$ from part (b)

From parts (a) and (b), $A=2$, $B=3$, and $\mathrm{f}(x) = 2x^3 - 7x^2 + 2x + 3 = (x-1)(2x+1)(x-3)$.

2. Recognise that $g(x)$ is $f(x-2)$ and substitute $x-2$ into the factorised form

Notice that $g(x) = \mathrm{f}(x-2)$.
Substituting $(x-2)$ for $x$ in the factorised form of $\mathrm{f}(x)$:
$g(x) = ((x-2) - 1)(2(x-2) + 1)((x-2) - 3)$

3. Simplify the factors to obtain the final answer

$g(x) = (x - 3)(2x - 4 + 1)(x - 5)$
$g(x) = (x - 3)(2x - 3)(x - 5)$