函数 $f$ 定义为 $f(x) = 3x^3 - 5cx^2 + kc^2x + 4c^3$，其中 $c$ 和 $k$ 是非零常数。当 $f(x)$ 除以 $x + 2c$ 时，余数为 $-32c^3$。

1. Apply the Remainder Theorem by substituting $x = -2c$ into $f(x)$.

By the Remainder Theorem, $f(-2c) = -32c^3$.
$3(-2c)^3 - 5c(-2c)^2 + kc^2(-2c) + 4c^3 = -32c^3$

2. Simplify the equation and solve for $k$.

$3(-8c^3) - 5c(4c^2) - 2kc^3 + 4c^3 = -32c^3$
$-24c^3 - 20c^3 - 2kc^3 + 4c^3 = -32c^3$
$-40c^3 - 2kc^3 = -32c^3$
$-2kc^3 = 8c^3$
Since $c \neq 0$, we can divide by $c^3$:
$-2k = 8$
$k = -4$

1. Find the value of $k$ from part (a)

From part (a), $f(-2c) = -32c^3$.
$3(-2c)^3 - 5c(-2c)^2 + kc^2(-2c) + 4c^3 = -32c^3$
$-24c^3 - 20c^3 - 2kc^3 + 4c^3 = -32c^3$
$-40c^3 - 2kc^3 = -32c^3$
$-2k = 8 \implies k = -4$
So $f(x) = 3x^3 - 5cx^2 - 4c^2x + 4c^3$.

2. Factorise the given quadratic expression

$x^2 - cx - 2c^2 = (x - 2c)(x + c)$

3. Apply the Factor Theorem to check if $(x - 2c)$ and $(x + c)$ are factors

$f(2c) = 3(2c)^3 - 5c(2c)^2 - 4c^2(2c) + 4c^3$
$= 24c^3 - 20c^3 - 8c^3 + 4c^3 = 0$

$f(-c) = 3(-c)^3 - 5c(-c)^2 - 4c^2(-c) + 4c^3$
$= -3c^3 - 5c^3 + 4c^3 + 4c^3 = 0$

4. Conclude based on the Factor Theorem

Since $f(2c) = 0$ and $f(-c) = 0$, both $(x - 2c)$ and $(x + c)$ are factors of $f(x)$.
Therefore, their product $x^2 - cx - 2c^2$ is a factor of $f(x)$.