已知表达式 $2x^3 + ax^2 + bx + c$ 除以 $(x - 5)$ 的余数与除以 $(2x + 1)$ 的余数相同，证明 $9a + 2b = -91$。 已知表达式 $2x^3 + ax^2 + bx + c$ 除以 $(x - 1)$ 的余数是 $-6$，且 $(x + 1)$ 是它的一个因式，求 $c$ 的值。

1. Define the polynomial and apply the Remainder Theorem for division by $(x - 5)$.

Let $f(x) = 2x^3 + ax^2 + bx + c$.
When divided by $(x - 5)$, the remainder is $f(5)$.
$f(5) = 2(5)^3 + a(5)^2 + b(5) + c = 250 + 25a + 5b + c$

2. Apply the Remainder Theorem for division by $(2x + 1)$.

When divided by $(2x + 1)$, the remainder is $f(-\frac{1}{2})$.
$f(-\frac{1}{2}) = 2(-\frac{1}{2})^3 + a(-\frac{1}{2})^2 + b(-\frac{1}{2}) + c$
$= 2(-\frac{1}{8}) + \frac{1}{4}a - \frac{1}{2}b + c$
$= -\frac{1}{4} + \frac{1}{4}a - \frac{1}{2}b + c$

3. Equate the two remainders and simplify to show the required equation.

Since the remainders are the same:
$250 + 25a + 5b + c = -\frac{1}{4} + \frac{1}{4}a - \frac{1}{2}b + c$
$250 + 25a + 5b = -\frac{1}{4} + \frac{1}{4}a - \frac{1}{2}b$
Multiply by 4 to clear denominators:
$1000 + 100a + 20b = -1 + a - 2b$
$100a - a + 20b + 2b = -1 - 1000$
$99a + 22b = -1001$
Divide by 11:
$9a + 2b = -91$ (Shown)

1. Apply the Remainder Theorem for division by $(x - 1)$.

Let $f(x) = 2x^3 + ax^2 + bx + c$.
Since the remainder is $-6$ when divided by $(x - 1)$:
$f(1) = -6$
$2(1)^3 + a(1)^2 + b(1) + c = -6$
$2 + a + b + c = -6$
$a + b + c = -8 \quad \text{--- (1)}$

2. Apply the Factor Theorem for $(x + 1)$.

Since $(x + 1)$ is a factor:
$f(-1) = 0$
$2(-1)^3 + a(-1)^2 + b(-1) + c = 0$
$-2 + a - b + c = 0$
$a - b + c = 2 \quad \text{--- (2)}$

3. Solve the simultaneous equations to find $b$.

Subtracting equation (2) from equation (1):
$(a + b + c) - (a - b + c) = -8 - 2$
$2b = -10$
$b = -5$

4. Use the result from part (a) to find $a$, and subsequently $c$.

From part (a), we have $9a + 2b = -91$.
Substitute $b = -5$:
$9a + 2(-5) = -91$
$9a - 10 = -91$
$9a = -81$
$a = -9$

Substitute $a = -9$ and $b = -5$ into equation (1):
$-9 - 5 + c = -8$
$-14 + c = -8$
$c = 6$