表达式 $f(x) = ax^3 + (a - 3b)x^2 + 3bx + c$ 能被 $x^2 + 3x$ 整除。当 $f(x)$ 除以 $x + 2$ 时，余数为 10。求 $a$、$b$ 和 $c$ 的值。由此将 $f(x)$ 完全因式分解。 已知对所有 $x$ 的值，$3x^3 - 2x^2 + x - 4 = A(x - 1) + B(x - 1)(x + 1) + Cx(x^2 - 1) + D$ 均成立，求 $A$、$B$、$C$ 和 $D$ 的值。

1. Use the fact that $f(x)$ is divisible by $x^2 + 3x$ to find $c$.

Since $x^2 + 3x = x(x + 3)$, $f(x)$ is divisible by $x$ and $x + 3$.
By the Factor Theorem, $f(0) = 0$.
$f(0) = a(0)^3 + (a - 3b)(0)^2 + 3b(0) + c = 0$
$\Rightarrow c = 0$

2. Use the Factor Theorem for $x + 3$ to form an equation in $a$ and $b$.

$f(-3) = 0$
$a(-3)^3 + (a - 3b)(-3)^2 + 3b(-3) + 0 = 0$
$-27a + 9(a - 3b) - 9b = 0$
$-27a + 9a - 27b - 9b = 0$
$-18a - 36b = 0$
$a = -2b$ --- (1)

3. Use the Remainder Theorem for $x + 2$ to form another equation in $a$ and $b$.

$f(-2) = 10$
$a(-2)^3 + (a - 3b)(-2)^2 + 3b(-2) + 0 = 10$
$-8a + 4(a - 3b) - 6b = 10$
$-8a + 4a - 12b - 6b = 10$
$-4a - 18b = 10$
$-2a - 9b = 5$ --- (2)

4. Solve the simultaneous equations to find $a$ and $b$.

Substitute (1) into (2):
$-2(-2b) - 9b = 5$
$4b - 9b = 5$
$-5b = 5 \Rightarrow b = -1$
Substitute $b = -1$ into (1):
$a = -2(-1) = 2$

5. Factorise $f(x)$ completely.

$f(x) = 2x^3 + (2 - 3(-1))x^2 + 3(-1)x + 0$
$f(x) = 2x^3 + 5x^2 - 3x$
Since $x^2 + 3x$ is a factor, $f(x) = (x^2 + 3x)(2x - 1)$
$f(x) = x(x + 3)(2x - 1)$

1. Substitute $x = 1$ to find $D$.

Let $x = 1$:
$3(1)^3 - 2(1)^2 + 1 - 4 = A(0) + B(0) + C(0) + D$
$3 - 2 + 1 - 4 = D$
$D = -2$

2. Substitute $x = -1$ to find $A$.

Let $x = -1$:
$3(-1)^3 - 2(-1)^2 + (-1) - 4 = A(-1 - 1) + B(0) + C(0) + D$
$-3 - 2 - 1 - 4 = -2A + D$
$-10 = -2A - 2$
$-2A = -8$
$A = 4$

3. Compare coefficients of $x^3$ to find $C$.

Comparing the coefficients of $x^3$ on both sides:
LHS coefficient of $x^3$ is $3$.
RHS coefficient of $x^3$ comes from $Cx(x^2 - 1) = Cx^3 - Cx$, which is $C$.
Therefore, $C = 3$.

4. Substitute $x = 0$ to find $B$.

Let $x = 0$:
$3(0)^3 - 2(0)^2 + 0 - 4 = A(-1) + B(-1)(1) + C(0) + D$
$-4 = -A - B + D$
$-4 = -4 - B - 2$
$B = -2$